1 Introduction
Let D be a ring and \(A= D[x; \sigma , \delta ]\) be a skew polynomial ring where \(\sigma \) is an automorphism of D and \(\delta \) is a \(\sigma \)-derivation of D (for all \(a, b\in D\), \(\delta (ab) = \delta (a)b+\sigma (a) \delta (b)\)). The ring A is generated by D and x subject to the defining relations \(xa=\sigma (a) x+ \delta (a)\) for all elements \(a\in D\). When D is a Dedekind domain, a classification of simple A-modules is given in [4]. This is a large class of rings. A machinery is developed in [4] to cover all possible situations (non-commutative valuations, etc).
The algebrais a particular example of the ring A where \(\sigma =\mathrm{id}\) is the identity automorphism of the polynomial ring K[X], \(f\in K[X]\) and \(\delta = f\frac{d}{dX}\) is a K-derivation of K[X] (\(\delta (X)=f\)). If \(f=1\) (or, more generally, \(f\in K^\times \backslash \{ 0\}\)) then the algebra \(\Lambda (1)\) is the Weyl algebraIn 1981, a classification of simple \(A_1\)-modules was obtained by Block (over the field of complex numbers) in [9] (see also [2, 3] for an alternative approach via generalized Weyl algebras in a more general situation).
$$\begin{aligned} \Lambda =K[X]\left[ Y; \delta := f\frac{d}{dX} \right] =\bigoplus _{i\ge 0} K[X]Y^i \end{aligned}$$
$$\begin{aligned} A_1=K\langle X, \partial \, | \, \partial X-X\partial =1\rangle \simeq K[X]\left[ Y; \frac{d}{dX} \right] . \end{aligned}$$
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Recently, classifications of simple weight modules are obtained for some classical algebras (the Euclidean algebra, the Schrödinger algebra, the universal enveloping algebra \(U(\mathrm{sl}_2\ltimes V_2)\)), see [5‐7]. In these classifications, classifications of all simple modules over certain subalgebras of the Weyl algebra \(A_1\) that contain the polynomial algebra K[X] (the, so-called, polyonic algebras) play a crucial role. The polyonic algebras are investigated in [8]. Each polyonic algebra contains the algebra \(\Lambda =\Lambda (f)\) for some non-scalar polynomial \(f\in K[X]\) which play an important role in studying of its properties. This is the main reason why we decided to collect main properties of the algebras \(\Lambda \) in this paper. In particular, a classification of simple \(\Lambda \)-modules is given in Sect. 2 (Lemma 2.1 and Theorem 2.10). This classification can be derived from [4] but we give different and simpler proofs which are based on generalized Weyl algebras rather than skew polynomial rings.
An ideal \(\mathfrak {p}\) of a ring R is called a completely prime ideal if the factor ring \(R/\mathfrak {p}\) is a domain. A completely prime ideal is a prime ideal. The sets of prime and completely prime ideals of the ring R are denoted by \(\mathrm{Spec}(R)\) and \(\mathrm{Spec}_c(R)\), respectively.
In Theorem 1.1, a classification of prime and completely prime ideals of the algebra \(\Lambda \) is given, the Krull and global dimensions of the algebra \(\Lambda \) are found. The algebra \(\Lambda \) is a Noetherian domain of Gelfand-Kirillov dimension 2.
Theorem 1.1
Let K be a field of characteristic zero, \(\Lambda =K[X][Y; \delta := f\frac{d}{dX} ]\) where \(f\in K[X]\backslash K\). Let \(f=p_1^{n_1}\cdots p_s^{n_s}\) be a unique (up to permutation) product of irreducible polynomials of K[X]. Then
1.
The Krull dimension of \(\Lambda \) is \(\mathrm{Kdim } (\Lambda )=2\).
2.
The global dimension of \(\Lambda \) is \(\mathrm{gldim}(\Lambda )=2\).
3.
The elements \(p_1, \ldots , p_s\) are regular normal elements of the algebra \(\Lambda \) (i.e. \(p_i\) is a non-zero-divisor of \(\Lambda \) and \(p_i\Lambda = \Lambda p_i\)).
4.
\(\mathrm{Spec}(\Lambda )=\mathrm{Spec}_c (\Lambda )=\{ 0 , \Lambda p_i, (p_i, q_i)\, | \, i=1, \ldots , s;\; q_i\in \mathrm{Irr}_m(F_i[Y])\}\) where \(F_i:=K[X]/(p_i)\) is a field and \(\mathrm{Irr}_m(F_i[Y])\) is the set of monic irreducible polynomials of the polynomial algebra \(F_i[Y]\) over the field \(F_i\) in the variable Y. If, in addition, the field K is an algebraically closed and \(\lambda _1, \ldots , \lambda _s\) are the roots of the polynomial f then \(\mathrm{Spec}(\Lambda )=\{ 0 , \Lambda (X-\lambda _i), (X-\lambda _i, Y-\mu )\, | \, i=1, \ldots , s;\; \mu \in K\}\).
