Lemma 2.1

Let \(P_{i}({\mathbb {U}}, {\mathbb {Y}})\in \mathcal {F}\langle {\mathbb {Y}}\rangle \{{\mathbb {U}}\}\) \((i=1, \ldots , m)\) where \({\mathbb {U}}\) and \({\mathbb {Y}}\) are sets of differential indeterminates. If \(\theta _{ij}\big (P_{i}({\mathbb {U}}, {\mathbb {Y}})\big )\,(i=1, \ldots , m; j=1,\ldots ,n_{i})\) are algebraically dependent over \(\mathcal {F}\langle {\mathbb {U}}\rangle \) for \(\theta _{ij}\in \Theta \), then for any differential specialization \({\mathbb {U}}^0\subset \mathcal {F}\) of \({\mathbb {U}}\) over \(\mathcal {F}\), \(\theta _{ij}\big (P_{i}({\mathbb {U}}^0, {\mathbb {Y}})\big )\) are algebraically dependent over \(\mathcal {F}\). In particular, if \(P_{i}({\mathbb {U}}, {\mathbb {Y}})\) \((i=1, \ldots , m)\) are differentially dependent over \(\mathcal {F}\langle {\mathbb {U}}\rangle \), then for any differential specialization \({\mathbb {U}}^0\subset \mathcal {F}\) of \({\mathbb {U}}\) over \({\mathcal {F}}\), \(P_{i}({\mathbb {U}}^0,{\mathbb {Y}})\) are differentially dependent over \(\mathcal {F}\).

Definition 2.2

Let \(\mathcal {I}\) be a differential ideal of \(\mathcal {F}\{{\mathbf {z}}\}\) and \(\mathcal {I}\hbox {{:}}{\mathbf {z}}=\{f\in \mathcal {F}\{{\mathbf {z}}\}|\,z_jf\in \mathcal {I}, j=0,\ldots ,l\}\). Call \(\mathcal {I}\) a differentially homogenous differential ideal of \(\mathcal {F}\{{\mathbf {z}}\}\) if \(\mathcal {I}\hbox {{:}}{\mathbf {z}}=\mathcal {I}\) and for every \(P\in \mathcal {I}\) and a differential indeterminate \(\lambda \) over \(\mathcal {F}\{{\mathbf {z}}\}\), \(P(\lambda {\mathbf {z}})\in \mathcal {F}\{\lambda \}\mathcal {I}\) in \(\mathcal {F}\{\lambda ,{\mathbf {z}}\}\).

Lemma 2.3

[42, p. 45] Let \(\mathcal {I}\) be a prime differential ideal of codimension one in \(\mathcal {F}\{{\mathbb {Y}}\}\). Then there exists an irreducible differential polynomial \(A\) such that \(\mathcal {I}=\mathrm{sat}(A)\) and \(\{A\}\) is the characteristic set of \(\mathcal {I}\) w.r.t. any ranking.

Definition 3.1

A Laurent differential monomial of order \(s\in {\mathbb N}\) is a Laurent monomial in variables \({\mathbb {Y}}^{[s]}=(y_i^{(k)})_{1\le i\le n;0\le k\le s}\). More precisely, it has the form \(\prod _{i=1}^n\prod _{k=0}^s(y_i^{(k)})^{m_{ik}}\), where \(m_{ik}\) are integers which can be negative. A Laurent differential polynomial is a finite linear combination of Laurent differential monomials with coefficients from \(\mathcal {E}\).

Remark 3.2

\(\mathcal {F}\{{\mathbb {Y}}^{\pm }\}=\mathcal {F}\{y_1,y_1^{-1},\ldots ,y_n,y_n^{-1}\}\) is only a notation for Laurent differential polynomial ring. It is not equal to \(\mathcal {F}[y_{i}^{(k)},(y_{i}^{-1})^{(k)}\,|\,k\ge 0]\).

Lemma 3.3

The above maps \(\phi \) and \(\psi \) are well defined. Moreover, \(\phi \circ \psi =\text { id}_{\mathcal {T}}\) and \(\psi \circ \phi =\text { id}_{\mathcal {S}}\).

Proof

\(\psi \) is obviously well defined. To show that \(\phi \) is well defined, it suffices to show that given another \((N_1,\ldots ,N_s)\in \mathbb {m}^s\) with \(N_iF_i\in \mathcal {F}\{{\mathbb {Y}}\}\,(i=0,\ldots ,n)\), \(([M_1F_1,\ldots ,M_sF_s]\hbox {{:}}\mathbb {m})_{\mathcal {F}\{{\mathbb {Y}}\}}=([N_1F_1,\ldots ,N_sF_s]\hbox {{:}}\mathbb {m})_{\mathcal {F}\{{\mathbb {Y}}\}}\). It follows from the fact that \(N_iF_i\in ([M_1F_1,\ldots ,M_sF_s]\hbox {{:}}\mathbb {m})_{\mathcal {F}\{{\mathbb {Y}}\}}\) and \(M_iF_i\in ([N_1F_1,\ldots ,N_sF_s]\hbox {{:}}\mathbb {m})_{\mathcal {F}\{{\mathbb {Y}}\}}\). For each \(\mathcal {I}=[F_1,\ldots ,F_s]_{ \mathcal {F}\{{\mathbb {Y}}^{\pm }\}}\in \mathcal {S}\), \(\psi \circ \phi (\mathcal {I})=\psi (([M_1F_1,\ldots ,\) \(M_sF_s]\hbox {{:}}\mathbb {m})_{ \mathcal {F}\{{\mathbb {Y}}\}})=[M_1F_1,\ldots ,M_sF_s]_{ \mathcal {F}\{{\mathbb {Y}}^{\pm }\}}= \mathcal {I}\) where \(M_iF_i\in \mathcal {F}\{{\mathbb {Y}}\}\). So we have \(\psi \circ \phi =\text { id}_{\mathcal {S}}\). And for each \(\mathcal {J}=([f_1,\ldots ,f_r]\hbox {{:}}\mathbb {m})_{ \mathcal {F}\{{\mathbb {Y}}\}}\in \mathcal {T}\), \(\phi \circ \psi (\mathcal {J})=\phi ([f_1,\ldots ,f_r]_{ \mathcal {F}\{{\mathbb {Y}}^{\pm }\}})=\mathcal {J}\). Thus, \(\phi \circ \psi =\text { id}_{\mathcal {T}}\) follows. \(\square \)

Definition 3.4

For every Laurent differential polynomial \(F\in \mathcal {E}\{{\mathbb {Y}}^{\pm }\}\), there exists a unique Laurent differential monomial \(M\) such that (1) \(M\cdot F\in \mathcal {E}\{{\mathbb {Y}}\}\) and (2) for any Laurent differential monomial \(T\) with \(T\cdot F\in \mathcal {E}\{{\mathbb {Y}}\}\), \(T\cdot F\) is divisible by \(M\cdot F\) as differential polynomials. This \(M\cdot F\) is defined to be the norm form of \(F\), denoted by \(F^{\text {N}}\). The order of \(F^{\text {N}}\) is defined to be the effective order of F, denoted by \(\hbox {{Eord}}(F)\). Clearly, \(\hbox {{Eord}}(F)\le {\mathrm{ord}}(F)\). And the degree of \(F\) is defined to be the degree of \(F^{\text {N}}\), denoted by \(\mathrm{deg}(F)\).

Definition 3.5

Let \(\mathcal {E}^{\wedge }=\mathcal {E}\backslash \{a\in \mathcal {E}\big |\,\exists k\in \mathbb {N}, \,\text { s.t.}\, a^{(k)}=0 \}\). Let \(F\) be a Laurent differential polynomial in \(\mathcal {F}\{{\mathbb {Y}}^{\pm }\}\). A point \((a_1,\ldots ,a_n)\in (\mathcal {E}^{\wedge })^n\) is called a non-polynomial differential zero of \(F\) if \(F(a_1,\ldots ,a_n)=0.\)

Definition 3.6

A set of Laurent differential polynomials of the form (3) is said to be a Laurent differentially essential system if there exist \(k_i\,(i=0,\ldots ,n)\) with \(1\le k_i\le l_i\) such that \(\hbox {{d.tr.deg}}\,\mathbb {Q}\langle \frac{M_{0k_0}}{M_{00}},\) \(\frac{M_{1k_1}}{M_{10}},\ldots ,\frac{M_{nk_n}}{M_{n0}}\rangle /\mathbb {Q}=n.\) In this case, we also say that \(\mathcal {A}_0,\ldots ,\mathcal {A}_n\) or \({\mathbb {S}}_0,\ldots ,\) \({\mathbb {S}}_n\) form a Laurent differentially essential system.

Lemma 3.7

Proof

It is trivial that 1) implies 2). For the other direction, assume 2) holds. Without loss of generality, suppose \(\frac{M_{1k_1}}{M_{1j_1}},\ldots ,\frac{M_{nk_n}}{M_{nj_n}} \) are differentially independent over \(\mathbb {Q}\). We need to show (1) holds. Suppose the contrary, then for any \(m_i\in \{1,\ldots ,l_{i}\}\), \(\frac{M_{1m_1}}{M_{10}},\ldots ,\frac{M_{nm_n}}{M_{n0}}\) are differentially dependent over \({\mathbb Q}\). Now we claim that \((*)\) suppose for each \(i\in \{1,2\}\), \(a\) and \(b_i\) are differentially dependent over \(\mathbb {Q}\), then \(a\) and \(b_1/b_2\) are differentially dependent over \(\mathbb {Q}\). Indeed, if \(a\) is differentially algebraic over \(\mathbb {Q}\), then \((*)\) follows. If \(a\) is differentially transcendental over \(\mathbb {Q}\), then each \(b_i\) is differentially algebraic over \(\mathbb {Q}\langle a\rangle \). Thus, \(b_1/b_2\) is differentially algebraic over \(\mathbb {Q}\langle a\rangle \) [29, p. 102] and the claim is proved. Since \(\frac{M_{ik_i}}{M_{ij_i}}=\frac{M_{ik_i}}{M_{i0}}\big /\frac{M_{ij_i}}{M_{i0}}\), by claim \((*)\), \(\frac{M_{ik_i}}{M_{ij_i}}\,(i=1,\ldots ,n)\) are differentially dependent over \({\mathbb Q}\), which leads to a contradiction. \(\square \)

Lemma 3.8

\(\mathcal {I}_{{\mathbb {Y}}^{\pm },{\mathbf {u}}}\cap \mathbb {Q}\{{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}=\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\cap \mathbb {Q}\{{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}\).

Proof

It is obvious that the right elimination ideal is contained in the left one. For the other direction, let \(G\) be any element in the left ideal. Then there exist \(H_{ij}\in \mathbb {Q}\{{\mathbb {Y}}^{\pm };{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}\) such that \(G=\sum _{i,j}H_{ij}{\mathbb {P}}_i^{(j)}\). So \(G=\sum _{i,j}H_{ij}\big (\frac{{\mathbb {P}}_i^{\text {N}}}{M_i}\big )^{(j)}=\sum _{i,j}\widetilde{H}_{ij}\big ({\mathbb {P}}_i^{\text {N}}\big )^{(j)}\) with \(\widetilde{H}_{ij}\in \mathbb {Q}\{{\mathbb {Y}}^{\pm };{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}\). Thus, there exists an \(M\in \mathbb {m}\) such that \(MG\in [{\mathbb {P}}_0^{\text {N}},\ldots ,{\mathbb {P}}_n^{\text {N}}]_{\mathbb {Q}\{{\mathbb {Y}},{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}}\) and \(G\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\cap \mathbb {Q}\{{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}\) follows. \(\square \)

Theorem 3.9

Proof

To prove 1), it suffices to show that \(\theta =(\eta ;\zeta )\) is a generic point of \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\). Clearly, \({\mathbb {P}}_i^{\text {N}}=M_i{\mathbb {P}}_i\) vanishes at \(\theta \) \((i=0,\ldots ,n)\). For any \(f\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\), there exists an \(M\in \mathbb {m}\) such that \(Mf\in [{\mathbb {P}}_0^{\text {N}},{\mathbb {P}}_1^{\text {N}},\ldots ,{\mathbb {P}}_n^{\text {N}}]_{{\mathbb Q}\{{\mathbb {Y}},{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}}\). It follows that \(f(\theta )=0\). Conversely, let \(f\) be any differential polynomial in \({\mathbb Q}\{{\mathbb {Y}},{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}\) satisfying \(f(\theta )=0\). Clearly, \({\mathbb {P}}_0^{\text {N}},{\mathbb {P}}_1^{\text {N}},\ldots ,{\mathbb {P}}_n^{\text {N}}\) constitute an auto-reduced set with \(u_{i0}\) as leaders. Let \(f_1\) be the differential remainder of \(f\) w.r.t. this auto-reduced set. Since \({\mathbb {P}}_i^\text {N}\) is linear in \(u_{i0}\), \(f_1\) is free from \(u_{i0}\,(i=0,\ldots ,n)\). By (2), there exist \(k_i\ge 0\) such that \(\prod _{i=0}^n(N_{i0})^{k_i}\cdot f\equiv f_1,\mathrm{{mod}}\,[{\mathbb {P}}_0^{\text {N}},{\mathbb {P}}_1^{\text {N}},\ldots ,{\mathbb {P}}_n^{\text {N}}]\). Hence, \(f_1(\theta )=0\). Since \(f_1\in {\mathbb Q}\{{\mathbf {u}},{\mathbb {Y}}\}\), \(f_1(\theta )=f_1(\eta ,{\mathbf {u}})=0\) means \(f_1=0\). Thus, \(f\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\). So \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\) is a prime differential ideal with \(\theta \) as its generic point.

Consequently, \(\mathcal {I}_{\mathbf {u}}=\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\cap {\mathbb Q}\{{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}\) is a prime differential ideal with a generic point \(\zeta =(\zeta _0,u_{01},\ldots ,u_{0l_0};\ldots ;\zeta _n,u_{n1},\ldots ,u_{nl_n})\). From (9), it is clear that \(\hbox {{d.tr.deg}}\,{\mathbb Q}\langle \zeta \rangle /{\mathbb Q}\) \(\le \sum _{i=0}^nl_i+n\). Suppose \({\mathbb {P}}_0,\ldots ,{\mathbb {P}}_n\) form a Laurent differentially essential system, that is, there exist pairs \((i_k,j_k)\)  \((k=1,\ldots ,n)\) with \(1\le j_k\le l_{i_k}\) and \(i_{k_1}\ne i_{k_2}\) \((k_1\ne k_2)\) such that \(\frac{N_{i_1j_1}}{N_{i_10}},\ldots ,\frac{N_{i_nj_n}}{N_{i_n0}}\) are differentially independent over \({\mathbb Q}\). Then by Lemma 2.1, \(\zeta _{i_1},\ldots ,\zeta _{i_n}\) are differentially independent over \({\mathbb Q}\langle {\mathbf {u}}\rangle \). For if not, by specializing \(u_{i_kj_k}\) to \(-1\) and the other \({\mathbf {u}}\) to \(0\), Lemma 2.1 guarantees that \(\frac{N_{i_1j_1}}{N_{i_10}},\ldots ,\frac{N_{i_nj_n}}{N_{i_n0}}\) are differentially dependent over \({\mathbb Q}\), a contradiction. Then it follows that \(\hbox {{d.tr.deg}}\,{\mathbb Q}\langle \zeta \rangle /{\mathbb Q}= \sum _{i=0}^nl_i+n\). Thus, \(\mathcal {I}_{\mathbf {u}}\) is of codimension 1.

