1 Skew Category Algebras, Examples and Constructions
In this paper, K is a commutative ring with 1, algebra means a K-algebra. In general, it is not assumed that a K-algebra has an identity element. Module means a left module. Missed definitions can be found in [1].
Let \( \mathcal{C}\) be a category, \(\mathrm{Ob} ( \mathcal{C})\) be the set of its objects and \(\mathrm{Mor}( \mathcal{C})\) be the set of its morphisms. For each objects \(i,j\in \mathrm{Ob} ( \mathcal{C})\), \( \mathcal{C}(i,j)\) is the set of morphisms \(f:i\rightarrow j\), the objects \(i=t(f)\) and \(j=h(f)\) are called the tail and head of the morphism f, respectively. For each object \(i\in \mathrm{Ob} ( \mathcal{C})\), \(e_i\) is the identity morphism \(i\rightarrow i\).
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Definition 1.1
([2]) Let \( \mathcal{C}\) be a category and \(\sigma \) be a functor from the category \( \mathcal{C}\) to the category of unital K-algebras over a commutative ring K (eg, \(K=\mathbb {Z}\) or K is a field). So, for each object \(i\in \mathrm{Ob} ( \mathcal{C})\), \(D_i:=\sigma (i)\) is a K-algebra and for each morphismis a K-algebra homomorphism, and \(\sigma _{fg}= \sigma _f\sigma _g\) for all morphisms f and g such that \(t(f) = h(g)\). The direct sum of left K-moduleswhere \(D_{h(f)}f\) is a free left \(D_{h(f)}\)-module of rank 1, is a K-algebra with multiplication given by the rule: For all \(f,g\in \mathrm{Mor}( \mathcal{C})\), \(a\in D_{h(f)}\) and \(b\in D_{h(g)}\),It is a trivial exercise to verify that the multiplication is associative. The K-algebra \( \mathcal{C}(\sigma )\) is called a skew category K-algebra. If \(K=\mathbb {Z}\), the \(\mathbb {Z}\)-algebra \( \mathcal{C}(\sigma )\) is called a skew category ring.
$$\begin{aligned} f: i\mapsto j, \;\; \sigma _f: D_i\rightarrow D_j \end{aligned}$$
$$\begin{aligned} \mathcal{C}(\sigma )=\bigoplus _{f\in \mathrm{Mor}( \mathcal{C})} D_{h(f)}f \end{aligned}$$
(1)
$$\begin{aligned} af\cdot bg={\left\{ \begin{array}{ll} a\sigma _f(b)fg&{} \text {if }t(f) = h(g),\\ 0&{} \text {otherwise}.\\ \end{array}\right. } \end{aligned}$$
(2)
Definition 1.2
If the direct sum (1) admits an associative product which is given by the rule: For all \(f,g\in \mathrm{Mor}( \mathcal{C})\), \(a\in D_{h(f)}\) and \(b\in D_{h(g)}\),wherethen it is called the twisted skew category K-algebra and is denoted by \( \mathcal{C}(\sigma , c)\).
$$\begin{aligned} af\cdot bg={\left\{ \begin{array}{ll} a\sigma _f(b)c(f,g)fg&{} \text {if }t(f) = h(g),\\ 0&{} \text {otherwise}, \\ \end{array}\right. } \end{aligned}$$
(3)
$$\begin{aligned} c(f,g)\in {\left\{ \begin{array}{ll} D_{h(f)}&{} \text {if }t(f) = h(g),\\ \{ 0\}&{} \text {otherwise,} \\ \end{array}\right. } \end{aligned}$$
(4)
The categorical nature of the above classes of rings especially the categorical/explicit nature of their multiplications makes these classes important as far as various computational aspects are concerned.
Let \(1_i\) be the identity of the algebra \(D_i\). Then \(1_ie_i\in D_ie_i\subseteq \mathcal{C}(\sigma )\) where \(i\in \mathrm{Ob}( \mathcal{C})\). Abusing the notation, we write \(e_i\) for \(1_ie_i\). Then \(e_i\in \mathcal{C}(\sigma )\).