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2 Classification of Simple \(\Lambda \)-Modules
In this section, ‘module’ means left module, K is an algebraically closed field of characteristic zero, \(\Lambda =K[X][Y, \delta =f\frac{d}{dX}]\) where \(f\in K[X]\backslash K\). The algebra \(\Lambda \) is a Noetherian domain. The aim of the section is to give a classification of simple \(\Lambda \)-modules (Lemma 2.1 and Theorem 2.10).
The elementfis a regular normal element of\(\Lambda \). It follows fromthat the element f is a normal element of \(\Lambda \) (i.e. \(\Lambda f=f\Lambda )\). It determines a K-automorphism \(\omega _f\) of the algebra \(\Lambda \):The algebra \(\Lambda \) can be identified with a subalgebra of the first Weyl algebra \(A_1\) by the mapThe Weyl algebra\(A_1\)is a generalized Weyl algebra. The Weyl algebra \(A_1\) is a simple Noetherian domain with restricted minimum condition, i.e. any proper left or right factor module of \(A_1\) has finite length, [10].
$$\begin{aligned} fY=Yf-f'f=(Y-f')f,\;\;\mathrm{where}\;\;f'=\frac{df}{dX}, \end{aligned}$$
$$\begin{aligned} fu=\omega _f(u)f,\;\; u\in \Lambda ,\\ \omega _f:X\mapsto X,\;\;Y\mapsto Y-f'. \end{aligned}$$
$$\begin{aligned} \Lambda \rightarrow A_1,\;\;X\mapsto X,\;\; Y\mapsto f\partial . \end{aligned}$$
(1)
Definition, [1, 2]. Let D be a ring, \(\sigma \) be an automorphisms of D and a be a central element of D. A generalized Weyl algebra (GWA) \(A=D(\sigma , a)\) of degree 1, is the ring generated by D and by two indeterminates X and Y subject to the relations [1, 2]: For all \(\alpha \in D\),The algebrais a \(\mathbb {Z}\)-graded algebra, where \(A_n=Dv_n,\)\(v_n=X^n\,\, (n>0), \,\,v_n=Y^{-n}\,\, (n<0), \,\,v_0=1.\)
$$\begin{aligned} X\alpha =\sigma (\alpha )X \;\; \mathrm{and}\;\; Y\alpha =\sigma ^{-1}(\alpha )Y,\;\; \ YX=a \;\; \mathrm{and}\;\; XY=\sigma (a). \end{aligned}$$
$$\begin{aligned} A={\bigoplus }_{n\in \mathbb {Z}}\, A_n \end{aligned}$$
The Weyl algebra \(A_1\) is a GWA,with coefficients from a polynomial ring K[H] where \(\sigma \in \mathrm{Aut}_K\,K[H]\) and \(\sigma :H\rightarrow H-1\).
$$\begin{aligned} A_1= D(\sigma ,a=H),\; X\leftrightarrow X,\; \partial \leftrightarrow Y,\; \partial X\leftrightarrow H,\;\;D=K[H], \end{aligned}$$
We denote by \(\Lambda _f\) (resp., \(A_{1,f}\)) the localization of the ring \(\Lambda \) (resp., \(A_1\)) at the powers of the element f, i.e.By (1), \(\Lambda _f\) is a subalgebra of \(A_{1,f}\) such thatThe algebras \(\Lambda \) and \(A_1\) can be considered as subalgebras of \(A_{1,f}\),The algebra \(A_{1,f}\) is a simple Noetherian domain with restricted minimum condition.
$$\begin{aligned} \Lambda _f=S^{-1}_f\Lambda ,\;\;(\mathrm{resp., }\;\;A_{1,f}=S^{-1}_fA_1)\;\;\mathrm{where}\;\; S_f=\{ f^i, i\ge 0\}. \end{aligned}$$
$$\begin{aligned} \Lambda _f=A_{1,f}. \end{aligned}$$
(2)
$$\begin{aligned} \Lambda \subseteq A_1\subseteq \Lambda _f=A_{1,f}. \end{aligned}$$
(3)
\(\hat{\Lambda }(f-\mathrm{torsion})\). The sets of isoclasses of \(\Lambda \)-modules \(\hat{\Lambda }\) and of \(A_1\)-modules \(\hat{A}_1\) are disjoint unions of f-torsion\((M_f=0)\) and f-torsionfree\((M_f\ne 0)\) simple \(\Lambda \)-modules and \(A_1\)-modules, respectively,Lemma 2.1 is a classification of simple f-torsion \(\Lambda \)-modules.