Conversely, assume that \(\mathcal {I}_{\mathbf {u}}\) is of codimension 1. That is, \(\hbox {{d.tr.deg}}\,{\mathbb Q}\langle \zeta \rangle /{\mathbb Q}= \sum _{i=0}^nl_i+n\). We need to show that there exist pairs \((i_k,j_k)\)  \((k=1,\ldots ,n)\) with \(1\le j_k\le l_{i_k}\) and \(i_{k_1}\ne i_{k_2}\) \((k_1\ne k_2)\) such that \(\frac{N_{i_1j_1}}{N_{i_10}},\ldots ,\frac{N_{i_nj_n}}{N_{i_n0}}\) are differentially independent over \({\mathbb Q}\). Suppose the contrary, i.e., \(\frac{N_{i_1j_1}(\eta )}{N_{i_10}(\eta )},\ldots ,\frac{N_{i_nj_n}(\eta )}{N_{i_n0}(\eta )}\) are differentially dependent for any \(n\) different \(i_k\) and \(j_k\in \{1,\ldots ,l_{i_k}\}\). Since each \(\zeta _{i_k}\) is a linear combination of \(\frac{N_{i_kj_k}(\eta )}{N_{i_k0}(\eta )}\) \((j_k=1,\ldots ,l_{i_k})\), it follows that \(\zeta _{i_1},\ldots ,\zeta _{i_n}\) are differentially dependent over \({\mathbb Q}\langle {\mathbf {u}}\rangle \). So \(\hbox {{d.tr.deg}}\,{\mathbb Q}\langle \zeta \rangle /{\mathbb Q}< \sum _{i=0}^nl_i+n\), a contradiction. \(\square \)

Definition 3.10

\({\mathbf {R}}({\mathbf {u}}_{0},\ldots ,{\mathbf {u}}_{n})\in {\mathbb Q}\{{\mathbf {u}}_{0},\ldots ,{\mathbf {u}}_{n}\}\) in (10) is defined to be the sparse differential resultant of the Laurent differentially essential system \({\mathbb {P}}_0,\ldots ,{\mathbb {P}}_n\), denoted by \(\hbox {Res}_{\mathcal {A}_0,\ldots ,\mathcal {A}_n}\) or \(\hbox {Res}_{{\mathbb {P}}_0,\ldots ,{\mathbb {P}}_n}\). And when all the \(\mathcal {A}_i\) are equal to the same \(\mathcal {A}\), we simply denote it by \(\hbox {Res}_\mathcal {A}\).

Corollary 3.11

\(\mathcal {I}_{\mathbf {u}}=\mathrm{sat}({\mathbf {R}})\) is a prime differential ideal in \( {\mathbb Q}\{{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}\) with a generic zero \(\zeta \), where \(\zeta \) is defined in (9).

Corollary 3.12

\(\mathcal {I}_{\mathbf {u}}=\mathrm{sat}({\mathbf {R}})\) is a prime differential ideal in \({\mathbb Q}\{{\mathbf {u}},u_{00},\ldots ,u_{n0}\}\) with a generic zero \(\zeta =({\mathbf {u}},\zeta _0,\ldots ,\zeta _n)\), where \(\zeta _i\) is defined in (9).

Corollary 3.13

For each \(i\), if \({\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_i)=h_i\ge 0\), then \({\mathrm{ord}}({\mathbf {R}},u_{ik})=h_i\,(k=0,\ldots ,l_i)\).

Proof

Firstly, we claim that \({\mathrm{ord}}({\mathbf {R}},u_{i0})=h_i\). For if not, suppose \({\mathrm{ord}}({\mathbf {R}},u_{ik})=h_i\ge 0\) for some \(k\ne 0\). By (11), \(\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}({\mathbf {u}};\zeta _0,\ldots ,\zeta _n)=0\), where \( \zeta _i\) are defined in (9). By Corollary 3.12, we have \(\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}\in \mathrm{sat}({\mathbf {R}})\), a contradiction since \({\mathbf {R}}\) is irreducible. Thus, \({\mathrm{ord}}({\mathbf {R}},u_{i0})=h_i\). For each \(k\ne 0\), \({\mathrm{ord}}({\mathbf {R}},u_{ik})\le h_i\). If \({\mathrm{ord}}({\mathbf {R}},u_{ik})< h_i\), by (11), we have \(\frac{\partial {\mathbf {R}}}{\partial u_{i0}^{(h_i)}}({\mathbf {u}};\zeta _0,\ldots ,\zeta _n)\cdot (-\frac{N_{ik}(\eta )}{N_{i0}(\eta )})=0\). So \(\frac{\partial {\mathbf {R}}}{\partial u_{i0}^{(h_i)}}({\mathbf {u}};\zeta _0,\ldots ,\zeta _n)=0\) and \(\frac{\partial {\mathbf {R}}}{\partial u_{i0}^{(h_i)}}\in \mathrm{sat}({\mathbf {R}})\), a contradiction. Thus, for each \(k=0,\ldots ,l_i\), \({\mathrm{ord}}({\mathbf {R}},u_{ik})=h_i\). \(\square \)

Corollary 3.14

For \(i=1,\ldots ,n\) and \(k\in {\mathbb N}\), \(y_i^{(k)}\not \in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\).

Proof

Assume the contrary, \(y_i^{(k)}\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\). Since \(\zeta \) in (9) is a generic point of \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\), we have \(\eta _i^{(k)}=0\), which contradicts to the fact that \(\eta =(\eta _1,\ldots ,\eta _n)\) is a generic point of \(([0])_{\mathbb {Q}\langle {\mathbf {u}}\rangle \{{\mathbb {Y}}\}}\). \(\square \)

Remark 3.15

Due to Lemma 3.8, the sparse differential resultant can also be defined as follows: \(\mathcal {I}_{{\mathbb {Y}}^{\pm },{\mathbf {u}}}\cap {\mathbb Q}\{{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}= \mathrm{sat}({\mathbf {R}})\). Although the sparse differential resultant is defined for Laurent differential polynomials \({\mathbb {P}}_0,\ldots ,{\mathbb {P}}_n\), it is more convenient to prove its properties using \({\mathbb {P}}_0^\text {N},\ldots ,{\mathbb {P}}_n^\text {N}\) instead of \({\mathbb {P}}_0,\ldots ,{\mathbb {P}}_n\), since \({\mathbb {P}}_i^\text {N}\) are differential polynomials, and we can thus use results from differential algebra freely.

Remark 3.16

The sparse differential resultant can be computed with characteristic set methods for differential polynomials [3, 8, 25, 42, 47], which is implemented in the diffalg package of Maple. In Sect. 6, we will give an algorithm to compute the sparse differential resultant, which has a better complexity bound.

Example 3.17

Example 3.18

Example 3.19

Let \(\mathcal {A}_0=\{\mathbf{1 },y_1y_2\}\), \(\mathcal {A}_1=\{\mathbf{1 },y_1y'_2\}\), and \(\mathcal {A}_2=\{\mathbf{1 },y'_1y'_2\}\). It is easy to verify that \(\mathcal {A}_0,\mathcal {A}_1,\mathcal {A}_2\) form a Laurent differentially essential system. And \(\hbox {Res}_{{\mathcal {A}}_0,{\mathcal {A}}_1,{\mathcal {A}}_2}=u_{10}u_{01}u_{21}u_{11}u'_{00}-u_{10}u_{00}u_{11}u_{21}u'_{01} -u_{01}^2u_{21}u_{10}^2-u_{01}u_{00}u_{11}^2u_{20}.\)

Example 3.20

Let \(n=1\) and \(\mathcal {A}_0=\mathcal {A}_1=\{y_1^2,(y_1')^2,y_1y_1'\}\). Clearly, \(\mathcal {A}_0,\mathcal {A}_1\) form a Laurent differentially essential system and \(\hbox {Res}_{\mathcal {A}}=u_{11}^2u_{00}^2-2u_{01}u_{10}u_{11}u_{00}+u_{01}^2u_{10}^2- u_{12}u_{02}u_{11}u_{00}-u_{12}u_{02}u_{01}u_{10}+u_{12}^2u_{01}u_{00}+u_{10}u_{11}u_{02}^2\).

Example 3.21

Let \(n=1\) and \(\mathcal {A}_0=\mathcal {A}_1=\{y_1,y_1',y_1^2\}\). Clearly, \(\mathcal {A}_0,\mathcal {A}_1\) form a Laurent differentially essential system and \(\hbox {Res}_{\mathcal {A}}=-u_{12}u_{01}u_{00}u_{10} -u_{12}u_{01} ^2u'_{10}+u_{12}u_{01}u'_{11}u_{00} + u_{12}u_{01}u_{11}u'_{00}-u_{11}u_{02}u_{00}u_{10}+u_{11}u_{02}u'_{10}u_{01} + u_{02}u_{01}u_{10}^2-u_{11}^2u_{02}u'_{00}+u_{11}u_{02}u'_{01}u_{10} +u_{11}u_{00}^2u_{12}\) \( + u_{11}^2u'_{02}u_{00}-u_{11}u'_{02}u_{01}u_{10} -u_{11}u_{01}u'_{12}u_{00} + u_{01}^2u'_{12}u_{10} -u_{11}u'_{01}u_{12}u_{00}-u'_{11}u_{02}u_{01}u_{10}. \)

Remark 3.22

When all the \(\mathcal {A}_i\,(i=0,\ldots ,n)\) are sets of differential monomials as in the above examples, unless explicitly mentioned, we always consider \({\mathbb {P}}_i\) as Laurent differential polynomials. In this paper, sometimes we regard \({\mathbb {P}}_i\) as differential polynomials where it will be indicated.

Definition 3.23

Let \(\mathcal {A}_i=\{M_{i0},M_{i1},\ldots ,M_{il_i}\}\,(i=0,\ldots ,n)\) be finite sets of Laurent differential monomials which form a Laurent differentially essential system. Consider \(n+1\) Laurent differential polynomials \((F_0,\ldots ,F_n)\in \prod _{i=0}^n\mathcal {L}(\mathcal {A}_i)\). The sparse differential resultant of \(F_0,\ldots ,F_n\), denoted as \(\hbox {Res}_{F_0,\ldots ,F_n}\), is obtained by replacing \({\mathbf {u}}_i\) by the corresponding coefficient vector of \(F_i\) in \(\hbox {Res}({\mathbf {u}}_{0},\ldots ,{\mathbf {u}}_{n})\) which is the sparse differential resultant of the \(n+1\) generic Laurent differential polynomials in (3).

Definition 4.1

A set of Laurent differential monomials \(B_1, B_2, \ldots , B_m\) or its symbolic support matrix \(\mathrm{{D}}\) is called reduced if for each \(i\le \min (m,n)\), \(-\infty \ne {\mathrm{ord}}(B_i,y_i) > {\mathrm{ord}}(B_{i+k},y_i)\), or equivalently \(-\infty \ne \mathrm{deg}(d_{ii},x_i)>\mathrm{deg}(d_{i+k,i},x_i)\), holds for all \(k>0\).

Example 4.2

Lemma 4.3

For \(\eta _1,\eta _2\) in an extension field of \(\mathbb {Q}\), \(\mathrm{d.tr.deg}\,\mathbb {Q}\langle \eta _1^{a_1},\eta _1^{a_2}\eta _2 \rangle /\mathbb {Q}=\mathrm{d.tr.deg}\,\mathbb {Q}\langle \eta _1,\) \(\eta _2 \rangle /\mathbb {Q}\), where \(a_1,a_2\) are nonzero rational numbers.

Proof

Theorem 4.4

Let \(B_1, B_2, \ldots , B_m\) be a set of reduced Laurent differential monomials in \({\mathbb {Y}}\). Then \(\mathrm{d.tr.deg}\,{\mathbb Q}\langle B_1, B_2, \ldots , B_m\rangle /{\mathbb Q}=\min (m,n)\).

Proof

It suffices to prove the case \(m = n\) by the following two facts. In the case \(m>n\), we need only to prove that \(B_1,\ldots , B_n\) are differentially independent. And in the case \(m<n\), we can treat \(y_{m+1},\ldots ,y_n\) as parameters, then \(B_1, B_2, \ldots , B_m\) are still reduced Laurent differential monomials. So if we have proved the result for \(m=n\), \(\hbox {{d.tr.deg}}\, {\mathbb Q}\langle B_1, B_2, \ldots , B_m\rangle /{\mathbb Q}\ge \hbox {{d.tr.deg}}\, {\mathbb Q}\langle y_{m+1},\ldots ,y_n\rangle \langle B_1, B_2, \ldots , B_m\rangle /{\mathbb Q}\langle y_{m+1},\ldots ,y_n\rangle \) \( = m\) follows.

Now, we prove that \(B_1,\ldots ,B_n\) are differentially independent over \(\mathbb {Q}\). Suppose the contrary, then there exists a nonzero differential polynomial \(P\in \mathbb {Q}\{z_1,\ldots ,z_n\}\) such that \(P(B_1,\ldots ,B_n) = 0\). Let \(P = \sum _k c_k P_k \), where \(P_k\) is a monomial and \(c_k\in {\mathbb Q}\backslash \{0\}\). Then, the leading monomial of \(P_k(B_1,\ldots ,B_n)\) is a product of \(LM_{it}\, (i=1,\ldots ,n; t\ge 0)\). We denote this product by \(LMP_k\), then \(LMP_k \ne LMP_j\) for \(k\ne j\) since these \(LM_{it}\) are algebraically independent. But there exists one and only one product which has the highest order, which cannot be eliminated by the others, which means that \(P(B_1,\ldots ,B_n)\ne 0\), a contradiction. \(\square \)

Definition 4.5

A set of generalized Laurent differential monomials \(B_1,\ldots ,B_m\) or their symbolic support matrix \(\mathrm{{D}}\) is said to be in T-shape with index \((i,j)\), if there exist \(1\le i\le \min (m,n), 0\le j \le \min (m,n)-i\) such that all elements except those in the first \(i\) rows and the \(i+1,\ldots ,(i+j)\)th columns of \(\mathrm{{D}}\) are zeros and the sub-matrix consisting of the first \(i+j\) columns of \(\mathrm{{D}}\) is reduced.

Lemma 4.6

The rank of a T-shape matrix with index \((i,j)\) equals to \(i+j\). Furthermore, a T-shape matrix is reduced if and only if it is of full rank, that is, \(i+j=\min {(m,n)}\).

Lemma 4.7

A T-shape matrix \(\mathrm{{D}}\) of index \((i,j)\) is not of full rank if and only if the \(0\)-rank \(r=m+n-i-j\) of its zero sub-matrix satisfies \(r \ge \mathrm{max}(m,n)+1\).

Proof

Note that the zero sub-matrix of \(\mathrm{{D}}\) is an \((m-i)\times (n-j)\) matrix with \(0\)-rank \(r=m+n-i-j\). By Lemma 4.6, \(\mathrm{{D}}\) is not of full rank if and only if \(i+j < \min (m,n)\), which is equivalent to \(r=m+n-i-j>m+n-\min (m,n)\) or \(r\ge \mathrm{max}(m,n)+1\). \(\square \)

Theorem 4.8

Let \(B_1,\ldots ,B_m\) be generalized Laurent differential monomials and \(\mathrm{{D}}\) their symbolic support matrix which is in T-shape with index \((i,j)\). Then \(\mathrm{d.tr.deg}\,\) \( {\mathbb Q}\langle B_1, \ldots , B_m\rangle /{\mathbb Q}=\mathrm{rank}(\mathrm{{D}})=i+j\).

Proof

Without loss of generality, each \(B_i\) is assumed to be a Laurent differential monomial. For otherwise, by Lemma 4.3, we may consider \(B_i^{k_i}\) for certain \(k_i\in {\mathbb N}\), which is a Laurent differential monomial.

Lemma 4.9

Let \(B_1,\ldots ,B_m\) be Laurent differential monomials and \(C_1,\ldots ,C_m\) generalized Laurent differential monomials obtained from \(B_1,\ldots ,B_m\) by a series of \({\mathbb Q}\)-elementary transformations. Then \(\mathrm{d.tr.deg}\,{\mathbb Q}\langle B_1, \ldots , B_m\rangle /{\mathbb Q}=\mathrm{d.tr.deg}\, {\mathbb Q}\langle C_1,\) \(\ldots , C_m\rangle /{\mathbb Q}\).

Theorem 4.10

The symbolic support matrix of any Laurent differential monomials \(B_1,\) \( \ldots , B_m\) can be reduced to a T-shape matrix by a finite number of \({\mathbb Q}\)-elementary transformations.

Theorem 4.11

Let \(B_1,\ldots , B_m\) be Laurent differential monomials in \({\mathbb {Y}}\) and \(\mathrm{{D}}\) their symbolic support matrix. Then \(\mathrm{d.tr.deg}\,{\mathbb Q}\langle B_1, \ldots , B_m\rangle /{\mathbb Q}=\mathrm{rank}(\mathrm{{D}})\).