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The\( \mathcal{C}\)-grading on\( \mathcal{C}(\sigma )\). By the very definition, the algebra \( \mathcal{C}(\sigma )\) is a \( \mathcal{C}\)-graded algebra, that isThe algebra \( \mathcal{C}(\sigma )\) is a direct sumand for all \(i,j,k,l\in \mathrm{Ob} ( \mathcal{C})\),where \(\delta _{jk}\) is the Kronecker delta. In particular, for each \(i\in \mathrm{Ob} ( \mathcal{C})\), \( \mathcal{C}(\sigma )_{ii}\) is a K-algebra without 1, in general. For each \(i,j\in \mathrm{Ob} ( \mathcal{C})\), \( \mathcal{C}(\sigma )_{ij}\) is a \(( \mathcal{C}(\sigma )_{ii}, \mathcal{C}(\sigma )_{jj})\)-bimodule.
$$\begin{aligned} D_{h(f)}f\cdot D_{h(g)}g\subseteq D_{h(fg)}fg\;\; \mathrm{for \; all}\;\; f,g\in \mathrm{Mor}( \mathcal{C}). \end{aligned}$$
$$\begin{aligned} \mathcal{C}(\sigma )= \bigoplus _{i,j\in \mathrm{Ob} ( \mathcal{C})} \mathcal{C}(\sigma )_{ij}\;\; \mathrm{where}\;\; \mathcal{C}(\sigma )_{ij}=\bigoplus _{f\in \mathcal{C}(j,i)}D_if \end{aligned}$$
(5)
$$\begin{aligned} \mathcal{C}(\sigma )_{ij} \mathcal{C}(\sigma )_{kl}\subseteq \delta _{jk} \mathcal{C}(\sigma )_{il} \end{aligned}$$
(6)
The next two examples show that even for two simplest categories that contain a single object, a single loop or a single invertible loop, the above construction gives apart from a skew polynomial ring or a skew Laurent polynomial ring, new classes of rings.
Example 1
Let \( \mathcal{C}\) be a category that contains a single object, say 1, and \(\mathrm{Mor}( \mathcal{C})= \{ x^i\, | \, i\in \mathbb {N}\}\) where \(e:=x^0\) is the identity morphism. So, \( \mathcal{C}(\sigma )= De\oplus Dx\oplus \cdots \oplus Dx^i\oplus \cdots \) where \(D= \sigma (1)\) and \(ed=\sigma _e(d)e\) and \(x^id= \sigma _x^i(d) x^i\) for all \(i\ge 1\) where \(\sigma _e\) and \(\sigma _x\) are ring endomorphisms of D such that \(\sigma _e \sigma _x= \sigma _x\sigma _e= \sigma _x\) and \(\sigma _e^2= \sigma _e\).
- If \(\sigma _e = \mathrm{id}_D\) then \( \mathcal{C}(\sigma )= D[x; \sigma _x]\) is a skew polynomial ring.
- If \(\sigma _e\ne \mathrm{id}_D\) then \( \mathcal{C}(\sigma )\) is not a skew polynomial ring since \(ed= \sigma _e(d)e\) and, in general, \(\sigma _e(d) e \ne de\) for all \(d\in D\) (since \(\sigma _e \ne \mathrm{id}_D\)). For example, let \(D= D_1\times D_2\times D_3\) and \(\sigma _e\) and \(\sigma _x\) are the projections onto \(D_1\times D_2\) and \(D_1\), respectively. Then \(eD_3=0\).