$$\begin{aligned} \hat{\Lambda }= & {} \hat{\Lambda }(f-\mathrm{torsion})\coprod \hat{\Lambda }(f-\mathrm{torsionfree}), \end{aligned}$$
(4)
$$\begin{aligned} \hat{A}_1= & {} \hat{A}_1 (f-\mathrm{torsion})\coprod \hat{A}_1(f-\mathrm{torsionfree}). \end{aligned}$$
(5)
Lemma 2.1
Let \(\lambda _1, \ldots ,\lambda _s\) be the roots of the polynomial f. ThenAll these \(\Lambda \)-modules are 1-dimensional and they are the only simple finite dimensional \(\Lambda \)-modules (by Theorem 2.10).
$$\begin{aligned} \hat{\Lambda }(f-\mathrm{torsion})= \{[ \Lambda /\Lambda (X-\lambda _i,Y-\mu )]\;\;|\;\; i=1,\ldots ,s; \; \mu \in K\}. \end{aligned}$$
Proof
Each simple f-torsion \(\Lambda \)-module M is annihilated by the (normal) element f\((fM=0)\). So, in fact, the \(\Lambda \)-module M is a simple module over the factor algebrawhich is isomorphic to the factor algebra of the polynomial ring K[X, Y] in two variables at the ideal (f) generated by f, and the equality in the lemma follows.
$$\begin{aligned} \Lambda /(f)=K[X, Y]/(f) \end{aligned}$$
Let N be a simple \(\Lambda \)-module. Then the mapis either 0 or a bijection (f is normal in \(\Lambda )\). In the second case N is, in fact, a simple \((\Lambda _f\equiv A_{1,f})\)-module, so \(\mathrm{dim }_K\,N=\infty \), since \(A_{1,f}\) is a simple infinite dimensional algebra. If the module N is finite dimensional, then \(fN=0\), i.e. \([N]\in \hat{\Lambda }(f-\mathrm{torsion})\). \(\square \)
$$\begin{aligned} f_N:\; N\rightarrow N, \; n\mapsto fn \end{aligned}$$
\(\hat{\Lambda }(f-\mathrm{torsionfree})\). The sum of all simple submodules of a \(\Lambda \)-module M is called the socle of M which is denoted by \(\mathrm{soc}_\Lambda M \). It is the largest semisimple submodule of M. A \(\Lambda \)-module N is called \(\Lambda \)-socle (or, socle, for short) provided \(\mathrm{soc}_\Lambda N\ne 0\). Denote by \(\hat{A_1}(\Lambda \)-socle) the set of isoclasses of simple \(\Lambda \)-socle \(A_1\)-modules. The proof of the following lemma is evident (see [2, Lemma 3.4] for details).
Lemma 2.2
1.
The canonical map is a bijection with inverse \([N]\rightarrow [\mathrm{soc}_\Lambda (N)]\).
$$\begin{aligned} (\cdot )_f :\hat{\Lambda } (f-\mathrm{torsionfree})\rightarrow \hat{A}_{1,f} (\Lambda -\mathrm{socle}),\;[M]\mapsto [ M_f:=A_{1,f}\otimes _\Lambda M] \end{aligned}$$
2.
Each simple f-torsionfree \(\Lambda \)-module has the form for some maximal left ideal \(\mathfrak {m}\) of the ring \(A_{1,f} \). Two such modules are isomorphic, \(M_\mathfrak {m}\simeq M_\mathfrak {n}\), iff the \(A_{1,f}\)-modules \(A_{1,f}/\mathfrak {m}\) and \(A_{1,f}/\mathfrak {n}\) are isomorphic. \(\square \)
$$\begin{aligned} M_\mathfrak {m}:= \Lambda /\Lambda \cap \mathfrak {m}\end{aligned}$$
(6)
Lemma 2.3
Let \(\lambda _1, \ldots ,\lambda _s\) be the roots of the polynomial f. Then
$$\begin{aligned} \hat{A}_1(f-\mathrm{torsion})=\{ [M_i:=A_1/A_1(X-\lambda _i)]\, | \, i=1,\ldots ,s \}. \end{aligned}$$
Proof
As a vector space the module \(M_i\) can be identified with the polynomial ring K[y] in a variable \(y=\partial +A_1(X-\lambda _i)\) andwhere by three dots we denote the sum of elements of smaller degree in the variable y. Thus the linear operatoris nilpotent if and only if \(\mu =\lambda _i\); otherwise, \(X-\mu \) is an isomorphism of the vector space \(M_i\). From this fact it follows that the \(A_1\)-modules \(\{ M_i\}\) are simple and non-isomorphic.
$$\begin{aligned} \partial y^j=y^{j+1}, \;\; Xy^j=\lambda _iy^j+\cdots \;\; \mathrm{for}\;\; j\ge 0 \end{aligned}$$
$$\begin{aligned} X-\mu : M_i\rightarrow M_i, \;\; m\mapsto (X-\mu )m, \;\;m\in M_i, \end{aligned}$$
Now, let \([M]\in \hat{A}_1(f-\mathrm{torsion})\). Then there exists i such that M is an epimorphic image of \(M_i\), hence \(M\simeq M_i\).