Proof

By Lemma 4.9, \({\mathbb Q}\)-elementary transformations keep the differential transcendence degree unchanged. The result follows from Theorems 4.8 and 4.10. \(\square \)

Corollary 4.12

The Laurent differential system (3) is Laurent differentially essential if and only if there exist \(M_{ij_i}\,(i=0,\ldots ,n)\) with \(1\le j_i\le l_{i}\) such that the symbolic support matrix of the Laurent differential monomials \(M_{0j_0}/M_{00},\ldots ,M_{nj_n}/M_{n0}\) is of rank \(n\).

Remark 4.13

Lemma 4.14

Let \(\mathrm{{D}}_{k_1,\ldots ,k_m}\) be the symbolic support matrix of the Laurent differential monomials \((M_{1k_1}/M_{10},\ldots ,M_{mk_m}/M_{m0})\). Then \(\mathrm{rank}(\mathrm{{D}}_{\mathbb {P}}) = \mathrm{max}_{ 1\le k_i \le l_i} \mathrm{rank}(\mathrm{{D}}_{k_1,\ldots ,k_m})\).

Proof

Let the rank of \(\mathrm{{D}}_{\mathbb {P}}\) be \(r\). Without loss of generality, we assume that the \(r\times r\) leading principal sub-matrix of \(\mathrm{{D}}_{\mathbb {P}}\), say \(\mathrm{{D}}_{{\mathbb {P}},r}\), is of full rank. By the properties of determinants, \(\det (\mathrm{{D}}_{{\mathbb {P}},r}) = \sum \limits _{k_1=1}^{ l_1}\cdots \sum \limits _{k_r=1}^{ l_r} \prod _{i=1}^r u_{ik_i}\det (k_1,\ldots ,k_r)\) where \(\det (k_1,\ldots ,k_r)\) is the determinant of the \(r\times r\) leading principal sub-matrix of \(\mathrm{{D}}_{k_1,\ldots ,k_m}\). So \(\det (\mathrm{{D}}_{{\mathbb {P}},r})\ne 0\) if and only if there exist \(k_1,\ldots ,k_r\) such that \(\det (k_1,\ldots ,k_r)\ne 0\). Hence, the rank of \(\mathrm{{D}}_{k_1,\ldots ,k_m}\) is no less than the rank of \(\mathrm{{D}}_{\mathbb {P}}\). On the other hand, let \(s= \mathrm{max}_{ 1\le k_i \le l_i} \mathrm{rank}(\mathrm{{D}}_{k_1,\ldots ,k_m})\). Without loss of generality, we assume \(\det (k_1,\ldots ,k_s)\ne 0\), then, \(\det (\mathrm{{D}}_{{\mathbb {P}},s})\ne 0\). Hence, \(s\) is no greater than the rank of \(\mathrm{{D}}_{\mathbb {P}}\). \(\square \)

Theorem 4.15

For \({\mathbb {P}}_i\) given in (12), \(\mathrm{d.tr.deg}\,{\mathbb Q}\langle \cup _{i=1}^m{\mathbf {u}}_i\rangle \langle {\mathbb {P}}_1/M_{10},\ldots , {\mathbb {P}}_m/M_{m0}\rangle /\) \({\mathbb Q}\langle \cup _{i=1}^m{\mathbf {u}}_i\rangle = \mathrm{rank}(\mathrm{{D}}_{\mathbb {P}})\).

Proof

By Lemma 2.1, \(\hbox {{d.tr.deg}}\,{\mathbb Q}\langle \cup _{i=1}^m{\mathbf {u}}_i\rangle \langle {\mathbb {P}}_1/M_{10},\ldots , {\mathbb {P}}_m/M_{m0}\rangle /{\mathbb Q}\langle \cup _{i=1}^m{\mathbf {u}}_i\rangle \) is no less than the maximal differential transcendence degree of \(M_{1k_1}/M_{10},\ldots ,M_{mk_m}/M_{m0}\) over \({\mathbb Q}\).

On the other hand, the differential transcendence degree will not increase by linear combinations, since for any differential polynomial \(a_i\) and \( \bar{a}_1\), \(\hbox {{d.tr.deg}}\,{\mathbb Q}\langle \lambda \rangle \langle a_1+\lambda \bar{a}_1,a_2,\ldots ,a_k\rangle /{\mathbb Q}\langle \lambda \rangle \le \mathrm{max}\{\hbox {{d.tr.deg}}\,{\mathbb Q}\langle a_1,a_2\ldots ,a_k\rangle /{\mathbb Q},\) \( \hbox {{d.tr.deg}}\,{\mathbb Q}\langle \bar{a}_1,a_2,\ldots ,a_k\rangle /{\mathbb Q}\}\). So, the differential transcendence degree of \({\mathbb {P}}_1/M_{10},\ldots , {\mathbb {P}}_m/M_{m0}\) over \({\mathbb Q}\langle \cup _{i=1}^m{\mathbf {u}}_i \rangle \) is no greater than the maximal differential transcendence degree of \(M_{1k_1}/M_{10},\ldots ,M_{mk_m}/M_{m0}\).

Thus, \(\hbox {{d.tr.deg}}\,{\mathbb Q}\langle \cup _{i=1}^m{\mathbf {u}}_i\rangle \langle {\mathbb {P}}_1/M_{10},\ldots , {\mathbb {P}}_m/M_{m0}\rangle /{\mathbb Q}\langle \cup _{i=1}^m{\mathbf {u}}_i\rangle =\mathrm{max}_{ k_i}\hbox {{d.tr.deg}}\,\mathbb {Q}\langle M_{1k_1}/M_{10},\) \(\ldots ,M_{mk_m}/M_{m0}\rangle /\mathbb {Q}\). By Theorem 4.11 and Lemma 4.14, the differential transcendence degree of \({\mathbb {P}}_1/M_{10},\ldots , {\mathbb {P}}_m/M_{m0}\) equals to the rank of \(\mathrm{{D}}_{\mathbb {P}}\).\(\square \)

Corollary 4.16

The Laurent differential system \({\mathbb {P}}=\{{\mathbb {P}}_0,\ldots ,{\mathbb {P}}_n\}\) defined in (3) is Laurent differentially essential if and only if \(\mathrm{rank}(\mathrm{{D}}_{\mathbb {P}})=n\).

Theorem 4.17

The above \(\mathcal {I}_u\) is a differential prime ideal with codimension \(m-\mathrm{rank}(\mathrm{{D}}_{{\mathbb {P}}})\).

Proof

Lemma 4.18

Let \({\mathbb {P}}=\{{\mathbb {P}}_0,\ldots ,{\mathbb {P}}_n\}\) given in (3) be a Laurent differentially essential system and \(I\subset \{0, 1,\ldots , n \}\). If \(|I|- \mathrm{rank}(\mathrm{{D}}_{{\mathbb {P}}_I}) = 1\), then \(([{\mathbb {P}}_I^{\text {N}}]\hbox {:}\mathbb {m})_{{\mathbb Q}\{{\mathbb {Y}},{\mathbf {u}}_I\}}\cap {\mathbb Q}\{{\mathbf {u}}_I\}=\mathrm{sat}({\mathbf {R}})\).

Proof

By Theorem 4.17, \(\mathcal {I}_1=([{\mathbb {P}}_I^{\text {N}}]\hbox {:}\mathbb {m})_{{\mathbb Q}\{{\mathbb {Y}},{\mathbf {u}}_I\}}\cap {\mathbb Q}\{{\mathbf {u}}_I\}\) is of codimension \(|I|-\mathrm{rank}(\mathrm{{D}}_{{\mathbb {P}}_I})=1\). Then \(\mathcal {I}_1=\mathrm{sat}({\mathbf {R}}_1)\subset \mathrm{sat}({\mathbf {R}})\) for an irreducible differential polynomial \({\mathbf {R}}_1\in {\mathbb Q}\{{\mathbf {u}}_I\}\). By Lemma 2.3, \({\mathbf {R}}\) can reduce \({\mathbf {R}}_1\) to zero under any ranking. If \(I = \{0,1,\ldots ,n\}\), then the lemma is proved. Otherwise, for any \(k\in \{0,1,\ldots ,n\}\setminus I\), we claim that \({\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_k)=-\infty \). Suppose the contrary, then under an arbitrary elimination ranking satisfying \({\mathbf {u}}_k > {\mathbf {u}}_i\) for \(i\ne k\), \({\mathbf {R}}_1\) cannot be reduced to zero w.r.t \({\mathbf {R}}\), a contradiction to \({\mathbf {R}}_1\in \mathrm{sat}({\mathbf {R}})\). So \({\mathbf {R}}\in {\mathbb Q}\{{\mathbf {u}}_I\}\) and it is easy to check that \({\mathbf {R}}\in ([{\mathbb {P}}_I^{\text {N}}]\hbox {:}\mathbb {m})_{{\mathbb Q}\{{\mathbb {Y}},{\mathbf {u}}_I\}}\cap {\mathbb Q}\{{\mathbf {u}}_I\}=\mathrm{sat}({\mathbf {R}}_1)\). Then \(\mathrm{sat}({\mathbf {R}})=\mathrm{sat}({\mathbf {R}}_1)\) and the lemma is proved. \(\square \)

Definition 4.19

Let \(I\subset \{0,1,\ldots ,n\}\). Then we say \(I\) or \({\mathbb {P}}_I\) is rank essential if the following conditions hold: (1) \(|I| - \mathrm{rank}(\mathrm{{D}}_{{\mathbb {P}}_I}) = 1\) and (2) \(|J| = \mathrm{rank}(\mathrm{{D}}_{{\mathbb {P}}_J})\) for each proper subset \(J\) of \(I\).

Theorem 4.20

Let \({\mathbb {P}}=\{{\mathbb {P}}_0,\ldots ,{\mathbb {P}}_n\}\) given in (3) be a Laurent differentially essential system. Then for any \(I\subset \{0, 1,\ldots , n \}\), \(|I|- \mathrm{rank}(\mathrm{{D}}_{{\mathbb {P}}_I}) \le 1\) and there exists a unique \(I\) which is rank essential. If \(I\) is rank essential, then \({\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_i)\ge 0\) if and only if \(i\in I\).

Proof

Let \(I\) be a rank essential set. By Lemma 4.18, the sparse differential resultant \({\mathbf {R}}\) of \({\mathbb {P}}\) involves only the coefficients of \({\mathbb {P}}_i\,({i\in I})\). For any \(i\in I\), let \(I_{\hat{i}}=I\setminus \{i\}\). Since \(I\) is rank essential, we have \(([{\mathbb {P}}_{I_{\hat{i}}}]\hbox {:}\mathbb {m})_{{\mathbb Q}\{{\mathbb {Y}},{\mathbf {u}}_{I_{\hat{i}}}\}}\cap {\mathbb Q}\{{\mathbf {u}}_{I_{\hat{i}}}\}=[0]\) and hence \({\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_i)\ge 0\) for any \(i\in I\). \(\square \)

Remark 4.21

Let \({\mathbb {P}}=\{{\mathbb {P}}_0,\ldots ,{\mathbb {P}}_n\}\) given in (3) be a Laurent differentially essential system. We can obtain a rank essential set \(I\subset \{0, 1,\ldots , n \}\) as follows. Let \(J=\{0, 1,\ldots , n \}\). If for all \(j\in J\), \(|J\setminus \{j\}| = \mathrm{rank}(\mathrm{{D}}_{{\mathbb {P}}_{J\setminus \{j\}}})\), then \(J\) is rank essential. Otherwise, by Theorem 4.20, there exists an \(j_0\in J\) such that \(|J\setminus \{j_0\}| = \mathrm{rank}(\mathrm{{D}}_{{\mathbb {P}}_{J\setminus \{j_0\}}})+1\). Repeating the procedure for \(J:=J\setminus \{j_0\}\), we will eventually obtain a rank essential system.

Example 4.22

In Example 3.18, \(\{{\mathbb {P}}_0,{\mathbb {P}}_1\}\) is a rank essential set since they involve \(y_1\) only.

Example 4.23

Theorem 4.24

Let \({\mathbb {P}}_i= u_{i0} + \sum \limits _{k=1}^{l_i}u_{ik} M_{ik}\,(i=1,\ldots ,m;\,m\le n)\) be generic differential polynomials in \(n\) differential indeterminates \({\mathbb {Y}}\) and \({\mathbf {u}}_i=(u_{i0},\ldots ,u_{il_i})\). Then \([{\mathbb {P}}_1,\ldots ,{\mathbb {P}}_m]_{{\mathbb Q}\langle {\mathbf {u}}_1,\ldots ,{\mathbf {u}}_m\rangle \{{\mathbb {Y}}\}}\) is either the unit ideal or a prime differential ideal of dimension \(n-m\).

Proof

Use the notation introduced in the proof of Theorem 4.17 with \(M_{i0}=1\). Let \(\mathcal {I}_0 = [{\mathbb {P}}_1,\ldots ,{\mathbb {P}}_m]_{{\mathbb Q}\{{\mathbf {u}}_1,\ldots ,{\mathbf {u}}_m, {\mathbb {Y}}\}}\) and \(\mathcal {I}_1 = [{\mathbb {P}}_1,\ldots ,{\mathbb {P}}_m]_{{\mathbb Q}\langle {\mathbf {u}}_1,\ldots ,{\mathbf {u}}_m\rangle \{{\mathbb {Y}}\}}\). Since \({\mathbb {P}}_i\) contains a nonvanishing degree zero term \(u_{i0}\), it is clear that \(\mathcal {I}_0=\mathcal {I}_0\hbox {{:}}\mathbb {m}=\mathcal {I}_1\cap {\mathbb Q}\{{\mathbf {u}}_1,\ldots ,{\mathbf {u}}_m, {\mathbb {Y}}\}\).

Lemma 5.1

Suppose the Laurent differential monomial sets \(\mathcal {A}_i\,(i=0,\ldots ,n)\) form a Laurent differentially essential system. Then \(\mathcal {Z}(\mathcal {A}_0,\ldots ,\mathcal {A}_n)\subseteq {\mathbb {V}}\big (\mathrm{sat}(\mathrm{Res}_{\mathcal {A}_0,\ldots ,\mathcal {A}_n})\big )\).

Proof

Example 5.2

Consider Example 3.17. Suppose \({\mathcal {F}}={\mathbb Q}(x)\) and \(\delta = \frac{d}{d x}\). In this example, we have \(\hbox {Res}_{{\mathbb {P}}_0,{\mathbb {P}}_1,{\mathbb {P}}_2}\ne 0\). But \(y_1=c_{11}x+c_{10}, y_2=c_{22}x^2+c_{21}x+c_{20}\) consist of a nonzero solution of \({\mathbb {P}}_0={\mathbb {P}}_1={\mathbb {P}}_2=0\) where \(c_{ij}\) are distinct arbitrary constants. This shows that Lemma 5.1 is not correct if we do not consider non-polynomial solutions. This example also shows why we need to consider non-polynomial differential solutions, or equivalently why we consider Laurent differential polynomials instead of the usual differential polynomials.

Theorem 5.3

Suppose the Laurent differential monomial sets \(\mathcal {A}_i\,(i=0,\ldots ,n)\) form a Laurent differentially essential system. Then \(\overline{\mathcal {Z}(\mathcal {A}_0,\ldots ,\mathcal {A}_n)}\!=\!{\mathbb {V}}\big (\mathrm{sat}(\mathrm{Res}_{\mathcal {A}_0,\ldots ,\mathcal {A}_n})\big )\).

Proof

Firstly, by Lemma 5.1, \(\mathcal {Z}(\mathcal {A}_0,\ldots ,\mathcal {A}_n) \subseteq {\mathbb {V}}\big (\mathrm{sat}(\hbox {Res}_{\mathcal {A}_0,\ldots ,\mathcal {A}_n})\big )\). So \(\overline{\mathcal {Z}(\mathcal {A}_0,\ldots ,\mathcal {A}_n)} \subseteq {\mathbb {V}}\big (\mathrm{sat}(\hbox {Res}_{\mathcal {A}_0,\ldots ,\mathcal {A}_n})\big )\).