Example 2
Let \( \mathcal{C}\) be a category that contains a single object, say 1, and \(\mathrm{Mor}( \mathcal{C})= \{ x^i\, | \, i\in \mathbb {Z}\}\) where \(e:=x^0\) is the identity morphism \((xx^{-1}=x^{-1}x=e\)). The functor \(\sigma \) is determined by the algebra \(D=\sigma (1)\) and its algebra endomorphisms \(\sigma _e\), \(\sigma _x\) and \(\sigma _{x^{-1}}\) such thatThen \( \mathcal{C}(\sigma )= \oplus _{i\in \mathbb {Z}}Dx^i\).
$$\begin{aligned} \sigma _e^2= \sigma _e, \;\; \sigma _e \sigma _{x^{\pm 1}}= \sigma _{x^{\pm 1}}\sigma _e= \sigma _{x^{\pm 1}}\;\; \mathrm{and}\;\; \sigma _x\sigma _{x^{-1}}=\sigma _{x^{-1}}\sigma _x=\sigma _e. \end{aligned}$$
- If \(\sigma _e = \mathrm{id}_D\) then \(\sigma _{x^{-1}}=\sigma _x^{-1}\) and \( \mathcal{C}(\sigma )= D[x^{\pm 1}; \sigma _x]\) is a skew Laurent polynomial ring.
- If \(\sigma _e\ne \mathrm{id}_D\) then \( \mathcal{C}(\sigma )\) is not a skew Laurent polynomial ring. For example, let \(D= D_1\times D_2\) be a direct product of algebras and \(\sigma _e=\sigma _x=\sigma _{x^{-1}}\) be the projection onto \(D_1\). Then \(eD_2=0\) and \(xD_2=x^{-1}D_2=0\).
Example 3
Let \( \mathcal{C}\) be a category that contains a single object, say 1, and the monoid \( \mathcal{C}(1, 1)\) is generated by elements x and y subject to the defining relation \(yx=e\). The functor \(\sigma \) is determined by the algebra \(D=\sigma (1)\) and its three algebra endomorphisms \(\sigma _x\), \(\sigma _y\) and \(\sigma _e\) such thatThe skew category algebra \( \mathcal{C}(\sigma )\) is called the skew semi-Laurent polynomial ring [2]. It is a new class of rings. Suppose, for simplicity, that \(\sigma _e=\mathrm{id}_D\). Then the ring \( \mathcal{C}(\sigma )\) is generated by a ring D and elements x and y subject to the defining relations:We denote this ring by \(D[x,y; \sigma _x, \sigma _y]\). In particular, \(D[x,y; \tau , \tau ^{-1}]\) where \(\tau \) is an automorphism of D.
$$\begin{aligned} \sigma _y\sigma _x=\sigma _e. \end{aligned}$$
$$\begin{aligned} yx=1, \;\; xd= \sigma _x(d) x \;\; \mathrm{and }\;\; yd= \sigma _y(d) y \;\; \mathrm{for \; all}\;\; d\in D. \end{aligned}$$
Example 4
Let \(n\ge 1\) be a natural number and \(\mathcal{M}_n\) be the matrix units category:Let D be a ring and \(f_1, \ldots , f_n\) be its automorphisms. Define \(\sigma \) by the rule \(\sigma (i) = D\) and \(\sigma (E_{ij})=f_if^{-1}_j\). The skew category algebrais called the skew matrix ring where the multiplication is given by the rule
$$\begin{aligned} \mathrm{Ob}(\mathcal{M}_n)= \{ 1, \ldots , n\}, \;\; \mathcal{M}_n(i,j)=\{ E_{ji}\}\;\; \mathrm{and}\;\; E_{ij}E_{jk}=E_{ik}\;\; \mathrm{for \; all}\;\; i,j,k. \end{aligned}$$
$$\begin{aligned} \mathcal{M}_n(\sigma ) = \oplus _{i,j=1}^n DE_{ij} \end{aligned}$$
$$\begin{aligned} dE_{ij}\cdot d'E_{kl}=\delta _{jk} df_if^{-1}_j(d')E_{jl}\;\; \mathrm{for \; all}\;\; d,d'\in D. \end{aligned}$$
The skew graph rings and the skew tree rings.