\(\square \)
Theorem 2.4
The mapis bijective.
$$\begin{aligned} \hat{A}_1 (f-\mathrm{torsionfree})=\hat{A}_1 \backslash \{ [M_1],\ldots ,[M_s]\}\rightarrow \hat{A}_{1,f}, \;\;[M]\mapsto [M_f] \end{aligned}$$
Proof
The map above is well defined and injective.
Let \([N]\in \hat{A}_{1,f}\). Then \(N\simeq A_{1,f}/J\) for some nonzero maximal left ideal J of \(A_{1,f}\). Then \(I=J\cap A_1\ne 0\) and \(A_1/I\) is a \(A_1\)-submodule of N. The \(A_1\)-module \(A_1/I\) has finite length [10], thus it contains a simple \(A_1\)-submodule, say M. Then \(N\simeq M_f\) which means that the map above is surjective.\(\square \)
Recall that \(D=K[H]\). The localization \(B=S^{-1} A_1\) of the Weyl algebra \(A_1\) at the Ore set \(S=D\backslash \{ 0\}\) is a skew Laurent polynomial ringwith coefficients from the field K(H) of rational functions. The algebra B is a right and left Euclidean domain with respect to the ‘length’ functionhence, it is a right and left principal ideal domain.
$$\begin{aligned} B=K(H)[X, X^{-1}; \sigma ], \;\; \sigma (H)=H-1, \end{aligned}$$
$$\begin{aligned} l: B\backslash \{ 0\}\rightarrow \mathbb {N}:=\{0, 1, 2, \ldots \},\;\; l(\alpha X^m+\beta X^{m+1}+\cdots +\gamma X^n)=n-m,\;\;\alpha \ne 0, \, \gamma \ne 0\in K(H), \end{aligned}$$
We havewhere a simple \(A_1\)-module M belongs to the first (resp., second) set if \(S^{-1}M=0\) (resp., \(S^{-1}M\ne 0\)).
$$\begin{aligned} \hat{A}_1=\hat{A}_1 (D-\mathrm{torsion})\coprod \hat{A}_1 (D-\mathrm{torsionfree}) \end{aligned}$$
For \(\lambda \in K\) set \(\mathcal{O}(\lambda ):=\lambda +\mathbb {Z}\). We say that scalars \(\lambda \) and \(\mu \) are equivalent, \(\lambda \sim \mu \), if either \(\mathcal{O}(\lambda )=\mathcal{O}(\mu )\ne \mathbb {Z}\) or both \(\lambda \) and \(\mu \) belong either to \((-\infty ,0]:=\{ i\in \mathbb {Z}\, | \, i\le 0\}\) or to \([1,\infty ):=\{ i\in \mathbb {Z}\, | \, i\ge 1\}\). Then \(\sim \) is an equivalence relation on K. Let \(K/\sim \) be the set of equivalence classes of K under \(\sim \). So, the elements of the set \(K/\sim \) are distinct sets \(\lambda +\mathbb {Z}\) where \(\lambda \not \in \mathbb {Z}\) and the two sets \((-\infty ,0]\) and \([1,\infty )\). Notice that \(\mathbb {Z}= (-\infty ,0]\coprod [1,\infty )\).
Proposition 2.5
([2, Theorem 3.1]) The mapis a bijection, where
$$\begin{aligned} K/\sim \;\rightarrow \;\hat{A}_1 (D-\mathrm{torsion}), \;\; [\Gamma ]\mapsto [L(\Gamma )], \end{aligned}$$
1.
If \(\Gamma =\mathcal{O}(\lambda )\ne \mathbb {Z}\), then \(L(\Gamma )= A_1/A_1(H-\lambda )\).
2.
If \(\Gamma =(-\infty ,0]\), then \(L(\Gamma )= A_1/A_1X\).
3.