For the other direction, let \(\eta ,\zeta \) be as defined in (9). By Theorem 3.9, \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\) is a prime differential ideal with a generic point \((\eta ;\zeta )\). Let \((F_0,\ldots ,F_n)\in \mathcal {L}(\mathcal {A}_0)\times \cdots \times \mathcal {L}(\mathcal {A}_n)\) be a set of Laurent differential polynomials represented by \(\zeta \). Clearly, \(\eta \) is a non-polynomial solution of \(F_i=0\). Thus, \(\zeta \in \mathcal {Z}(\mathcal {A}_0,\ldots ,\mathcal {A}_n) \subset \overline{\mathcal {Z}(\mathcal {A}_0,\ldots ,\mathcal {A}_n)}\). By Corollary 3.11, \(\zeta \) is a generic point of \(\mathrm{sat}(\hbox {Res}_{\mathcal {A}_0,\ldots ,\mathcal {A}_n})\). It follows that \({\mathbb {V}}\big (\mathrm{sat}(\hbox {Res}_{\mathcal {A}_0,\ldots , \mathcal {A}_n})\big )\subseteq \overline{\mathcal {Z}(\mathcal {A}_0, \ldots ,\mathcal {A}_n)}\). As a consequence, \({\mathbb {V}}\big (\mathrm{sat}(\hbox {Res}_{\mathcal {A}_0,\ldots ,\mathcal {A}_n})\big )= \overline{\mathcal {Z}(\mathcal {A}_0,\ldots ,\mathcal {A}_n)}\). \(\square \)

Theorem 5.4

Proof

Let \(\mathcal {J}=\big ([{\mathbf {R}},(Q_{ik})_{0\le i\le n;1\le k\le l_i}]\hbox {{:}}S^\infty \big )_{\mathbb {Q}\{{\mathbb {Y}},{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}}\). By Theorem 3.9, \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\) is a prime differential ideal with a generic point \(\theta =(\eta ;\zeta )\) given in (9). By Corollary 3.12, \(\zeta =({\mathbf {u}},\zeta _0,\ldots ,\zeta _n)\) is a generic zero point of \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\cap \mathbb {Q}\{{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}=\mathrm{sat}({\mathbf {R}})\). Since \(\frac{M_{ik}(\eta )}{M_{i0}(\eta )}=\frac{N_{ik}(\eta )}{N_{i0}(\eta )}\), by (11), \(Q_{ik}({\mathbf {u}},\zeta _0,\ldots ,\zeta _n)=0\). So \(Q_{ik}\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\).

Since \({\mathbb {P}}\) is rank essential, \({\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_i)\ge 0\). Substituting \(\big (Q_{ik}+N_{i0}\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}\big )/\frac{\partial {\mathbf {R}}}{\partial u_{i0}^{(h_i)}}\) for \(N_{ik}\) in each \({\mathbb {P}}_i^\text {N}\), we obtain \({\mathbb {P}}_i^\text {N}=u_{i0}N_{i0}+\sum _{k=1}^{l_i}u_{ik}\big (Q_{ik}+N_{i0}\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}\big )/\frac{\partial {\mathbf {R}}}{\partial u_{i0}^{(h_i)}}\). So \(\frac{\partial {\mathbf {R}}}{\partial u_{i0}^{(h_i)}}{\mathbb {P}}_i^\text {N}=\sum _{k=1}^{l_i} u_{ik}Q_{ik}+(\sum _{k=0}^{l_i}u_{ik}\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}})N_{i0}\). Since \(Q_{ik}\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\), \(R_i=\sum _{k=0}^{l_i}u_{ik}\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\). Since \(R_i\) and \({\mathbf {R}}\) have the same degree and \({\mathbf {R}}\) is irreducible, there exists some \(a\in \mathbb {Q}\) such that \(R_i=a{\mathbf {R}}\). It follows that \({\mathbb {P}}_i^\text {N}\in \mathcal {J}\). For any differential polynomial \(f\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\), there exists a differential monomial \(M\in \mathbb {m}\) such that \(Mf\in [{\mathbb {P}}_0^\text {N},\ldots ,{\mathbb {P}}_n^\text {N}]\subset \mathcal {J}\). Thus, \(f\in \mathcal {J}\) and \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\subseteq \mathcal {J}\) follows. Conversely, for any differential polynomial \(g\in \mathcal {J}\), there exist some differential monomial \(M\) and some \(b\in \mathbb {N}\) such that \(M(\prod _{i}\frac{\partial {\mathbf {R}}}{\partial u_{i0}^{(h_i)}})^bg\in [{\mathbf {R}},Q_{ik}]\subset \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\). Since \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\) is a prime differential ideal, \(g\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\). Hence, \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}=\mathcal {J}.\) \(\square \)

Definition 5.5

\({\mathbb {P}}=\{{\mathbb {P}}_0,\ldots ,{\mathbb {P}}_n\}\) given in (3) is called normal rank essential if \({\mathbb {P}}\) is rank essential and for each \(j=1,\ldots ,n\), \(\mathbf e _j\in \hbox {Span}_{\mathbb {Z}}(\{\alpha _{ik}-\alpha _{i0}\,|\,i=0,\ldots ,n;k=1,\ldots ,l_i\})\).

Lemma 5.6

Proof

Let \({\mathcal {J}}=\mathrm{sat}({\mathbf {R}},S_1y_1-T_1,\ldots ,S_ny_n-T_n)_{\mathbb {Q}\{{\mathbb {Y}},{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}}\). It is easy to verify that \({\mathcal {J}}\) is a prime differential ideal. Since \({\mathbb {P}}\) is rank essential, \(h_i={\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_i)\ge 0\) for each \(i\). By equation (11), we have \(\frac{N_{ik}(\eta )}{N_{i0}(\eta )}= \frac{M_{ik}(\eta )}{M_{i0}(\eta )}=\overline{\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}}\big /\overline{\frac{\partial {\mathbf {R}}}{\partial u_{i0}^{(h_i)}}}\). Since \(\mathbf e _j\in \hbox {Span}_{\mathbb {Z}}(\{\alpha _{ik}-\alpha _{i0}\,|\,k=1,\ldots ,l_i;i=0,\ldots ,n\})\), for \(j=1,\ldots ,n\), there exist \(t_{jik}\in \mathbb {Z}\) such that \(\sum _{i,k}t_{jik}(\alpha _{ik}-\alpha _{i0})=\mathbf e _j\). So, \(\prod _{i,k}\Big (\frac{N_{ik}}{N_{i0}}\Big )^{t_{jik}}=y_j\). Thus, \(\prod _{i,k}\Big (\frac{N_{ik}(\eta )}{N_{i0}(\eta )}\Big )^{t_{jik}}=\eta _j=\prod _{i,k}\big (\overline{\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}}\big /\overline{\frac{\partial {\mathbf {R}}}{\partial u_{i0}^{(h_i)}}}\big )^{t_{jik}}\). By Theorem 3.9, there exist \(S_j\) and \(T_j\) which are nonnegative power products of \(\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}\) such that \(S_jy_j-T_j\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\). Since \(\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}\not \in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\) and \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\) is prime, \(S_j\not \in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\) follows. Thus, \({\mathcal {J}}\subset \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\). To prove \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\subset {\mathcal {J}}\), for each \(k=0,\ldots ,n\), let \(R_k\) be the differential remainder of \({\mathbb {P}}_k^\text {N}\) w.r.t. \({\mathbf {R}}, S_1y_1-T_1,\ldots ,S_ny_n-T_n\) under the given ranking. Then \(R_k\in {\mathbb Q}\{{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}\). And by (2), \(R_k\in [{\mathbf {R}}, S_1y_1-T_1,\ldots ,S_ny_n-T_n,{\mathbb {P}}_k^\text {N}]\subset \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\). So \(R_k\in {\mathbb Q}\{{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}\cap \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}=\mathrm{sat}({\mathbf {R}})\). Since \(R_k\) is reduced w.r.t. \({\mathbf {R}}\), \(R_k=0\) and \({\mathbb {P}}_k^\text {N}\in {\mathcal {J}}\) follows. By Corollary 3.14, \(y_i^{(j)}\not \in {\mathcal {J}}\subset \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\) for each \(i\) and \(j\). Thus, \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\subset {\mathcal {J}}\). \(\square \)

Theorem 5.7

Let \({\mathbb {P}}=\{{\mathbb {P}}_0,\ldots ,{\mathbb {P}}_n\}\) given in (3) be normal rank essential. Let \(\overline{{\mathbb {P}}}_{i}\) be a specialization of \({\mathbb {P}}_i\) with coefficient vector \({\mathbf {v}}_i\,(i=0,\ldots ,n)\). Then there exists a differential polynomial set \(\mathcal {S}\subset {\mathbb Q}\{{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}\) such that \({\mathbb {V}}({\mathbf {R}})\backslash \bigcup \limits _{S\in \mathcal {S}}{\mathbb {V}}(S)\ne \emptyset \) and whenever \(({\mathbf {v}}_0,\ldots ,{\mathbf {v}}_n)\in {\mathbb {V}}({\mathbf {R}}) \backslash \bigcup \limits _{S\in \mathcal {S}}{\mathbb {V}}(S)\), \(\overline{{\mathbb {P}}}_{i}=0\,(i=0,\ldots ,n)\) have a unique common non-polynomial solution.

Proof

By Lemma 5.6, \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}} = \mathrm{sat}({\mathbf {R}},A_1,\ldots ,A_n)\), where \(A_l=S_ly_l-T_l(l=1,\ldots ,n)\). Let \(\mathcal {S}=\{\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}},(S_j)^{m+1}\big (\frac{T_j}{S_j}\big )^{(m)}\,\big |\,\) \(0\le i\le n;0\le k\le l_i;1\le j\le n;m\in \mathbb {N}\}\). Firstly, we show that \({\mathbb {V}}({\mathbf {R}})\backslash \bigcup \limits _{S\in \mathcal {S}}{\mathbb {V}}(S)\ne \emptyset \). Suppose the contrary, viz. \({\mathbb {V}}({\mathbf {R}})\subset \bigcup \limits _{S\in \mathcal {S}}{\mathbb {V}}(S)\). In particular, there exists one \(S\in \mathcal {S}\) such that \(S\) vanishes at the generic point \(\zeta \) of \(\mathrm{sat}({\mathbf {R}})\). It is obvious that \(\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}\) does not vanish at \(\zeta \). If \((S_j)^{m+1}\big (\frac{T_j}{S_j}\big )^{(m)}\) vanishes at \(\zeta \) for some \(m\), \((S_j)^{m+1}\big (\frac{T_j}{S_j}\big )^{(m)}\in \mathrm{sat}({\mathbf {R}})\). Replacing \(\frac{T_j}{S_j}\) by \(y_j -\frac{A_j}{S_j}\), we have \(S_j^{m+1}y_j^{(m)}\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\). Since \(S_j\) is a power product of certain \(\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}\), \(S_j^{m+1}\not \in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\). Then, \(y_j^{(m)}\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\), contradicting to Corollary 3.14.

Suppose \(({\mathbf {v}}_0,\ldots ,{\mathbf {v}}_n)\in {\mathbb {V}}({\mathbf {R}})\backslash \bigcup \limits _{S\in \mathcal {S}}{\mathbb {V}}(S)\). Let \(\bar{T}_j=T_j({\mathbf {v}}_0,\ldots ,{\mathbf {v}}_n)\) and \(\bar{S_j}=S_j({\mathbf {v}}_0,\ldots ,{\mathbf {v}}_n)\). Since \(\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}({\mathbf {v}}_0,\ldots ,{\mathbf {v}}_n)\ne 0\) for each \(i\) and \(k\), \(\bar{T}_j\bar{S}_j\ne 0\). Let \(\bar{y}_j=\frac{\bar{T}_j}{\bar{S}_j}\) and denote \(\bar{y}=(\bar{y}_1,\ldots ,\bar{y}_n)\). For each \(m\in \mathbb {N},\) \(\bar{y}_j^{(m)}=(\frac{\bar{T}_j}{\bar{S}_j})^{(m)}\ne 0\). Thus, \(\bar{y}\in (\mathcal {E}^\wedge )^n.\) Since \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}=\mathrm{sat}({\mathbf {R}},A_1,\ldots ,A_n)\), \(H\cdot {\mathbb {P}}_i^\text {N}\in [{\mathbf {R}},A_1,\ldots ,A_n]\) where \(H\) is a product of powers of \(\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}\). Hence, \(\overline{{\mathbb {P}}}_{i}^\text {N}(\bar{y})= M_i(\bar{y})\cdot \overline{{\mathbb {P}}}_{i}(\bar{y})=0\), which implies that \(\overline{{\mathbb {P}}}_{i}(\bar{y})=0\). Thus, \(\bar{y}\) is a non-polynomial common solution of \(\overline{{\mathbb {P}}}_{i}\). On the other hand, if \(\xi \) is a non-polynomial common solution of \(\overline{{\mathbb {P}}}_{i}\), then \(\bar{S}_jy_j-\bar{T}_j\) vanishes at \(\xi \) for each \(i\). Hence, \(\xi =\bar{y}\). As a consequence, \(\overline{{\mathbb {P}}}_{i}=0\) have a unique common non-polynomial solution. \(\square \)

Corollary 5.8

Let \(\mathcal {Z}_1(\mathcal {A}_0,\ldots ,\mathcal {A}_n) \subset \mathcal {E}^{l_0+1}\times \cdots \times \mathcal {E}^{l_n+1}\) be the set consisting of \(({\mathbf {v}}_0,\ldots ,{\mathbf {v}}_n)\) for which the corresponding Laurent differential polynomials \(F_i=0\,(i=0,\ldots ,n)\) have a unique non-polynomial common solution and \(\overline{\mathcal {Z}_1(\mathcal {A}_0,\ldots ,\mathcal {A}_n)}\) the Kolchin closure of \(\mathcal {Z}_1(\mathcal {A}_0,\ldots ,\mathcal {A}_n)\). Then if \(\mathcal {A}_0,\ldots ,\mathcal {A}_n\) are normal rank essential, \(\overline{\mathcal {Z}_1(\mathcal {A}_0,\ldots ,\mathcal {A}_n)}=\) \({\mathbb {V}}\big (\mathrm{sat}(\) \(\mathrm{Res}_{\mathcal {A}_0,\ldots ,\mathcal {A}_n})\big )\).

Example 5.9

In Example 3.18, the sparse differential resultant \({\mathbf {R}}\) of \({\mathbb {P}}_0,{\mathbb {P}}_1,{\mathbb {P}}_2\) is free from the coefficients of \({\mathbb {P}}_2.\) The system can be solved as follows: \(y_1\) can be solved from \({\mathbb {P}}_0={\mathbb {P}}_1=0\) and \({\mathbb {P}}_2=u_{10}+u_{11}y'_2\) is of order one in \(y_2\) which leads to an infinite number of solutions. Thus, the system cannot have a unique solution. This shows the importance of rank essential condition.

Example 5.10

In Example 3.19, the characteristic set of \([{\mathbb {P}}_0,{\mathbb {P}}_1,{\mathbb {P}}_2]\) w.r.t. the elimination ranking \(u_{ik}\prec y_2\prec y_1\) is \({\mathbf {R}},u_{11}u_{00}y'_2-u_{01}u_{10}y_2,u_{01}y_2y_1+u_{00}\). Here \(\mathcal {A}_0,\mathcal {A}_1,\mathcal {A}_2\) are rank essential but not normal rank essential, and the system \(\{{\mathbb {P}}_0,{\mathbb {P}}_1,{\mathbb {P}}_2\}\) does not have a unique solution under the condition \({\mathbf {R}}=0\).

Definition 5.11

A differential polynomial \(f \in \mathcal {F}\{z_{0},\ldots ,z_{n}\}\) is called differentially homogenous of degree \(m\) if for a new differential indeterminate \(\lambda \), we have \(f(\lambda z_{0},\lambda z_{1}\ldots ,\lambda z_{n})=\lambda ^{m}f(z_{0},z_{1},\ldots ,z_{n}) \).

Theorem 5.12

Theorem 5.13

The sparse differential resultant is differentially homogenous in each \({\mathbf {u}}_i\) which is the coefficient vector of \({\mathbb {P}}_i\).

Proof

Suppose \({\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_i)=h_i\ge 0\). Follow notations in Sect. 3.2. By Corollary 3.12, \({\mathbf {R}}({\mathbf {u}};\zeta _0,\ldots ,\zeta _n)=0\). Differentiating this identity w.r.t. \(u_{ij}^{(k)}\,(j=1,\ldots ,l_i)\), respectively, we have

In the above equations, \(\overline{\frac{\partial {\mathbf {R}}}{\partial u_{i j}^{(k)}}}\) \((k=0,\ldots ,h_i; j=0,\ldots ,l_i)\) are obtained by replacing \(u_{i0}\) by \(\zeta _{i}\,(i=0, 1, \ldots , n)\) in each \(\frac{\partial {\mathbf {R}}}{\partial u_{i j}^{(k)}}\), respectively.