Definition 1.3
([2]) Let \(\Gamma = (\Gamma _0, \Gamma _1)\) be a non-oriented graph without cycles where \(\Gamma _0\) is the set of vertices and \(\Gamma _1\) is the set of edges. If, in addition, \(\Gamma \) is connected then it is called a tree. So, any non-oriented graph without cycles is a disjoint union of its connected components which are trees. Let \({\varvec{\Gamma }}\) be the category groupoid associated with \(\Gamma \): \(\mathrm{Ob}({\varvec{\Gamma }})=\Gamma _0\), for each \(i\in \mathrm{Ob}(\Gamma )\), \({\varvec{\Gamma }}(i,i)=\{ e_{ii} \}\), for distinct \(i,j\in \mathrm{Ob}({\varvec{\Gamma }})\) such that \((i,j)\in \Gamma _1\), \({\varvec{\Gamma }} (i,j) = \{ e_{ji}\}\) and \({\varvec{\Gamma }} (j,i) = \{ e_{ij}\}\), \(e_{ij}e_{ji}=e_{ii}\) and \(e_{ji}e_{ij}=e_{jj}\). Let \(\sigma \) be a functor from \({\varvec{\Gamma }}\) to the category of rings. Then \({\varvec{\Gamma }}(\sigma )\) is called the skew graph ring. If \(\Gamma \) is a tree then \({\varvec{\Gamma }}(\sigma )\) is called the skew tree ring. We say that the functor \(\sigma \) is of isomorphism type if \(\sigma (e_{ij}):\sigma (i) \rightarrow \sigma (j)\) is a unital ring isomorphism for all \((i,j) \in \Gamma _1\).
Theorem 1.4
Let \(\Gamma \) be a finite tree, \(n= |\Gamma _0|\) and the functor \(\sigma \) be of isomorphism type. Suppose that for some \(i\in \Gamma _0\) the ring \(D_i=\sigma (i)\) is a semiprime, left (resp., right) Goldie ring and \(Q_l(D_i)\) (resp., \(Q_r(D_i))\) is its left (resp., right) quotient ring. Then \(\mathbf{\Gamma }(\sigma )\) is a semiprime, left (resp., right) Goldie ring and \(Q_l(\mathbf{\Gamma }(\sigma ))\simeq M_n(Q_l(D_i))\) (resp., \(Q_r(\mathbf{\Gamma }(\sigma ))\simeq M_n(Q_r(D_i))\)) where \(M_n(R)\) is a matrix ring over a ring R. In particular, the left (resp., right) uniform dimension of \({\varvec{\Gamma }}(\sigma )\) is \(nd_l\) (resp., \(nd_r\)) where \(d_l\) (resp., \(d_r\)) is a left (resp., right) uniform dimension of \(D_i\).
Proof
(Sketch). Let \( \mathcal{C}_{D_j}\) be the set of regular elements of the ring \(D_j=\sigma (j)\). All the rings \(D_j\) are isomorphic. The set of regular elements \(S=\oplus _{j=1}^n \mathcal{C}_{D_j}e_{jj}\) is a left Ore set of \(\mathbf{\Gamma }(\sigma )\) such that \(S^{-1}{} \mathbf{\Gamma }(\sigma )\) is a semisimple Artinian ring. Furthermore, \(S^{-1}{} \mathbf{\Gamma }(\sigma )\simeq M_n(Q_l(D_i))\). Hence, \(Q_l(\mathbf{\Gamma }(\sigma ))\simeq M_n(Q_l(D_i))\), and so \(\mathbf{\Gamma }(\sigma )\) is a semiprime, left Goldie ring. The rest is obvious. \(\square \)
As a result we have the following corollary.
Corollary 1.5
Let \(\Gamma \) be a finite non-orientable graph, i.e., \(\Gamma =\coprod _{s=1}^\nu \Gamma ^{(s)}\) is a disjoint union of finite trees \(\Gamma ^{(s)}\). Then
1.
The skew graph ring \(\mathbf{\Gamma }(\sigma )\) is a direct product \(\prod _{s=1}^\nu \mathbf{\Gamma }^{(s)}(\sigma _s)\) of skew tree rings where \(\sigma _s\) is the restriction of the functor \(\sigma \) to \(\Gamma ^{(s)}(\sigma _s)\).
2.