If \(\Gamma =[1,\infty )\), then \(L(\Gamma )= A_1/A_1(H-1, Y)\). \(\Box \)
Corollary 2.6
$$\begin{aligned} \hat{A}_1 (D-\mathrm{torsion}, f-\mathrm{torsion})= {\left\{ \begin{array}{ll} \{ [L((-\infty ,0])= A_1/A_1X]\} &{} \text {if 0 is a root of} \; f(X),\\ \emptyset &{} \text {if 0 is not a root of}\; f(X).\\ \end{array}\right. } \end{aligned}$$
Proof
Straightforward. \(\square \)
Corollary 2.7
Let \([M]\in \hat{A}_1 (D-\mathrm{torsion}, f-\mathrm{torsionfree})\).
1.
If \(M=A_1/A_1X\) (i.e. 0 is not a root of f, by Corollary 2.6) then M is a simple f-torsionfree \(\Lambda \)-module with \(M=M_f\).
2.
If \(M\ne A_1/A_1X\) then \(\mathrm{soc}_\Lambda \, M=\mathrm{soc}_\Lambda \, M_f=0\). The set \(\hat{\Lambda }\) (\(D-\)torsion, \(f-\)torsionfree) is equal to \(\{A_1/A_1/X\}\) if 0 is not a root of f and \(\emptyset \), otherwise.
Proof
1. As a vector space the module \(M=A_1/A_1X\) has the basis \(\{ y^i=\partial ^i+A_1X, \; i\ge 0\} \), andfor some scalars \(\mu _j \in K\). Now, it is obvious that the \(\Lambda \)-module M is a simple f-torsionfree \(\Lambda \)-module \((f(0)\ne 0)\). Moreover, the linear map \(f_M:M\rightarrow M\), \(m\mapsto fm\) is a bijection, hence, \(M=M_f\).
$$\begin{aligned} \partial y^i=y^{i+1}, \;\;Xy^i=-iy^{i-1} \;\; \mathrm{and}\;\; Yy^i=f(0)y^{i+1}+\sum _{0\le j\le i}\; \mu _jy^j, \end{aligned}$$
2. Since \(\Lambda _f=A_{1,f}\), \(\mathrm{soc}_\Lambda \, M=\mathrm{soc}_\Lambda \, M_f\). Let M belongs to the first (resp., third) class of modules from Proposition 2.5 , i.e.The element \(\bar{1}=1+A_1(H-\lambda )\) (resp., \(\bar{1}=1+A_1(H-1, Y )\)) is a canonical generator of the \(A_1\)-module M. In both cases, for \(i\ge 0\), set \(x^i=X^i\bar{1}\). In the first case, for \(i<0\), set \(x^i=\mu _i\partial ^{-i}\bar{1}\), \(\mu _i\in K\). The scalars \(\mu _i\) can be chosen in such a way that (in both cases) \( Xx^i=x^{i+1}\) for all possible i. Degree argument shows that the module M contains a strictly descending chain of \(\Lambda \)-submodulesSuppose that \(N:=\mathrm{soc}_\Lambda M\ne 0\), then, in a view of Lemma 2.2 and Theorem 2.4, N is an essential simple \(\Lambda \)-submodule of both \(M_f\) and M, hence \(0\ne N\subseteq \cap _{n\ge 0}\; f^nM=0\), a contradiction. \(\square \)
$$\begin{aligned} M=L(\Gamma ), \;\; \Gamma =\mathcal{O}(\lambda )\ne \mathbb {Z}\;\; (\mathrm{resp}., \;\; \Gamma =[1, \infty )). \end{aligned}$$
$$\begin{aligned} M\supset fM \supset \cdots \supset f^nM\supset \cdots \;\;\mathrm{with}\;\; \bigcap _{n\ge 0}\; f^nM=0. \end{aligned}$$
An element of a ring is called regular if it is not a zero divisor. Given a ring A and a multiplicatively closed subset S of A which consists of regular normal elements. Let \(B=S^{-1} A\) be the localization of A at S.
Theorem 2.8
Let A, B, and S be as above and let \(\mathfrak {m}\) be a maximal left ideal of B. The following are equivalent.
1.
The A-module \(M_\mathfrak {m}:=A/A\cap \mathfrak {m}\) is simple.
2.
The socle \(\mathrm{soc}_A(M_\mathfrak {m})\ne 0\).
3.
\(A=As+A\cap \mathfrak {m}\) for all \(s\in S\). \(\square \)
Remark. If \(S=\{ f^n, n\ge 0\}\) for some regular normal element f of A, then the last condition of this lemma is equivalent to \(A=Af+A\cap \mathfrak {m}\). We shall use this fact in what follows. In general situation, it suffices to check whether the third condition holds only for generators of the monoid S.
Proof
The implications \((1\Rightarrow 2)\) and \((1\Rightarrow 3)\) are obvious.