Lemma 5.14

\((\mathbb {Q}_\tau , \delta _\tau )\) defined above is a finitely generated differential extension field of \(\mathbb {Q}\langle \tilde{{\mathbf {u}}}\rangle \), which is differentially \(\mathbb {Q}\langle \tilde{{\mathbf {u}}}\rangle \)-isomorphic to a subfield of \(\mathcal {E}\).

Lemma 5.15

Let \(G\) be a differential polynomial in \(\mathbb {Q}\{\tilde{{\mathbf {u}}},u_{00}\}\). Then \(G\in \mathrm{sat}({\mathbf {R}})\) if and only if \(G\) vanishes at \(u_{00}^{(h_0)}=\gamma _{\tau }\).

Theorem 5.16

Proof

Theorem 5.17

Proof

Since \({\mathbb {P}}\) is rank essential, each \({\mathbf {u}}_i\) effectively occurs in \({\mathbf {R}}\), so each \(h_i\ge 0.\) By Theorem 3.9, \(\theta =(\eta ;\zeta _0,u_{01},\ldots ,u_{0l_0};\ldots ; \zeta _n,u_{n1},\ldots ,u_{nl_n})\) is a generic point of \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\). By Lemma 5.6, there exist \(S_j\) and \(T_j\) which are nonnegative power products of \(\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}\) such that \(S_jy_j-T_j\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\). That is, \(\eta _j=\overline{T_j}/\overline{S_j}\) for \(j=1,\ldots ,n\), where \(\overline{T_j} \) and \( \overline{S_j}\) are obtained by substituting \((u_{00},\ldots ,u_{n0})=(\zeta _0,\ldots ,\zeta _n)\) in \(T_j\) and \(S_j\), respectively. Since \({\mathbf {R}}\) is an irreducible polynomial, every \(\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}\) does not vanishes at \(u_{00}^{(h_0)}=\gamma _\tau \). Let \(\eta _{\tau j}=\frac{T_j}{S_j}\big |_{u_{00}^{(h_0)}=\gamma _\tau }\) and \(\eta _\tau =(\eta _{\tau 1},\ldots ,\eta _{\tau n})\). By (11), \(\frac{N_{0k}(\eta )}{N_{00}(\eta )}= \prod \limits _{j=1}^n\prod \limits _{k=0}^{s_0}(\eta _j^{(k)})^{(\alpha _{0k}-\alpha _{00})_{jk}} =\overline{\frac{\partial {\mathbf {R}}}{\partial u_{0k}^{(h_0)}}}\Big /\overline{\frac{\partial {\mathbf {R}}}{\partial u_{00}^{(h_0)}}}\). So \(\prod \limits _{j=1}^n\prod \limits _{k=0}^{s_0}\Big [\Big (\frac{\overline{T_j}}{\overline{S_j}}\Big )^{(k)}\Big ]^{(\alpha _{0k}-\alpha _{00})_{jk}} =\overline{\frac{\partial {\mathbf {R}}}{\partial u_{0k}^{(h_0)}}}\Big /\overline{\frac{\partial {\mathbf {R}}}{\partial u_{00}^{(h_0)}}}\). Let \(\mathcal {S}\) be the differential polynomial set consisting of \(\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}\) and \((S_j)^{m+1}\big (\frac{T_j}{S_j}\big )^{(m)}\) for all \(i=0,\ldots ,n;k=0,\ldots ,l_i;j=1,\ldots ,n\) and \(m\in \mathbb {N}\). By Corollary 3.12, there exists a finite set \(\mathcal {S}_1\) of \(\mathcal {S}\) and \(a\in \mathbb {N}\) such that \(H=\big (\prod \limits _{S\in \mathcal {S}_1}S\big )^a \Big (\prod \limits _{j=1}^n\prod \limits _{k=0}^{s_0} \big [\big (T_j/S_j\big )^{(k)}\big ]^{(\alpha _{0k}-\alpha _{00})_{jk}} -\frac{\partial {\mathbf {R}}}{\partial u_{0k}^{(h_0)}}\Big /\frac{\partial {\mathbf {R}}}{\partial u_{00}^{(h_0)}}\Big )\in \mathrm{sat}({\mathbf {R}})\). By Lemma 5.15, \(H\) vanishes at \(u_{00}^{(h_0)}=\gamma _\tau .\) And by the proof of Theorem 5.7, \(\mathcal {S}\cap \mathrm{sat}({\mathbf {R}})=\emptyset \). So \(\xi _{\tau k}=\frac{N_{0k}(\eta _\tau )}{N_{00}(\eta _\tau )}\). By Theorem 5.16, \({\mathbf {R}}=A\prod _{\tau =1}^{t_0} (u_{00}+\sum \limits _{k=1}^{l_0} u_{0 k}\xi _{\tau k})^{(h_0)}\). Thus, (20) follows.

To prove the second part of this theorem, we need first to show that \(\delta _\tau ^{k}\eta _{\tau j}\ne 0\) for each \(k\ge 0.\) Suppose the contrary, that is, there exists some \(k\) such that \(\delta _\tau ^{k}\eta _{\tau j}=0\). From \(\eta _{\tau j}=\frac{T_j}{S_j}\big |_{u_{00}^{(h_0)}=\gamma _\tau }\), \(\delta _\tau ^{k}\eta _{\tau j}=\big (\frac{T_j}{S_j}\big )^{(k)}\big |_{u_{00}^{(h_0)}=\gamma _\tau }=0\). Thus, \(S_j^{k+1}\big (\frac{T_j}{S_j}\big )^{(k)}\in \mathrm{sat}({\mathbf {R}})\). It follows that \(\eta _j^{(k)}=\big (\frac{\overline{T_j}}{\overline{S_j}}\big )^{(k)}=0\), a contradiction to the fact that \(\eta _j\) is a differential indeterminate.

Following the above procedure, we can show that \(\frac{N_{ik}(\eta _\tau )}{N_{i0}(\eta _\tau )}=\widehat{\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}}\big /\widehat{\frac{\partial {\mathbf {R}}}{\partial u_{i0}^{(h_i)}}}\) where \(\widehat{\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}}=\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}\Big |_{u_{00}^{(h_0)}=\gamma _\tau }\). Similar to (19), we have \(\sum _{k=0}^{l_i}u_{ik}\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}=b{\mathbf {R}}\) for some \(b\) in \({\mathbb Q}\). So, for each \(i\ne 0\), \(\sum _{k=0}^{l_i}u_{ik}\widehat{\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}}=0.\) It follows that for each \(i\ne 0\), \({\mathbb {P}}_i(\eta _\tau )=\sum _{k=0}^{l_i}u_{ik}N_{ik}(\eta _\tau ) =\frac{N_{i0}(\eta _\tau )}{\widehat{\frac{\partial {\mathbf {R}}}{\partial u_{i0}^{(h_i)}}}}\bigg (\sum _{k=0}^{l_i}u_{ik}\widehat{\frac{\partial {\mathbf {R}}}{\partial u_{ik}^{(h_i)}}}\bigg )=0\). So each \(\eta _\tau \) is a common non-polynomial differential zero of \({\mathbb {P}}_1,\ldots ,{\mathbb {P}}_n\). \(\square \)

Theorem 5.18

The elements \(\eta _\tau \,(\tau =1,\ldots ,t_0)\) defined in Theorem 5.17 are generic points of the prime ideal \(([{\mathbb {P}}_1^{\text {N}},\ldots ,{\mathbb {P}}_n^{\text {N}}]\hbox {{:}}\mathbb {m})_{ \mathbb {Q}\langle \hat{{\mathbf {u}}}\rangle \{{\mathbb {Y}}\}}\), where \(\hat{{\mathbf {u}}}=\cup _{i=1}^n{\mathbf {u}}_i\).

Proof

Let \({\mathcal {J}}=([{\mathbb {P}}_1^{\text {N}},\ldots ,{\mathbb {P}}_n^{\text {N}}]\hbox {{:}}\mathbb {m})_{\mathbb {Q}\langle \hat{{\mathbf {u}}}\rangle \{{\mathbb {Y}}\}}\) and \({\mathcal {J}}_0=([{\mathbb {P}}_1^{\text {N}},\ldots ,{\mathbb {P}}_n^{\text {N}}]\hbox {{:}}\mathbb {m})_{\mathbb {Q}\{{\mathbb {Y}},\hat{{\mathbf {u}}}\}}\). Similar to the proof of Theorem 3.9, it is easy to show that \({\mathcal {J}}_0\) is a prime differential ideal. Since \({\mathbb {P}}\) is rank essential, \({\mathcal {J}}_0\cap \mathbb {Q}\{\hat{{\mathbf {u}}}\}=[0]\). Thus, \({\mathcal {J}}=([{\mathcal {J}}_0])_{\mathbb {Q}\langle \hat{{\mathbf {u}}}\rangle \{{\mathbb {Y}}\}}\) is a prime differential ideal and \({\mathcal {J}}\cap \mathbb {Q}\{{\mathbb {Y}},\hat{{\mathbf {u}}}\}={\mathcal {J}}_0\). Let \(\xi =(\xi _1,\ldots ,\xi _n)\) be a generic point of \({\mathcal {J}}\). Then \((\xi ;\hat{{\mathbf {u}}})\) is a generic point of \({\mathcal {J}}_0\). Let \(\beta =-\sum _{k=1}^{l_0}u_{0k}N_{0k}(\xi )/N_{00}(\xi )\). Then \((\xi ;\beta ,u_{01},\ldots ,u_{0l_0};\hat{{\mathbf {u}}})\) is a generic point of \(\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}=([{\mathbb {P}}_0^{\text {N}},{\mathbb {P}}_1^\text {N},\ldots ,{\mathbb {P}}_n^{\text {N}}]\hbox {{:}}\mathbb {m})_{\mathbb {Q}\{{\mathbb {Y}};{\mathbf {u}}_0,\hat{{\mathbf {u}}}\}}\). Since \(\mathrm{sat}({\mathbf {R}})=\mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\cap \mathbb {Q}\{{\mathbf {u}}_0,\hat{{\mathbf {u}}}\}\), \(\gamma =(\beta ,u_{01},\ldots ,u_{0l_0};\hat{{\mathbf {u}}})\) is a generic point of \(\mathrm{sat}({\mathbf {R}})\). By Lemma 5.6, for \(j=1,\ldots ,n\), \(S_jy_j-T_j\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\). Then, \(\xi _j=\frac{T_j}{S_j}(\gamma )\).

By Theorem 5.17, \(\eta _\tau \) is a common non-polynomial solution of \({\mathbb {P}}_i^{\text {N}}=0\,(i=1,\ldots ,n)\) and thus also a differential zero of \({\mathcal {J}}\). Recall \(\eta _{\tau j}=\frac{T_j}{S_j}\big |_{u_{00}^{(h_0)}=\gamma _\tau }\). If \(f\) is any differential polynomial in \(\mathbb {Q}\langle \hat{{\mathbf {u}}}\rangle \{{\mathbb {Y}}\}\) such that \(f(\eta _\tau )=0\), then \(f(\frac{T_1}{S_1},\ldots ,\frac{T_n}{S_n})\big |_{u_{00}^{(h_0)}=\gamma _\tau }=0\). There exist \(a_j\in \mathbb {N}\) such that \(g=\prod _jS_j^{a_j}f(\frac{T_1}{S_1},\ldots ,\frac{T_n}{S_n})\in \mathbb {Q}\{{\mathbf {u}}_0,\hat{{\mathbf {u}}}\}.\) Then \(g|_{u_{00}^{(h_0)}=\gamma _\tau }=0\). By Lemma 5.15, \(g\in \mathrm{sat}({\mathbf {R}})\) while \(S_j\not \in \mathrm{sat}({\mathbf {R}})\). As a consequence, \(g(\gamma )=0\) and \(S_j(\gamma )\ne 0\). It follows that \(f(\xi )=f(\xi _1,\ldots ,\xi _n)=f(\frac{T_1}{S_1}(\gamma ),\ldots ,\frac{T_n}{S_n}(\gamma ))=0\) and hence \(f\in {\mathcal {J}}\), since \(\xi \) is a generic point of \({\mathcal {J}}\). Thus, \(\eta _\tau \) is a generic point of \({\mathcal {J}}\). \(\square \)

Definition 5.19

The Kolchin projective differential closure of the image of \(\phi _{\mathcal {A}}\) is defined to be the differential toric variety w.r.t. \(\mathcal {A}\), denoted by \(X_\mathcal {A}\). That is, \(X_\mathcal {A}=\overline{\phi _{\mathcal {A}}\big ((\mathcal {E}^\wedge )^n\big )}\).

Theorem 5.20

\(X_\mathcal {A}\) is an irreducible projective differential variety over \(\mathbb {Q}\) of dimension \(n\).

Proof

Let \({\mathcal {J}}_1={\mathcal {J}}\cap \mathbb {Q}\{z_0,z_1,\ldots ,z_l\}\). Then \({\mathcal {J}}_1\) is a prime differential ideal with a generic point \(\theta \). Denote \({\mathbf {z}}=(z_0,z_1,\ldots ,z_l)\). For any \(f\in {\mathcal {J}}_1\hbox {{:}}{\mathbf {z}}\), since \(z_0f\in {\mathcal {J}}_1\), \(z_0f\) vanishes at \(\theta \) and \(f(\theta )=0\) follows. So \(f\in {\mathcal {J}}_1\), and it follows that \({\mathcal {J}}_1\hbox {{:}}{\mathbf {z}}={\mathcal {J}}_1\). And for any \(f\in {\mathcal {J}}_1\subset {\mathcal {J}}\) and any differential indeterminate \(\lambda \) over \({\mathbb Q}\langle \eta ,v\rangle \), let \(f(\lambda {\mathbf {z}})=\sum \phi (\lambda )f_\phi ({\mathbf {z}})\) where \(\phi (\lambda )\) are distinct differential monomials in \(\lambda \) and \(f_\phi ({\mathbf {z}})\in {\mathbb Q}\{{\mathbf {z}}\}\). Then \(f(\lambda \theta )=0=\sum \phi (\lambda )f_\phi (\theta )\). So each \(f_\phi (\theta )=0\) and \(f_\phi \in {\mathcal {J}}_1\) follows. Thus, \(f(\lambda {\mathbf {z}})\in {\mathbb Q}\{\lambda \}{\mathcal {J}}_1\). By Definition 2.2, \({\mathcal {J}}_1\) is a differentially homogenous differential ideal. Then \(V={\mathbb {V}}({\mathcal {J}}_1)\) is an irreducible projective differential variety in \(\mathbf P (l)\). Since \(\theta \) is a generic point of \(V\) and \({\mathcal A}\) is differentially essential, \(\hbox {{dim}}(V)=\hbox {{d.tr.deg}}\,\mathbb {Q}\langle \frac{N_1(\eta )}{N_0(\eta )},\ldots ,\frac{N_l(\eta )}{N_0(\eta )}\rangle /\mathbb {Q}=n\). If we can show \(X_\mathcal {A}=V\), then it follows that \(X_\mathcal {A}\) is an irreducible projective differential variety of dimension \(n\).