If the trees \(\Gamma ^{(s)}\) (\(s=1, \ldots , \nu \)) satisfy the conditions of Theorem 1.4 then \(Q_l(\mathbf{\Gamma }(\sigma ))\simeq \prod _{s=1}^\nu Q_l(\mathbf{\Gamma }^{(s)}(\sigma _s))\) (resp., \(Q_r(\mathbf{\Gamma }(\sigma ))\simeq \prod _{s=1}^\nu Q_r(\mathbf{\Gamma }^{(s)}(\sigma _s))\)) is a direct product of semiprime, left (resp., right) Goldie rings, and so it is a semiprime, left (resp., right) Goldie ring.
2 Properties of Skew Category Algebras
In this section, criteria are given for a skew category algebra \( \mathcal{C}(\sigma )\) to be left/right Noetherian or semiprime or simple.
The ideal\(\mathfrak {a}\)and the algebra\(\overline{ \mathcal{C}(\sigma )}\).
Lemma 2.1
Let D be a ring and \(\sigma '\) be its ring endomorphism such that \(\sigma '^2= \sigma '\). Then \(D=\sigma ' (D)\oplus \mathrm{ker } (\sigma ' )\) and the restriction homomorphism \(\sigma ' |_{\sigma ' (D)}: \sigma ' (D) \rightarrow \sigma ' (D)\), \(d\mapsto d\) is the identity automorphism.
Proof
Straightforward. \(\square \)
By (5), the formal sumdetermines two well-defined maps:Clearly, the map \(\cdot e\) is the identity map \(\mathrm{id}\) on \( \mathcal{C}(\sigma )\) but the kernel \(\mathfrak {a}\) of the map \(e\cdot \) is equal towhere \(\mathfrak {a}_i:= \mathrm{ker } (\sigma _{e_i})\) and \(\sigma _i:=\sigma _{e_i}:D_i\rightarrow D_i\) is a K-algebra endomorphism, and \((e\cdot )^2=e\cdot \). Since \(\sigma _i^2=\sigma _i\),by Lemma 2.1.is a K-subalgebra of \( \mathcal{C}(\sigma )\) such that the maps \((e\cdot )|_{\overline{ \mathcal{C}(\sigma )}}:\overline{ \mathcal{C}(\sigma )}\rightarrow \overline{ \mathcal{C}(\sigma )}\), \( c\mapsto c\) and \((\cdot e)|_{\overline{ \mathcal{C}(\sigma )}}:\overline{ \mathcal{C}(\sigma )}\rightarrow \overline{ \mathcal{C}(\sigma )}\), \( c\mapsto c\) are the identity map on \(\overline{ \mathcal{C}(\sigma )}\).
$$\begin{aligned} e=\sum _{i\in \mathrm{Ob} ( \mathcal{C})}e_i \end{aligned}$$
$$\begin{aligned} e\cdot : \mathcal{C}(\sigma )\rightarrow \mathcal{C}(\sigma ),\;\; a\mapsto ea\;\; \mathrm{and}\;\; \cdot e : \mathcal{C}(\sigma )\rightarrow \mathcal{C}(\sigma ), \;\; a\mapsto ae. \end{aligned}$$
$$\begin{aligned} \mathfrak {a}( \mathcal{C}(\sigma )) := \mathfrak {a}:=\bigoplus _{f\in \mathrm{Mor}( \mathcal{C})} \mathfrak {a}_{h(f)}f \end{aligned}$$
$$\begin{aligned} D_i= \sigma _i (D)\oplus \mathfrak {a}_i\;\; \mathrm{for \; all}\;\; i\in \mathrm{Ob}( \mathcal{C}), \end{aligned}$$
(7)
$$\begin{aligned} \mathcal{C}(\sigma )=\overline{ \mathcal{C}(\sigma )}\oplus \mathfrak {a}\;\; \mathrm{where} \;\; \overline{ \mathcal{C}(\sigma )}:=\bigoplus _{f\in \mathrm{Mor}( \mathcal{C})} \sigma _{h(f)}(D_{h(f)})f \end{aligned}$$
(8)
Lemma 2.2
The set \(\mathfrak {a}\) is an ideal of the algebra \( \mathcal{C}(\sigma )\) such that \( \mathcal{C}(\sigma )\mathfrak {a}=0\), \(\mathfrak {a}\, \mathcal{C}(\sigma )=\mathfrak {a}\) and \(\mathfrak {a}^2 = 0\).