\((2\Rightarrow 1)\) If \(\mathrm{soc}_A(M_\mathbf{m})\ne 0\) then it is a simple A-module which for some \(s\in S\) is equal towhere \({}^{\omega _s^{-1} }M_{\mathfrak {m}} \) is the twisted A-module \(M_{\mathfrak {m}} \) by the automorphism \(\omega _s^{-1}\) of A (the element s is regular and normal). Since the A-module \({}^{\omega _s^{-1} }M_{\mathfrak {m}} \) is simple, so is \(M_{\mathfrak {m}}\).
$$\begin{aligned} (As+A\cap \mathfrak {m})/A\cap \mathfrak {m}\simeq As/As\cap \mathfrak {m}\simeq A/A\cap \mathfrak {m}s^{-1}=\omega _s(A)/\omega _s(A\cap \mathfrak {m})\simeq {}^{\omega _s^{-1} }M_{\mathfrak {m}}, \end{aligned}$$
\((3\Rightarrow 1)\) If J is a left ideal of A which contains \(A\cap \mathfrak {m}\) but does not coincide with it, then, by the maximality of \(\mathfrak {m}\), \(S^{-1} J=B\). Therefore \(J\cap S\ne \emptyset \). Let \(s\in J\cap S\). Then \(J\supseteq As+A\cap \mathfrak {m}=A\), that is \(M_\mathfrak {m}\) is a simple A-module.
\(\square \)
\(\hat{A}_1(D-\mathrm{torsionfree})\). Let us recall a description of \(\hat{A}_1(D-\mathrm{torsionfree})\) from [2]. In the set \(S=K[H]\backslash \{ 0\}\) consider the relation <: \(\alpha <\beta \) if there are no roots \(\lambda \) and \(\mu \) of the polynomial \(\alpha \) and \(\beta \) respectively and such that \(\lambda -\mu \) is non-negative integer.
Definition, [2]. An element \(b=Y^m\beta _{-m}+\cdots +\beta _0\in A_1\), \(m>0\), all \(\beta _i\in D\), is called l-normal if \(\beta _0<\beta _{-m}\) and \(\beta _0<H\), (i.e. the polynomial \(\beta _0\) has no root from \(\{ 0, 1, 2, \ldots \}\) and there are no roots \(\lambda \) and \(\mu \) of the polynomials \(\beta _0\) and \(\beta _m \) respectively with \(\lambda -\mu \in \{ 0, 1, 2, \ldots \}\)).
Theorem 2.9
([2, Theorem 3.8]) Let \(b=Y^m\beta _{-m}+\cdots +\beta _0\in A_1\), \(m>0\), all \(\beta _i\in D\), be an l-normal and irreducible element in B. Thenis a simple D-torsionfree \(A_1\)-module. Two such \(A_1\)-modules are isomorphic, \( \mathcal{M}_b\simeq \mathcal{M}_c\), iff \(B/Bb\simeq B/Bc\) as B-modules. Each simple D-torsionfree \(A_1\)-module is isomorphic to some \( \mathcal{M}_b\). \(\square \)
$$\begin{aligned} \mathcal{M}_b:=A_1/A_1\cap Bb \end{aligned}$$
Setfor the localization of the (left)\(\Lambda \)-moduleB at \(S_f\). Then the algebra \(A_{1,f}= \Lambda _f\) can be considered as a (\(A_{1,f}= \Lambda _f\))-submodules of \(B_f\). For any nonzero \(b\in B\), \((Bb)_f=B_fb\).
$$\begin{aligned} B_f:=S_f^{-1}B=A_{1,f}\otimes _\Lambda \,B=\Lambda _f\otimes _\Lambda \,B \end{aligned}$$
Theorem 2.10
Let \(b=Y^m\beta _{-m}+\cdots +\beta _0\in A_1\), \(m>0\), all \(\beta _i\in D\), be an l-normal and irreducible element in B such thatThenis a simple f-torsionfree \(\Lambda \)-module. Two such \(\Lambda \)-modules are isomorphic, \( \mathcal {M}_b\simeq \mathcal {M}_c\), iff \(B/Bb\simeq B/Bc\) as B-modules.
1.
\(\Lambda =\Lambda f+\Lambda \cap B_fb \; (= \Lambda f+\Lambda \cap Bb )\), and
2.
the simple B-module B / Bb is not isomorphic to any of modules \(B/B(X-\lambda )\) where \(\lambda \) runs through the nonzero roots of f.
$$\begin{aligned} \mathcal {M}_b:=\Lambda /\Lambda \cap Bb\; \, (=\Lambda /\Lambda \cap B_fb) \end{aligned}$$
Each simple f-torsionfree \(\Lambda \)-module is isomorphic either to some \( \mathcal {M}_b\) or to the module \(M=A_1/A_1X\) from Corollary 2.7, if 0 is not a root of f (the \(\Lambda \)-module M is not isomorphic to any \(\mathcal {M}_b\)). The condition 1 above is equivalent to the condition that \(\Lambda = \Lambda (X-\lambda _i)+\Lambda \cap B_fb \; (= \Lambda (X-\lambda _i)+\Lambda \cap Bb )\) for all roots \(\lambda _i\) of the polynomial f.