For any point \(\xi \in (\mathcal {E}^\wedge )^n\), it is clear that \((\xi ;N_0(\xi ),N_1(\xi ),\ldots ,N_l(\xi ))\) is a differential zero of \({\mathcal {J}}\) and consequently \((N_0(\xi ),N_1(\xi ),\ldots ,N_l(\xi ))\in {\mathbb {V}}({\mathcal {J}}_1)=V\). So \(\phi _{\mathcal {A}}(\xi )=(N_0(\xi ),N_1(\xi ),\ldots ,\) \(N_l(\xi ))\in V\). Thus, \(\phi _{\mathcal {A}}\big ((\mathcal {E}^\wedge )^n\big )\subseteq V\) and \(X_{\mathcal {A}}=\overline{\phi _{\mathcal {A}}\big ((\mathcal {E}^\wedge )^n\big )} \subseteq V\) follows. Conversely, since \(\phi _{\mathcal {A}}(\eta )=(1,\frac{N_1(\eta )}{N_0(\eta )},\ldots ,\frac{N_l(\eta )}{N_0(\eta )})\in X_\mathcal {A}\) is a generic point of \(V\), \(V\subseteq X_\mathcal {A}\). Thus, \(V= X_\mathcal {A}\). \(\square \)

Theorem 5.21

[33, Theorem4.7] Let \(F({\mathbf {u}}_{0},{\mathbf {u}}_{1},\ldots ,{\mathbf {u}}_{d})\) be the differential Chow form of \(V\) with \({\mathrm{ord}}(F)=h\) and \(S_{F}=\frac{\partial F}{\partial u_{00}^{(h)}}\) . Suppose that \({\mathbf {u}}_i\) are differentially specialized over \(\mathbb {Q}\) to sets \({\mathbf {v}}_i\subset \mathcal {E}\) and \(\overline{{\mathbb {P}}}_{i}\) are obtained by substituting \({\mathbf {u}}_i\) by \({\mathbf {v}}_i\) in \({\mathbb {P}}_i\,(i=0,\ldots ,d)\). If  \(\overline{{\mathbb {P}}}_i=0\,(i=0,\ldots ,d)\) meet \(V\), then \(\mathrm{sat}(F)\) vanishes at \(({\mathbf {v}}_{0},\ldots ,{\mathbf {v}}_{d})\). Furthermore, if \(F({\mathbf {v}}_{0},\ldots ,{\mathbf {v}}_{d})=0\) and \(S_{F}({\mathbf {v}}_{0},\ldots ,{\mathbf {v}}_{d})\ne 0\), then the \(d+1\) differential hyperplanes \(\overline{{\mathbb {P}}}_{i}=0\) \(\,(i=0,\ldots ,d)\) meet \(V\).

Theorem 5.22

Let \(\mathrm{Res}_{\mathcal {A}}\) be the sparse differential resultant of \({\mathbb {P}}_0,\ldots ,{\mathbb {P}}_n\) given in (21). Then \(\mathrm{Res}_{\mathcal {A}}\) is the differential Chow form of \(X_\mathcal {A}\) with respect to the generic hyperplanes \({\mathbb {L}}_0,\ldots ,{\mathbb {L}}_n\) given in (23).

Proof

By the proof of Theorem 5.20, \(X_\mathcal {A}\) is an irreducible projective differential variety of dimension \(n\) with a generic point \((1,\frac{N_1(\eta )}{N_0(\eta )},\ldots ,\frac{N_l(\eta )}{N_0(\eta )})\). Let \(\zeta _i=-\sum _{k=1}^lu_{ik}\frac{N_k(\eta )}{N_0(\eta )}\,(i=0,\ldots ,n)\) and \(\zeta =(\zeta _0,u_{01},\ldots ,u_{0l};\ldots ;\zeta _n,u_{n1},\ldots ,u_{nl})\). Then \(\mathrm{sat}(\hbox {{Chow}}(X_\mathcal {A}))=\mathbb {I}(\zeta )\), which is the vanishing differential ideal of \(\zeta \) in \({\mathbb Q}\{{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}\). And by the definition of sparse differential resultant, \(\mathrm{sat}(\hbox {Res}_{\mathcal {A}})=\mathbb {I}(\zeta )\). By Lemma 2.3, \(\hbox {{Chow}}(X_\mathcal {A})\) and \(\hbox {Res}_{\mathcal {A}}\) can only differ at most by a nonzero element in \(\mathbb {Q}\). Thus, \(\hbox {Res}_{\mathcal {A}}\) is just the differential Chow form of \(X_\mathcal {A}\). \(\square \)

Corollary 5.23

Let \(\overline{{\mathbb {L}}}_i=v_{i0}z_0+v_{i1}z_1+\cdots +v_{il}z_l=0\,(i=0,\ldots ,n)\) be projective differential hyperplanes with \({\mathbf {v}}_i=(v_{i0},\ldots ,v_{il})\in \mathcal {E}^{l+1}\). Denote \({\mathrm{ord}}(\mathrm{Res}_{\mathcal {A}})=h\) and \(S_{\mathbf {R}}=\frac{\partial \mathrm{Res}_{\mathcal {A}}}{\partial u_{00}^{(h)}}\). If \(X_\mathcal {A}\) meets \(\overline{{\mathbb {L}}}_i=0\,(i=0,\ldots ,n)\), then \(\mathrm{Res}_{\mathcal {A}}({\mathbf {v}}_0,\ldots ,{\mathbf {v}}_n)=0\). And if \(\mathrm{Res}_{\mathcal {A}}({\mathbf {v}}_0,\ldots ,{\mathbf {v}}_n)=0\) and \(S_{\mathbf {R}}({\mathbf {v}}_0,\ldots ,{\mathbf {v}}_n)\ne 0\), then \(X_\mathcal {A}\) meets \(\overline{{\mathbb {L}}}_i=0\,(i=0,\ldots ,n)\).

Proof

It follows directly from Theorems 5.22 and 5.21. \(\square \)

Example 5.24

Let \({\mathcal A}={\mathcal A}_0\), where \({\mathcal A}_0\) is given in Example 3.21. Following the proof of Theorem 5.20, let \({\mathcal {J}}=[y_1z_1-y_1'z_0,y_1z_2-y_1^2z_0]\hbox {{:}}\mathbb {m}\). It is easy to show that \(X_\mathcal {A}\) is the general component of \(z_1z_2-(z_0z_2'-z_0'z_2)\), that is, \(X_\mathcal {A}={\mathbb {V}}(\mathrm{sat}(z_1z_2-(z_0z_2'-z_0'z_2)))\). And \(\mathrm{Res}_\mathcal {A}\) is equal to the differential Chow form of \(X_\mathcal {A}\).

Lemma 6.1

Let \({\mathbb {P}}\) be a Laurent differentially essential system and \({\mathbf {k}}=(k_0,k_1,\ldots ,k_n)\in \mathbb {Z}_{\ge 0}^{n+1}\) be a vector satisfying 26. Then \(\mathrm{ord}({\mathbf {R}},{\mathbf {u}}_i)\le k_i\) for each \(i=0,\ldots ,n\).

Proof

Denote \(\mathbb {m}^{[{\mathbf {k}}]}\) to be the set of all monomials in variables \({\mathbb {Y}}^{[\widetilde{\mathbf {k}}]}\). Suppose \({\mathcal {I}}=(\widehat{{\mathbb {P}}}^{[{\mathbf {k}}]})\hbox {{:}}\mathbb {m}^{[{\mathbf {k}}]}=\)

\(\{f\in \mathbb {Q}[{\mathbb {Y}}^{[\widetilde{\mathbf {k}}]},{\mathbf {u}}^{[{\mathbf {k}}]}]\big | \exists M\in \mathbb {m}^{[{\mathbf {k}}]}, Mf\in (\widehat{{\mathbb {P}}}^{[{\mathbf {k}}]}) \}\). Denote \(U={\mathbf {u}}^{[{\mathbf {k}}]}\backslash \cup _{i=0}^nu_{i0}^{[k_i]}\). Assume \({\mathbb {P}}_i^{\text {N}}=\sum _{k=0}^{l_i}u_{ik} N_{ik}\,(i=0,\ldots ,n)\). Let \(\zeta _{il}=-(\sum _{k=1}^{l_i}u_{ik}N_{ik}/N_{i0})^{(l)}\) for \(i=0,1,\ldots ,n;l=0,1,\ldots ,k_i\). Denote \(\bar{\zeta } =(U,\zeta _{0k_0},\ldots ,\zeta _{00},\ldots ,\zeta _{nk_n},\ldots ,\zeta _{n0})\). It is easy to show that \(({\mathbb {Y}}^{[\widetilde{\mathbf {k}}]},\bar{\zeta } )\) is a generic point of \({\mathcal {I}}\). Indeed, it is clear that each polynomial in \({\mathcal {I}}\) vanishes at \(({\mathbb {Y}}^{[\widetilde{\mathbf {k}}]},\bar{\zeta } )\). And if \(f\) is an arbitrary polynomial in \(\mathbb {Q}[{\mathbb {Y}}^{[\widetilde{\mathbf {k}}]},{\mathbf {u}}^{[{\mathbf {k}}]}]\) such that \(f({\mathbb {Y}}^{[\widetilde{\mathbf {k}}]},\bar{\zeta } )=0\), substitute \(u_{i0}^{(l)}=\big (({\mathbb {P}}_i^{\text {N}}-\sum _{k=1}^{l_i}u_{ik}N_{ik})/N_{i0}\big )^{(l)}\) into \(f\), then we have \(\prod _{i=0}^nN_{i0}^{a_i}f\,\equiv \,f_1,\mathrm{{mod}}\,(\widehat{{\mathbb {P}}}^{[{\mathbf {k}}]}),\) where \(f_1\in \mathbb {Q}[{\mathbb {Y}}^{[\widetilde{\mathbf {k}}]},U]\). Clearly, \(f_1=0\) and \(f\in {\mathcal {I}}\) follows.

Let \({\mathcal {I}}_1={\mathcal {I}}\cap \mathbb {Q}[{\mathbf {u}}^{[{\mathbf {k}}]}]\). Then \({\mathcal {I}}_1\) is a prime ideal with \(\bar{\zeta } \) as its generic point. Since \(\mathbb {Q}(\bar{\zeta } )\subset \mathbb {Q}({\mathbb {Y}}^{[\widetilde{\mathbf {k}}]},U)\), \(\mathrm{Codim}({\mathcal {I}}_1)=|U|+\sum _{i=0}^n(k_i+1)-\hbox {{tr.deg}}\,\mathbb {Q}(\bar{\zeta } )/\mathbb {Q}\ge |U|+|\widehat{{\mathbb {P}}}^{[{\mathbf {k}}]}|-\hbox {{tr.deg}}\,\mathbb {Q}({\mathbb {Y}}^{[\widetilde{\mathbf {k}}]},U)/\mathbb {Q}=|\widehat{{\mathbb {P}}}^{[{\mathbf {k}}]}|-|{\mathbb {Y}}^{[\widetilde{\mathbf {k}}]}|\ge 1\). Thus, \({\mathcal {I}}_1\ne (0)\). Suppose \(f\) is any nonzero polynomial in \({\mathcal {I}}_1\). Clearly, \({\mathrm{ord}}(f,{\mathbf {u}}_{i})\le k_i\). Since \({\mathcal {I}}_1\subset \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\cap \mathbb {Q}\{{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}=\mathrm{sat}({\mathbf {R}})\), \(f\in \mathrm{sat}({\mathbf {R}})\). Note that \({\mathbf {R}}\) is a characteristic set of \(\mathrm{sat}({\mathbf {R}})\) w.r.t. any ranking by Lemma 2.3. Thus, \({\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_{i})\le {\mathrm{ord}}(f,{\mathbf {u}}_i)\le k_i\). \(\square \)

Lemma 6.2

Let \({\mathbb {P}}\) be a Laurent differentially essential system and \(\text { J}_i\ge 0\) for each \(i=0,\ldots ,n\). Then \(\sum _{j=1}^n\mathrm{max}(e_{0j}+\text { J}_0,\cdots ,e_{nj}+\text { J}_{n})=\sum _{i=0}^n\text { J}_{i}\).

Proof

Let \(E = (e_{ij})\) be the \((n+1)\times n\) order matrix of \(\widehat{{\mathbb {P}}}\), where \(e_{ij} = {\mathrm{ord}}({\mathbb {P}}_i^{\text {N}},y_j)\). Without loss of generality, suppose \(\text { J}_{0} = e_{11} + e_{22} + \cdots + e_{nn}\).

Firstly, we will show that for each \(k\ne 1\), \(e_{11}+\text { J}_1 \ge e_{k1}+\text { J}_k\). Since \(\text { J}_k\) is the Jacobi number of \(\widehat{{\mathbb {P}}}_{\hat{k}}\) and \(k\ne 1\), \(\text { J}_k\) has a summand of the form \(e_{1p_1}\). Let \(m\) be the biggest \(s\) such that \(e_{1p_1}+e_{p_1p_2}+ \cdots + e_{p_{s-1}p_s}\) is a partial sum of successive summands in \(\text { J}_k\) and denote \(T_0=e_{1p_1}+e_{p_1p_2}+ \cdots + e_{p_{m-1}p_m}\). Suppose \(\text { J}_k = T_0 + T_1\). Since \(\text { J}_k\) is a diagonal sum, \(p_i\ne p_j\) for \(1\le i < j\). For otherwise, \(\text { J}_k\) contains \(e_{p_{i-1}p_i}\) and \(e_{p_{j-1}p_i}\) as summands \((p_0=1)\), a contradiction. Also note that \(p_i\ne 0\) for \(1\le i \le m\). Now we claim that \(p_m\) is either equal to \(1\) or equal to \(k\). Indeed, if \(p_m=1\) or \(p_m=k\), \(T_0\) cannot be any longer and these two cases may happen. But if \(p_m\ne 1\) and \(p_m\ne k\), then we can add another summand \(e_{p_{m}p_{m+1}}\) to \(T_0\), which contradicts the fact that \(T_0\) is the longest one. So \(p_m=1\) or \(k.\) Now three cases are considered.

Case 1) If \(p_1=1\), \(\text { J}_k = e_{11}+T_1\) and \(e_{k1}+\text { J}_k = e_{11} + e_{k1} + T_1\). Since \(e_{k1} + T_1\) is a diagonal sum of \(\widehat{{\mathbb {P}}}_{\hat{1}}\), \(e_{k1} + T_1 \le \text { J}_1\). Thus, \(e_{11}+\text { J}_1 \ge e_{k1}+\text { J}_k\).

Case 2) If \(p_m=1\) for \(m>1\), \(T_0 = e_{1p_1}+e_{p_1p_2}+\cdots +e_{p_{m-1}1}\). Since \(\text { J}_{0} = e_{11}+\cdots + e_{nn}\), \(T_0 \le e_{11} + e_{p_1p_1} + \cdots + e_{p_{m-1}p_{m-1}}\). For otherwise, since \(p_i\ne 0\), \(T_0+\sum _{k\in \{2,\ldots ,n\}\backslash \{p_1,\ldots ,p_{m-1}\}}e_{kk}\) is a diagonal sum of \(\widehat{{\mathbb {P}}}_{\hat{0}}\) which is greater than \(\text { J}_0\), a contradiction. Then \(e_{k1}+\text { J}_k = e_{k1}+T_0 + T_1 \le e_{k1}+e_{11} + e_{p_1p_1} + \cdots + e_{p_{m-1}p_{m-1}} + T_1\le e_{11}+\text { J}_1\), where the last inequality follows from the fact that \(e_{k1}+e_{p_1p_1} + \cdots + e_{p_{m-1}p_{m-1}} + T_1\) is a diagonal sum of \(\widehat{{\mathbb {P}}}_{\hat{1}}\).

Corollary 6.3

Let \({\mathbb {P}}\) be a Laurent differentially essential system and \(\text { J}_i\ge 0\) for each \(i=0,\ldots ,n\). Then \({\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_i)\le \text { J}_i\,(i=0,\ldots ,n)\).

Proof

It is a direct consequence of Lemma 6.1 and Lemma 6.2. \(\square \)

Lemma 6.4

[9, 32] Let \(E\) be an \(m\times n\) matrix whose entries are \(0\)’s and \(1\)’s. Let \(\mathrm{Jac}(E)=\text { J}<\min \{m,n\}\). Then \(E\) contains an \(a\times b\) zero sub-matrix with \(a+b=m+n-\text { J}\).

Lemma 6.5

Proof

Theorem 6.6

Proof

Corollary 6.7

Let \({\mathbb {P}}\) be rank essential. Then \(\text { J}_i\ge 0\) for \(i=0,\ldots ,n\) and \({\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_i) \le \text { J}_i\).