Proof
\( \mathcal{C}(\sigma )\mathfrak {a}= \mathcal{C}(\sigma )\cdot e\cdot \mathfrak {a}=0\), the rest is obvious. \(\square \)
The next theorem shows that the algebra \(\overline{ \mathcal{C}(\sigma )}\) is also a skew category algebra.
Theorem 2.3
1.
The subalgebra \(\overline{ \mathcal{C}(\sigma )}\) of \( \mathcal{C}(\sigma )\) is also a skew category algebra \(\overline{ \mathcal{C}(\sigma )}= \mathcal{C}(\overline{\sigma } )\) where for each \(i\in \mathrm{Ob}( \mathcal{C})\), \(\overline{\sigma }(i) := \sigma _i(D_i)\) and for each \(f\in \mathcal{C}(i,j)\), \(\overline{\sigma }_f:=\sigma _f|_{\sigma _i(D_i)}: \sigma _i (D_i) \rightarrow \sigma _i (D_i)\), \(d\mapsto \sigma _f(d)\).
2.
For all \(i\in \mathrm{Ob}( \mathcal{C})\), \(\overline{\sigma }_i=\mathrm{id}_{\overline{\sigma }(i)}\).
3.
\(\mathfrak {a}( \mathcal{C}( \overline{\sigma }))=0\).
4.
The maps \(e\cdot \) and \(\cdot e\) are the identity maps on \( \mathcal{C}(\overline{\sigma })\).
Proof
1.
Statement 1 follows from (8) and the fact that \(\sigma _j\sigma _f=\sigma _f = \sigma _f\sigma _i\) for all elements \(f\in \mathcal{C}(i,j)\).
2–4.
Statement 2 is obvious. Then statements 3 and 4 follow from statement 2. \(\square \)
The ideal \(\mathfrak {a}\) is a \( \mathcal{C}\)-graded ideal of the algebra \( \mathcal{C}(\sigma )\). Furthermore,where \(\mathfrak {a}_{ij}=\bigoplus _{f\in \mathcal{C}(j,i)}\mathfrak {a}_i f\subseteq \mathcal{C}(\sigma )_{ij}\), \(0=\mathfrak {a}_{ij}\mathfrak {a}_{kl}\subseteq \delta _{jk}\mathfrak {a}_{il}\) for all \(i,j,k,l\in \mathrm{Ob} ( \mathcal{C})\). Since \(\overline{ \mathcal{C}(\sigma )}= \mathcal{C}(\overline{\sigma })\) (Theorem 2.3.(1)), the factor algebrais a \( \mathcal{C}\)-graded algebra where \(\overline{D}_i=D_i/\mathfrak {a}_i=\mathrm{im}(\sigma _i)\). Furthermore,and \(\overline{ \mathcal{C}(\sigma )}_{ij}\overline{ \mathcal{C}(\sigma )}_{kl}\subseteq \delta _{jk}\overline{ \mathcal{C}(\sigma )}_{il}\) for all \(i,j,k,l\in \mathrm{Ob} ( \mathcal{C})\).