Each simple f-torsionfree \(\Lambda \)-module is infinite dimensional.
Proof
By Lemma 2.2,and \(M=\mathrm{soc}_\Lambda (M_f)\simeq \Lambda /\Lambda \cap \mathfrak {m}\) for some maximal left ideal \(\mathfrak {m}\) of \(A_{1,f}\). By Corollary 2.7, either \(M_f\simeq A_1/A_1X\) (0 is not a root of f) or \(M_f\in \hat{A}_{1,f}(D-\mathrm{torsionfree}, \Lambda -\mathrm{socle})\). In the first case, \(M=\mathrm{soc}_\Lambda (M_f)=M_f=A_1/A_1X\) (Corollary 2.7).
$$\begin{aligned}{}[M]\in \hat{\Lambda }(f-\mathrm{torsionfree}) \Leftrightarrow [M_f]\in \hat{A}_{1,f}(\Lambda -\mathrm{socle}) \end{aligned}$$
In the second case, by Theorems 2.4 and 2.9,for some l-normal irreducible element b from Theorem 2.9. Note that the left ideal \(\mathfrak {m}=A_{1,f}\cap B_fb\) of \(A_{1,f}\) is maximal. By Lemma 2.3 and Theorem 2.9, \([\mathcal{M}_b]\in \hat{A}_1(D-\mathrm{torsionfree}, f-\mathrm{torsionfree})\) iff the second condition of the theorem holds. Now,By Theorem 2.8 and by the Remark after it,In this case,Let us show that (in this case) the natural \(\Lambda \)-module epimorphismis an isomorphism. Note thatThe \(A_1\)-module \(\mathcal{M}_b\) is a submodule of \((\mathcal{M}_b)_f\simeq M_f\). So,By assumption \(\mathrm{soc}_\Lambda (M_f)\ne 0\), then it is a simple essential f-torsionfree \(\Lambda \)-submodule of \(M_f\). If \(\mathrm{ker}\, \varphi \ne 0\), then \(\mathrm{soc}_\Lambda (M_f)\subseteq \mathrm{ker}\, \varphi \), but \(\mathrm{ker}\, \varphi \) is an f-torsion \(\Lambda \)-module, a contradiction.
$$\begin{aligned} M_f\simeq (\mathcal{M}_b)_f=A_{1,f}/A_{1,f}\cap B_fb \end{aligned}$$
$$\begin{aligned} \mathrm{soc}_\Lambda (M_f)=\mathrm{soc}_\Lambda (\mathcal{M}_b)_f=\mathrm{soc}_\Lambda (\Lambda /\Lambda \cap A_{1,f}\cap B_fb)=\mathrm{soc}_\Lambda (\Lambda /\Lambda \cap B_fb). \end{aligned}$$
(7)
$$\begin{aligned} \mathrm{soc}_\Lambda (M_f)\ne 0\;\; \mathrm{iff}\;\; \Lambda =\Lambda f+\Lambda \cap (A_{1,f}\cap B_fb)= \Lambda f+\Lambda \cap B_fb. \end{aligned}$$
$$\begin{aligned} \mathrm{soc}_\Lambda (M_f)=\Lambda /\Lambda \cap (A_{1,f}\cap B_fb)=\Lambda /\Lambda \cap B_fb. \end{aligned}$$
$$\begin{aligned} \varphi :M_b=\Lambda /\Lambda \cap Bb\rightarrow \Lambda /\Lambda \cap B_fb, \;\; \lambda +\Lambda \cap Bb\rightarrow \lambda +\Lambda \cap B_fb, \end{aligned}$$
$$\begin{aligned} \mathrm{ker}\, \varphi =\Lambda \cap B_fb/\Lambda \cap Bb. \end{aligned}$$
$$\begin{aligned} \mathrm{soc}_\Lambda (M_f)=\mathrm{soc}_\Lambda (\mathcal{M}_b)=\mathrm{soc}_\Lambda (\Lambda /\Lambda \cap Bb). \end{aligned}$$
Let \(\mathcal{M}_b \) and \(\mathcal{M}_c\) be as in the theorem. By Lemma 2.2, \(\mathcal{M}_b \simeq \mathcal{M}_c\) as \(\Lambda \)-modules \(\Leftrightarrow \)\(A_{1,f}\otimes _\Lambda \,\mathcal{M}_b \simeq A_{1,f}\otimes _\Lambda \, \mathcal{M}_c\) as \(A_{1,f} \)-modules. Sinceby Theorem 2.4, the above \(A_{1,f}\)-modules are isomorphic iff \(\mathcal{M}_b \simeq \mathcal{M}_c\) as \(A_1\)-modules, so, by Theorem 2.9, \(B/Bb\simeq B/Bc\) as B-modules.