Proof

From the proof of Theorem 6.6, if \(\hbox {J}_i=-\infty \) for some \(i\), then \({\mathbb {P}}\) contains a proper differentially essential sub-system, which contradicts Theorem 4.20. Therefore, \(\text { J}_i\ge 0\) for \(i=0,\ldots ,n\). \(\square \)

Example 6.8

Theorem 6.9

Proof

Let \(\tilde{{\mathbb {P}}}_i\) be obtained from \({\mathbb {P}}_i\) by replacing \(y_j^{(k)}\) by \(y_j^{(k-\underline{o}_j)}\,\) \((j=1,\ldots ,n;k\ge \underline{o}_j)\) in \({\mathbb {P}}_i\) for \(i=0,\ldots ,n\) and denote \(\tilde{{\mathbb {P}}}=\{\tilde{{\mathbb {P}}}_0,\ldots ,\tilde{{\mathbb {P}}}_n\}\). Since

we obtain \(\hbox {{rk}}(\mathrm{{D}}_{\tilde{{\mathbb {P}}}})=\hbox {{rk}}(\mathrm{{D}}_{{\mathbb {P}}})=n.\) Thus, \({\mathcal {I}}=[\tilde{{\mathbb {P}}}]\cap \mathbb {Q}\{{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}\) is a prime differential ideal of codimension \(1\). We claim that \({\mathcal {I}}=\mathrm{sat}({\mathbf {R}})\). Suppose \({\mathbb {P}}_i=u_{i0}M_{i0}+T_i\) and \(\tilde{{\mathbb {P}}}_i=u_{i0}\tilde{M}_{i0}+\tilde{T}_i\). Let \(\zeta _i=-T_i/M_{i0}\) and \(\theta _i=-\tilde{T}_i/\tilde{M}_{i0}\). Denote \({\mathbf {u}}=\cup _{i=0}^n{\mathbf {u}}_i\backslash \{u_{i0}\}\). Then \(\zeta =({\mathbf {u}},\zeta _0,\ldots ,\zeta _n)\) is a generic point of \(\mathrm{sat}({\mathbf {R}})\) and \(\theta =({\mathbf {u}}, \theta _0,\ldots ,\theta _n)\) is a generic point of \({\mathcal {I}}\). For any differential polynomial \(G\in \mathrm{sat}({\mathbf {R}})\), \(G(\zeta )=0=(\sum _\phi \phi ({\mathbb {Y}})F_\phi ({\mathbf {u}}))/N({\mathbb {Y}})\) where \(\phi ({\mathbb {Y}})\) are distinct differential monomials in \({\mathbb {Y}}\) and so is \(N({\mathbb {Y}})\). Then \(F_\phi ({\mathbf {u}})\equiv 0\) for each \(\phi \). Thus, \(G(\theta )=(\sum _\phi \tilde{\phi }({\mathbb {Y}})F_\phi ({\mathbf {u}}))/\tilde{N}({\mathbb {Y}})=0\) and \(G\in {\mathcal {I}}\) follows. So \(\mathrm{sat}({\mathbf {R}})\subseteq {\mathcal {I}}\). Similarly, we can show that \({\mathcal {I}}\subseteq \mathrm{sat}({\mathbf {R}})\). Hence, \({\mathbf {R}}\) is the sparse differential resultant of \(\tilde{{\mathbb {P}}}\). Since \(\mathrm{Jac}(\tilde{{\mathbb {P}}}_{\hat{i}})=\mathrm{Jac}({\mathbb {P}}_{\hat{i}})-\underline{\gamma }\), by Theorem 6.6, the theorem is proved. \(\square \)

Remark 6.10

Let \({\mathbf {k}}=(e-e_0,e-e_1,\ldots ,e-e_n)\) where \(e=\sum _{i=0}^n e_i\). Clearly, \(|\widehat{{\mathbb {P}}}^{[{\mathbf {k}}]}|=ne+n+1=|{\mathbb {Y}}^{[e]}|+1 \ge |{\mathbb {Y}}^{[\widetilde{\mathbf {k}}]}|+1\). Then by Lemma 6.1, \({\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_i) \le e-e_i\le s-s_i\). Here \(s_i\) is the order of \({\mathbb {P}}_i\,(i=0,\ldots ,n)\) and \(s=\sum _{i=0}^ns_i\). If \(L_i = e-e_i-\gamma ({\mathbb {P}})\) where \(\gamma ({\mathbb {P}})=\sum _{j=1}^n (\underline{o}_j + \overline{e}_j)\) and \(\overline{e}_j = \min _i \{e_i-{\mathrm{ord}}({\mathbb {P}}^{\text {N}}_i,y_j) | {\mathrm{ord}}({\mathbb {P}}^{\text {N}}_i,y_j)\ne -\infty \}\). By [43], \((L_0,\ldots ,L_n)\) also consists of a solution to 26. Then \(\mathrm{deg}({\mathbf {R}},{\mathbf {u}}_i)\le L_i\). One can easily check that \(\bar{\text { J}}_i\le L_i \le e-e_i\) for each \(i\), and the modified Jacobi bound is better than the other two bounds as shown by the following example.

Example 6.11

Theorem 6.12

Proof

It suffices to show that \({\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_i) \le \mathrm{Jac}((E_{I})_{\hat{i}})\) for \(i=0,\ldots ,r\). Let \({\mathbb {L}}_i = u_{i0} + \sum _{j=1}^n u_{ij}y_j\) for \(i=r+1,\ldots ,n\). Since \({\mathbb {P}}_{I}\) is rank essential, there exist \(\frac{N_{ik_i}}{N_{i0}}\,( i= 1,\ldots ,r)\) such that their symbolic support matrix \(B\) is of full rank. Without loss of generality, we assume that the \(r\)th principal sub-matrix of \(B\) is of full rank. Consider a new Laurent differential polynomial system \(\widetilde{\mathbb {P}} = {\mathbb {P}}_{I} \cup \{{\mathbb {L}}_{r+1},\ldots ,{\mathbb {L}}_{n}\}\). This system is also Laurent differentially essential since the symbolic support matrix of \(\frac{N_{1k_1}}{N_{10}},\ldots , \frac{N_{rk_r}}{N_{r0}},y_{r+1},\ldots ,y_n\) is of full rank. And \({\mathbf {R}}\) is also the sparse differential resultant of \(\widetilde{\mathbb {P}}\), for \({\mathbb {P}}_I\) is the rank essential sub-system of \(\widetilde{\mathbb {P}}\). The order vector of \({\mathbb {L}}_i\) is \((0,\ldots ,0)\) for \(i=r+1,\ldots ,n\). So \(\mathrm{Jac}(\widetilde{\mathbb {P}}_{\hat{i}})=\mathrm{Jac}((E_{I})_{\hat{i}})\) for \(i=0,\ldots ,r\). By Theorem 6.6, \({\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_i) \le \mathrm{Jac}((E_{I})_{\hat{i}})\) for \(i=0,\ldots ,r\).\(\square \)

Example 6.13

Lemma 6.14

Let \(\mathcal {I}\) be a prime ideal in \(\mathcal {K}[{\mathbb {X}}]\) and \(\mathcal {I}_r=\mathcal {I}\cap \mathcal {K}[x_1,\ldots ,x_r]\) for any \(1\le r\le n\). Then \(\mathrm{deg}(\mathcal {I}_r)\le \mathrm{deg}(\mathcal {I})\).

Lemma 6.15

[21, Theorem1] Let \(V_1,\ldots ,V_r\,(r\ge 2)\) be a finite number of constructible sets in \(\overline{\mathcal {K}}^n\). Then \(\mathrm{deg}(V_1\cap \cdots \cap V_r)\le \prod _{i=1}^r\mathrm{deg}(V_i).\)

Theorem 6.16

Proof

2) In \({\mathbf {R}}\), let \(u_{i0}\,(i=0,\ldots ,n)\) be replaced, respectively, by \(\big ({\mathbb {P}}_{i}^{\text {N}}-\sum _{k=1}^{l_i}u_{ik}N_{ik}\big )/N_{i0}\) and let \({\mathbf {R}}\) be expanded as polynomials in \({\mathbb {P}}_{i}^{\text {N}}\) and their derivatives with coefficients in \(\mathbb {Q}\{{\mathbb {Y}}^{\pm };{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}\). We obtain \({\mathbf {R}}=\sum \limits _{M}q_M\cdot M({\mathbf {u}};u_{00},\ldots ,u_{n0})=\sum \limits _{M}q_M\cdot M\big ({\mathbf {u}};\frac{ {\mathbb {P}}_{0}^{\text {N}}-\sum _{k=1}^{l_0}u_{0k}N_{0k}}{N_{00}},\ldots ,\frac{{\mathbb {P}}_{n}^{\text {N}}-\sum _{k=1}^{l_n}u_{nk}N_{nk}}{N_{n0}}\big ) =\big (\sum \limits _{i=0}^n\sum \limits _{k=0}^{h_i}G_{ik}({\mathbb {P}}_{i}^{\text {N}})^{(k)}+T\big )\big /\prod \limits _{i=0}^nN_{i0}^{a_i}\), where \(q_M\in {\mathbb Q}\), \(a_i\in \mathbb {N}\), \(G_{ik}\in {\mathbb Q}[{\mathbf {u}}_0^{[h_0]},\ldots ,{\mathbf {u}}_n^{[h_n]},{\mathbb {Y}}^{[\mathbf t ]}]\) and \(T\in {\mathbb Q}\{{\mathbf {u}},{\mathbb {Y}}\}\) is free from \(u_{i0}\). So \(\prod \limits _{i=0}^nN_{i0}^{a_i}{\mathbf {R}}=\sum \limits _{i=0}^n\sum \limits _{k=0}^{h_i}G_{ik}({\mathbb {P}}_{i}^{\text {N}})^{(k)}+T\), and \(T\in \mathcal {I}_{{\mathbb {Y}},{\mathbf {u}}}\cap {\mathbb Q}\{{\mathbf {u}},{\mathbb {Y}}\}=\{0\}\) follows. Thus, \(T=0\) and we obtain a representation for \({\mathbf {R}}\) of the form (27).

Example 6.17

In Example 3.19, \(\text { J}_0=2\), \(\text { J}_1=\text { J}_2=1\) and \(m_0=m_1=m_2=2\). The expression of \({\mathbf {R}}\) shows that \(h_0={\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_0)=1< \text { J}_0\), \(h_i={\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_i)=0<\text { J}_i\,(i=1,2)\) and \(\mathrm{deg}({\mathbf {R}})=5<<3^4=\prod _{i=0}^2(m_i+1)^{h_i+1}.\)

Theorem 6.18

Proof

The algorithm finds a \(P\in {\mathbb Q}\{{\mathbf {u}}_0,\ldots ,{\mathbf {u}}_n\}\) satisfying (27), which has the smallest order and the smallest degree among those with the same order. Existence for such a differential polynomial is guaranteed by Theorem 6.16. Such a \(P\) is in \(\mathrm{sat}({\mathbf {R}})\) by (10). Since each differential polynomial in \(\mathrm{sat}({\mathbf {R}})\) not equal to \({\mathbf {R}}\) either has greater order than \({\mathbf {R}}\) or has the same order but greater degree than \({\mathbf {R}}\), \(P\) must be \({\mathbf {R}}\) (up to a factor in \(\mathbb {Q}\)).

We will estimate the complexity of the algorithm below. Denote \(D\) to be the degree bound of \({\mathbf {R}}.\) By Theorem 6.16, \(D\le (m+1)^{\sum _{i=0}^n(\text { J}_i+1)}=(m+1)^{\text { J}+n+1}\), where \(\text { J}=\sum _{i=0}^n\text { J}_i\). In each loop of Step 3, the complexity of the algorithm is clearly dominated by Step 3.1.2, where we need to solve a system of linear equations \(\mathcal {P}=0\) over \({\mathbb Q}\) in \({\mathbf {c}}_{0}\) and \({\mathbf {c}}_{ij}\). It is easy to show that \(|{\mathbf {c}}_{0}|={d+L_1-1\atopwithdelims ()L_1-1}\) and \(|{\mathbf {c}}_{ij}|={d_1-m_i-1+L_1+L_2\atopwithdelims ()L_1+L_2}\), where \(L_1=|U|=\sum \limits _{i=0}^n (h_i+1)(l_i+1)\), \(L_2=|{\mathbb {Y}}^{[\mathbf t ]}|=\sum \limits _{j=1}^n(\mathrm{max}_{i}\{h_i+e_{ij}\}+1)\) and \(d_1=[m+1+\sum _{i=0}^n(h_i+1)m_{i0}]d\). Then \(\mathcal {P}=0\) is a linear equation system with \(W_1={d+L_1-1\atopwithdelims ()L_1-1}+\sum _{i=0}^n(h_i+1){d_1-m_i-1+L_1+L_2\atopwithdelims ()L_1+L_2}\) variables and \(W_2={d_1+L_1+L_2\atopwithdelims ()L_1+L_2}\) equations. To solve it, we need at most \((\mathrm{max}\{W_1, W_2\})^{\omega }\) arithmetic operations over \({\mathbb Q}\), where \(\omega \) is the matrix multiplication exponent and the currently best known \(\omega \) is 2.376.

The iteration in Step 3.1.2 may go through \(1\) to \(\prod _{i=0}^n(m_i+1)^{h_i+1}\le (m+1)^{\sum _{i=0}^n(\text { J}_i+1)}\), and the iteration in Step 3.1 at most will repeat \(\prod _{i=0}^n(\text { J}_i+1)\) times. And by Theorem 6.16, Step 3 may loop from \(o=0\) to \(\sum _{i=0}^n\text { J}_i\). In the whole algorithm, \(L_1\le \sum \limits _{i=0}^n (\text { J}_i+1)(l_i+1)\le l\text { J}+l\), \(L_2=|{\mathbb {Y}}^{[\mathbf t ]}|\le \sum \limits _{j=1}^n(\mathrm{max}_{i}\{\text { J}_i+e_{ij}\}+1)=\text { J}+n\) by Lemma 6.2, and \(d_1\le [m+1+\sum _{i=0}^n(\text { J}_i+1)m_{i0}]D=\big (m(\text { J}+n+2)+1\big )D\). Thus, \(W_1\le {D+l\text { J}+l-1\atopwithdelims ()l\text { J}+l-1}+\sum _{i=0}^n(\text { J}_i+1){ (m(\text { J}+n+2)+1)D-m_i-1+l\text { J}+l+\text { J}+n \atopwithdelims ()l\text { J}+l+\text { J}+n}\) and \(\mathrm{max}\{W_1,W_2\}\le (\text { J}+n+2){[m(\text { J}+n+2)+1]D+l\text { J}+l+\text { J}+n \atopwithdelims ()l\text { J}+l+\text { J}+n}\).

Since \(l\ge 2(n+1)\), the complexity bound is \(O\big ({\big [(m(\text { J}+n+2)+1)D\big ]^{O(l\text { J}+l)}}/{n^{n}}\big )\). Our complexity assumes an \(O(1)\)-complexity cost for all field operations over \(\mathbb {Q}\). Thus, the complexity follows. Now 1) is proved. To prove 2), we just need to replace \(D\) by the degree bound for \({\mathbf {R}}\) in Theorem 6.16 in the complexity bound in 1). \(\square \)

Example 6.19

Let \(n=1\), \({\mathbb {P}}_0=u_{00}+u_{01}y' \), and \( {\mathbb {P}}_1=u_{10}+u_{11}y'\). We use this simple example to illustrate Algorithm SDResultant. Here, \(m_{i0}=0, m=1\), \(J_0=J_1=1\). In step 3.1, \(o=0\) and \((h_0,h_1)=(0,0)\). So \(U=\{u_{00},u_{01},u_{10},u_{11}\}\) and \({\mathbb {Y}}^{[\mathbf t ]}=\{y,y'\}\). We first execute steps 3.1.2.1 to 3.1.2.7 for \(d=1\). Set \({\mathbf {R}}_0=c_{01}u_{00}+c_{02}u_{01}+c_{03}u_{10}+c_{04}u_{11}\) and \({\mathbf {c}}_0=(c_{01},c_{02},c_{03},c_{04})\). Set \(G_{i0}=c_{i01}\) and \({\mathbf {c}}_{i0}=(c_{i01})\) for \(i=0, 1\). In step 3.1.2.5, since \({\mathbf {R}}_0-G_{00}{\mathbb {P}}_0-G_{10}{\mathbb {P}}_1=(c_{01}-c_{001})u_{00} +c_{02}u_{01}+(c_{03}-c_{101})u_{10}+c_{04}u_{11}-c_{001}u_{01}y' -c_{101}u_{11}y'\), \(\mathcal {P}=0\) consists of equations \(\{c_{01}-c_{001}=0, c_{02}=0, c_{03}-c_{101}=0, c_{04}=0, c_{001}=0, c_{101}=0\}\). \(\mathcal {P}=0\) has a unique solution \({\mathbf {c}}=(0,0,0,0)\) and \(c_{i01}=0\). Then \({\mathbf {R}}=0\).

Next, we execute steps 3.1.2.1 to 3.1.2.7 for \(d=2\). Set \({\mathbf {R}}_0=c_{01}u_{00}u_{10}+c_{02}u_{00}u_{11} +c_{03}u_{01}u_{10}+c_{04}u_{01}u_{11}+ c_{05}u_{00}^2+c_{06}u_{00}u_{01}+c_{07}u_{01}^2 +c_{08}u_{10}^2+c_{09}u_{10}u_{11}+ c_{0,10}u_{11}^2\), and \({\mathbf {c}}_0=(c_{01},\ldots ,c_{0,10})\). Set \(\{M_1,\ldots ,M_{28}\} \) to be the set of monomials in \(U\) and \({\mathbb {Y}}^{[\mathbf t ]}\) of degree not greater than 2. Let \(G_{i0}=\sum _{j=1}^{28}c_{i0j}M_{j}\) and \({\mathbf {c}}_{i0}=(c_{i01},\ldots ,c_{i0,28})\) for \(i=0, 1\). Regarding \(T={\mathbf {R}}_0-G_{00}{\mathbb {P}}_0-G_{10}{\mathbb {P}}_1\) as polynomials in \(U\) and \({\mathbb {Y}}^{[\mathbf t ]}\), let \(\mathcal {P}\) be the set of coefficients of \(T\), which are linear polynomials in \({\mathbb Q}[{\mathbf {c}}_{0},{\mathbf {c}}_{00},{\mathbf {c}}_{10}]\). Then \(\mathcal {P}=0\) consists of at most \({10 \atopwithdelims ()4}\) linear equations in \(66\) variables \({\mathbf {c}}_{0},{\mathbf {c}}_{00}\), and \({\mathbf {c}}_{10}\) with integral coefficients. Solving \(\mathcal {P}=0\), we obtain \({\mathbf {c}}_0=(0,q,-q,0,0,0,0,0,0,0)\), where \(q\in {\mathbb Q}\). Thus, the algorithm returns \({\mathbf {R}}=u_{00}u_{11}-u_{01}u_{10}\).

Remark 6.20

By Remark 4.21, we can compute a rank essential set \(I\) and the algorithm can be improved by only considering the Laurent differential polynomials \({\mathbb {P}}_i\,(i\in I)\) in the linear combination of the sparse differential resultant.

Remark 6.21

If the given system is algebraic, then the complexity bound given in 1) of Theorem 6.18 is essentially the same as that given in [45, p. 288] since \(D\gg m\) and \(D\gg n\).

Definition 6.22

Theorem 6.23

Lemma 6.24

Let \(f_1,\ldots ,f_n\) be polynomials in \(\mathbb {C}[x_1,\ldots ,x_n]\) and \(\mathcal {Q}_i\) be the Newton polytope of \(f_i\) in \(\mathbb {R}^n\). Then \(\mathcal {M}(\mathcal {Q}_1,\ldots ,\mathcal {Q}_n)\le \prod _{i=1}^n\mathrm{deg}(f_i)\).

Proof

Let \(\Delta \) be the standard unitary simplex of \(\mathbb {R}^n\). Then for each \(i\), \(\mathcal {Q}_i\subset d_i\Delta \), where \(d_i=\mathrm{deg}(f_i)\). By the monotonicity of the mixed volume, \(\mathcal {M}(\mathcal {Q}_1,\ldots ,\mathcal {Q}_n)\le \mathcal {M}(d_1\Delta ,\ldots ,d_n\Delta )=\prod _{i=1}^nd_i\cdot \mathcal {M}(\Delta ,\ldots ,\Delta )=\prod _{i=1}^nd_i\). \(\square \)

Theorem 6.25

Proof

By [17, Theorem6.8], \({\mathrm{ord}}({\mathbf {R}},{\mathbf {u}}_i)=s-s_i\,(i=0,\ldots ,n)\) and \(({\mathbf {R}})=({\mathbb {P}}_0^{[s-s_0]},\) \(\ldots ,\) \({\mathbb {P}}_n^{[s-s_n]})\cap {\mathbb Q}[{\mathbf {u}}_0^{[s-s_0]},\) \(\ldots ,{\mathbf {u}}_n^{[s-s_n]}]\). Regard each \({\mathbb {P}}_i^{(k)}~(i=0,\ldots ,n,k=0,\ldots ,s-s_i)\) as a polynomial in the \(n(s+1)\) variables \({\mathbb {Y}}^{[s]}=\{y_1,\ldots ,y_n,y'_1,\ldots ,y'_n,\ldots ,y^{(s)}_1,\) \(\ldots ,y^{(s)}_n\}\), and we denote its support by \(\mathcal {B}_{ik}\). Let \({\mathbb {F}}_{ik}\) be a generic sparse polynomial with support \(\mathcal {B}_{ik}\). Denote \({\mathbf {c}}_{ik}\) to be the set of coefficients of \({\mathbb {F}}_{ik}\), and in particular, suppose \(c_{ik0}\) is the coefficient of the monomial \(1\) in \({\mathbb {F}}_{ik}\). Now we claim that

C1) \(\overline{\mathcal {B}}=\{\mathcal {B}_{ik}\,|\,0\le i\le n; 0\le k \le s-s_i\}\) is an essential set.

C2) \(\overline{\mathcal {B}}=\{\mathcal {B}_{ik}\,|\,0\le i\le n; 0\le k \le s-s_i\}\) jointly span the affine lattice \(\mathbb {Z}^{n(s+1)}\).

Claim C2) follows from the fact that \(\{1,y_j^{[s]}\,|\,1\le j\le n\}\) is contained in the support of \({\mathbb {F}}_{0,s-s_0}\). From C1) and C2), the sparse resultant of \(({\mathbb {F}}_{ik})_{0\le i\le n; 0\le k\le s-s_i}\) exists and we denote it by \(G\). Then \((G)=\big (({\mathbb {F}}_{ik})_{0\le i\le n; 0\le k\le s-s_i}\big )\bigcap \) \(\mathbb {Q}[({\mathbf {c}}_{ik})_{0\le i\le n; 0\le k\le s-s_i}]\), and by Theorem 6.23, \(\mathrm{deg}(G,{\mathbf {c}}_{ik})=\mathcal {M}\big ((\mathcal {Q}_{jl})_{j\ne i,0\le l\le s-s_j},\mathcal {Q}_{i0},\ldots ,\) \(\mathcal {Q}_{i,k-1},\mathcal {Q}_{i,k+1},\ldots ,\mathcal {Q}_{i,s-s_i}\big )\).

Now suppose \(\xi \) is a generic point of the zero ideal \((0)_{\mathbb {Q}({\mathbf {c}})[{\mathbb {Y}}^{[s]}]}\). Let \(\zeta _{ik}=-{\mathbb {F}}_{ik}(\xi )+c_{ik0}\) and \(\overline{\zeta }_{ik}=-{\mathbb {P}}_i^{(k)}(\xi )+u_{i0}^{(k)}\) (\(i=0,\ldots ,n;k=0,\ldots ,s-s_i\)). Clearly, \(\zeta _{ik}\) and \(\overline{\zeta }_{ik}\) are free of \(c_{ik0}\) and \(u_{i0}^{(k)}\), respectively. It is easy to see that \((\xi ;{\mathbf {c}},\zeta _{00},\ldots ,\zeta _{0,s-s_0},\ldots ,\zeta _{n0},\ldots ,\zeta _{n,s-s_n})\) is a generic point of the algebraic prime ideal \(\big (({\mathbb {F}}_{ik})_{0\le i\le n; 0\le k\le s-s_i}\big )_{{\mathbb Q}[{\mathbb {Y}}^{[s]},({\mathbf {c}}_{ik})_{0\le i\le n; 0\le k\le s-s_i}]}\), while \((\xi ;\cup _{i=0}^n({\mathbf {u}}_i\backslash \{u_{i0}\})^{[s-s_i]},\) \(\overline{\zeta }_{00},\ldots ,\overline{\zeta }_{0,s-s_0},\) \(\ldots , \overline{\zeta }_{n0},\) \(\ldots ,\overline{\zeta }_{n,s-s_n})\) is a generic point of the algebraic prime ideal \(\big (({\mathbb {P}}_{i}^{(k)})_{0\le i\le n; 0\le k\le s-s_i}\big )_{{\mathbb Q}[{\mathbb {Y}}^{[s]},{\mathbf {u}}_0^{[s-s_0]},\ldots ,{\mathbf {u}}_n^{[s-s_n]}]}\). If we regard \(G\) as a polynomial in \(c_{ik0}\) over \(\mathbb {Q}({\mathbf {c}})\), then \(G\) is the vanishing polynomial of \((\zeta _{00},\ldots ,\zeta _{0,s-s_0},\ldots ,\zeta _{n0},\) \(\ldots ,\zeta _{n,s-s_n})\) over \({\mathbb Q}({\mathbf {c}})\). Now specialize the coefficients \({\mathbf {c}}_{ik}\) of \({\mathbb {F}}_{ik}\) to the corresponding coefficients of \({\mathbb {P}}_i^{(k)}\). Then each \(\zeta _{ik}\) is specialized to \(\overline{\zeta }_{ik}\). In particular, \(c_{ik0}\) are specialized to \(u_{i0}^{(k)}\) which are algebraically independent over the field \({\mathbb Q}(\xi ,\cup _{i=0}^n{\mathbf {u}}_i^{[s-s_i]}\backslash u_{i0}^{[s-s_i]})\). We claim that there exists a nonzero polynomial \(H(\cup _{i=0}^n{\mathbf {u}}_i^{[s-s_i]}\backslash u_{i0}^{[s-s_i]};\) \(u_{00},\ldots ,u_{00}^{(s-s_0)},\ldots ,\) \(u_{n0},\ldots ,u_{n0}^{(s-s_n)})\in {\mathbb Q}[{\mathbf {u}}_0^{[s-s_0]},\ldots ,\) \({\mathbf {u}}_n^{[s-s_n]}]\) such that

C3) \(H(\cup _{i=0}^n{\mathbf {u}}_i^{[s-s_i]}\backslash u_{i0}^{[s-s_i]};\overline{\zeta }_{00},\ldots ,\overline{\zeta }_{0,s-s_0},\ldots ,\overline{\zeta }_{n0},\ldots ,\overline{\zeta }_{n,s-s_n})=0\) and

C4) For each \(i\), \(\mathrm{deg}(H,{\mathbf {u}}_i^{[s-s_i]})\le \mathrm{deg}(G,\cup _{k=0}^{s-s_i}{\mathbf {c}}_{ik})\).

In the following, we construct \(H\) by specializing elements of \({\mathbf {c}}\) one by one in \(G\). For each \(v\in {\mathbf {c}}\), denote \(u\) to be its corresponding coefficient in \({\mathbb {P}}_i^{(k)}\). First specialize \(v \) to \(u \) and suppose \(\zeta _{ik}\) is specialized to \(\tilde{\zeta }_{ik}\) correspondingly. Clearly, \(G({\mathbf {c}}\backslash \{v\},u;\tilde{\zeta }_{00},\ldots ,\tilde{\zeta }_{0,s-s_0},\tilde{\zeta }_{n0},\ldots ,\tilde{\zeta }_{n,s-s_n})=0.\) If \(\bar{G}=G({\mathbf {c}}\backslash \{v\},u;c_{000},c_{010},\ldots ,c_{0,s-s_0,0},\ldots ,c_{n00},c_{n10}\ldots ,c_{n,s-s_n,0})\ne 0\), denote \(\bar{G}\) by \(H_1\). Otherwise, there exists some \(a\in \mathbb {N}\) such that \(G=(v-u)^aG_1\) with \(G_1|_{v=u}\ne 0\). But \(G({\mathbf {c}}\backslash \{v\},u;\tilde{\zeta }_{00},\) \(\ldots ,\tilde{\zeta }_{0,s-s_0},\tilde{\zeta }_{n0},\) \(\ldots ,\tilde{\zeta }_{n,s-s_n})=0= (v-u)^aG_1({\mathbf {c}}\backslash \{v\},u;\tilde{\zeta }_{00},\ldots ,\tilde{\zeta }_{0,s-s_0},\tilde{\zeta }_{n0},\) \(\ldots ,\tilde{\zeta }_{n,s-s_n})\), so \(G_1({\mathbf {c}}\backslash \{v\},\) \(u;\tilde{\zeta }_{00},\ldots ,\tilde{\zeta }_{0,s-s_0},\tilde{\zeta }_{n0},\ldots ,\tilde{\zeta }_{n,s-s_n})=0\). Denote \(G_1|_{v=u}\) by \(H_1\). Clearly, \(\mathrm{deg}(H_1,{\mathbf {u}}_i^{[s-s_i]}\bigcup \cup _{k}{\mathbf {c}}_{ik})\le \mathrm{deg}(G,\cup _{k}{\mathbf {c}}_{ik})\) for each \(i\). Continuing this process for \(|{\mathbf {c}}|\) times until each \(v\in {\mathbf {c}}\) is specialized to its corresponding element \(u\), we will obtain a nonzero polynomial \(H_{|{\mathbf {c}}|}(\cup _{i=0}^n({\mathbf {u}}_i\backslash \{u_{i0}\})^{[s-s_i]};c_{000},c_{010},\) \(\ldots ,c_{0,s-s_0,0},\ldots ,c_{n00},c_{n10},\) \(\ldots ,c_{n,s-s_n,0})\) satisfying \(H_{|{\mathbf {c}}|}(\cup _{i=0}^n({\mathbf {u}}_i\backslash \{u_{i0}\})^{[s-s_i]};\overline{\zeta }_{00},\ldots ,\overline{\zeta }_{0,s-s_0},\) \(\overline{\zeta }_{n0},\) \(\ldots ,\overline{\zeta }_{n,s-s_n})=0\) and moreover, for each \(i\), \(\mathrm{deg}(H_{|{\mathbf {c}}|},\) \({\mathbf {u}}_i^{[s-s_i]}\bigcup \cup _{k}\{c_{ik0}\})\le \mathrm{deg}(G,\cup _{k}{\mathbf {c}}_{ik})\). Since the \(u_{i0}^{(k)}\) are algebraically independent over the field \({\mathbb Q}(\xi ,\cup _{i=0}^n({\mathbf {u}}_i\backslash \{u_{i0}\})^{[s-s_i]})\), \(H=H_{|{\mathbf {c}}|}\big |_{c_{ik0}=u_{i0}^{(k)}}\in {\mathbb Q}[{\mathbf {u}}_0^{[s-s_0]},\ldots ,\) \({\mathbf {u}}_n^{[s-s_n]}]\) is a polynomial satisfying C3) and C4).

From C3), \(H\in ({\mathbb {P}}_0^{[s-s_0]},\ldots ,\) \({\mathbb {P}}_n^{[s-s_n]}).\) Since \(({\mathbb {P}}_0^{[s-s_0]},\ldots ,{\mathbb {P}}_n^{[s-s_n]})\cap {\mathbb Q}[{\mathbf {u}}_0^{[s-s_0]},\ldots ,{\mathbf {u}}_n^{[s-s_n]}]=({\mathbf {R}})\) and \({\mathbf {R}}\) is irreducible, \({\mathbf {R}}\) divides \(H\). Then \(\mathrm{deg}({\mathbf {R}},{\mathbf {u}}_i^{[s-s_i]})\le \mathrm{deg}(H,{\mathbf {u}}_i^{[s-s_i]})\) \(\le \mathrm{deg}(G,\cup _{k}{\mathbf {c}}_{ik})=\sum \limits _{k=0}^{s-s_i}\mathrm{deg}(G,{\mathbf {c}}_{ik})=\sum \limits _{k=0}^{s-s_i}\mathcal {M}\big ((\mathcal {Q}_{jl})_{j\ne i,0\le l\le s-s_j},\mathcal {Q}_{i0},\ldots ,\mathcal {Q}_{i,k-1},\mathcal {Q}_{i,k+1},\ldots ,\) \(\mathcal {Q}_{i,s-s_i}\big )\). \(\square \)

Corollary 6.26

Let \({\mathbb {P}}_i\,(i=0,\ldots ,n)\) be defined in (29) and \(s=\sum _{i=0}^n s_i\). Then for each \(i\in \{0,1,\ldots ,n\}\), \(\mathrm{deg}({\mathbf {R}},{\mathbf {u}}_i)\le \frac{s-s_i+1}{m_i}\prod _{j=0}^n m_j^{s-s_j+1}\).

Proof

Example 6.27