$$\begin{aligned} A_{1,f}\otimes _\Lambda \,\mathcal{M}_b \simeq A_{1,f}/A_{1,f}\cap B_fb\simeq (\mathcal{M}_b)_f, \end{aligned}$$
The condition 1 of the theorem is equivalent to the condition that \(\Lambda = \Lambda (X-\lambda _i)+\Lambda \cap B_fb \; (= \Lambda (X-\lambda _i)+\Lambda \cap Bb )\) for all roots \(\lambda _i\) of the polynomial f (since the elements \(X-\lambda _i\) are regular normal elements of \(\Lambda \) and \(\lambda _i\) are the roots of f).
By Lemma 2.1, each simple f-torsionfree \(\Lambda \)-module is infinite dimensional. If 0 is not a root of f, then the modules \(M=A_1/A_1X\) and \(\mathcal{M}_b\) (from the theorem) are not isomorphic, since the linear map \(X_M: M\rightarrow M\), \(m\mapsto Xm\) is locally nilpotent but \(\mathrm{ker}\, X_{\mathcal{M}_b}=0\). \(\square \)
3 The Prime Ideals, the Krull and Global Dimensions of the Algebra \(\Lambda \)
In this section, K is a field of characteristic zero (not necessarily algebraically closed) and \(f=p_1^{n_1}\cdots p_s^{n_s}\) is a nonscalar polynomial of K[X] where \(p_1, \ldots , p_s\) are irreducible, co-prime divisors of f (i.e. \(K[X]p_i+K[X]p_j=K[X]\) for all \(i\ne j\)). The aim of this section is to give a proof of Theorem 1.1.
Proof of Theorem 1.1
3. The elements \(p_1, \ldots , p_s\) are regular normal elements of the algebra \(\Lambda \) since4. The algebra \(\Lambda \) is a domain, hence \(0\in \mathrm{Spec}_c (\Lambda )\).
$$\begin{aligned} Yp_i=p_i(Y-p_i^{-1}f)\; \; \mathrm{and}\;\; Xp_i=p_iX. \end{aligned}$$
Sinceis a polynomial algebra with coefficients in the field \(F_i\) (since \(YX-XY=f\in \Lambda p_i\)), the ideal \(\Lambda p_i\) is a completely prime ideal of \(\Lambda \).
$$\begin{aligned} \Lambda / \Lambda p_i\simeq F_i[Y] \end{aligned}$$
(8)
By (3), \(\Lambda _f=A_{1,f}\) is a simple algebra (as a localization of a simple Noetherian algebra). If \(\mathfrak {p}\) is a nonzero prime ideal of the algebra \(\Lambda \) then \(f^n\in \mathfrak {p}\) for some natural number \(n\ge 1\). Hence, \(p_i\in \mathfrak {p}\) for some i, by statement 3. By (8), \(\mathfrak {p}= (p_i, g_i)\) for some monic irreducible polynomial \(g_i\) of the polynomial algebra \(F_i[Y]\).
1. By [11, Theorem 6.5.4.(i)], \(\mathrm{Kdim } (\Lambda ) \le \mathrm{Kdim } (K[X])+1=1+1=2\).
Since \(p_i\) is a regular normal element of the algebra \(\Lambda \),by [11, Theorem 6. 5.9]. Therefore, \(\mathrm{Kdim } (\Lambda )=2\).
$$\begin{aligned} \mathrm{Kdim } (\Lambda ) \ge \mathrm{Kdim } (\Lambda / \Lambda p_i)+1{\mathop {=}\limits ^{(8)}} \mathrm{Kdim } ( F_i[Y])+1=1+1=2, \end{aligned}$$
2. By [11, Theorem 7.5.3.(i)], \(\mathrm{gldim}(\Lambda ) \le \mathrm{gldim}(K[X])+1=1+1=2\).
By (8), \(\mathrm{gldim}(\Lambda / \Lambda p_i)= \mathrm{gldim}( F_i[Y])=1<\infty \). Now, by [11, Theorem 7.3.5.(i)],Therefore, \(\mathrm{gldim}(\Lambda )=2\). \(\square \)
$$\begin{aligned} \mathrm{gldim}(\Lambda ) \ge \mathrm{gldim}(\Lambda / \Lambda p_i)+1{\mathop {=}\limits ^{(8)}} \mathrm{gldim}( F_i[Y])+1=1+1=2. \end{aligned}$$
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