$$\begin{aligned} \mathfrak {a}= \bigoplus _{i,j\in \mathrm{Ob} ( \mathcal{C})}\mathfrak {a}_{ij} \end{aligned}$$
$$\begin{aligned} \overline{ \mathcal{C}(\sigma )}= \mathcal{C}(\sigma )/ \mathfrak {a}= \bigoplus _{f\in \mathrm{Mor}( \mathcal{C})} \overline{D}_{h(f)}f\subseteq \mathcal{C}(\sigma )\end{aligned}$$
$$\begin{aligned} \overline{ \mathcal{C}(\sigma )}= \bigoplus _{\mathbf{i},j\in \mathrm{Ob} ( \mathcal{C})}\overline{ \mathcal{C}(\sigma )}_{ij}\;\; \mathrm{where}\;\; \overline{ \mathcal{C}(\sigma )}_{ij}= \mathcal{C}(\sigma )_{ij}/ \mathfrak {a}_{ij} \end{aligned}$$
(9)
Theorem 2.4
(Criterion for \( \mathcal{C}(\sigma )\) to be a left Noetherian algebra) The algebra \( \mathcal{C}(\sigma )\) is a left Noetherian algebra iff the following conditions hold
1.
the set \(\mathrm{Ob} ( \mathcal{C})\) is a finite set,
2.
the ideal \(\mathfrak {a}\) is a finitely generated abelian group,
3.
for every object \(i\in \mathrm{Ob} ( \mathcal{C})\), the K-algebra \(\overline{ \mathcal{C}(\sigma )}_{ii}\) is a left Noetherian algebra, and
4.
for all objects \(i,j\in \mathrm{Ob} ( \mathcal{C})\) such that \(i\ne j\), the left \(\overline{ \mathcal{C}(\sigma )}_{ii}\)-module \(\overline{ \mathcal{C}(\sigma )}_{ij}\) is finitely generated.
Proof
The algebra \( \mathcal{C}(\sigma )= \bigoplus _{j\in \mathrm{Ob} ( \mathcal{C})} \mathcal{C}(\sigma )_{*j}\) is a direct sum of nonzero left ideals whereSo, the algebra \( \mathcal{C}(\sigma )\) is a left Noetherian algebra iff the set \(\mathrm{Ob} ( \mathcal{C})\) is a finite set and all the left ideals \( \mathcal{C}(\sigma )_{*j}\) are Noetherian left \( \mathcal{C}(\sigma )\)-modules iff \(|\mathrm{Ob} ( \mathcal{C})|<\infty \), the left \( \mathcal{C}(\sigma )\)-module \(\mathfrak {a}\) is Noetherian and all the left \(\overline{ \mathcal{C}(\sigma )}\)-modulesare Noetherian (since \( \mathcal{C}(\sigma )=\overline{ \mathcal{C}(\sigma )}\oplus \mathfrak {a}\) is a direct sum of left \( \mathcal{C}(\sigma )\)-modules) iff conditions 1 and 2 hold (since \( \mathcal{C}(\sigma )\mathfrak {a}=0\), Lemma 2.4) and the left \(\overline{ \mathcal{C}(\sigma )}_{ii}\)-module \(\overline{ \mathcal{C}(\sigma )}_{ij}\) is Noetherian for all \(i,j\in \mathrm{Ob} ( \mathcal{C})\) (since each left \(\overline{ \mathcal{C}(\sigma )}\)-submodule M of \(\overline{ \mathcal{C}(\sigma )}_{*j}\) is a direct sumwhere \(e_iM\) is a left \(\overline{ \mathcal{C}(\sigma )}_{ii}\)-submodule of \(\overline{ \mathcal{C}(\sigma )}_{ij}\) and the functor from the category of all \(\overline{ \mathcal{C}(\sigma )}_{ii}\)-submodules of \(\overline{ \mathcal{C}(\sigma )}_{ij}\) to the category of all \(\overline{ \mathcal{C}(\sigma )}\)-submodules of \(\overline{ \mathcal{C}(\sigma )}_{*j}\),is faithful since \(e_i\overline{ \mathcal{C}(\sigma )}N=\overline{ \mathcal{C}(\sigma )}_{ii}N=N\)) iff statements 1–4 hold. \(\square \)
$$\begin{aligned} \mathcal{C}(\sigma )_{*j}:=\bigoplus _{i\in \mathrm{Ob} ( \mathcal{C})} \mathcal{C}(\sigma )_{ij}. \end{aligned}$$
$$\begin{aligned} \overline{ \mathcal{C}(\sigma )}_{*j}=\bigoplus _{i\in \mathrm{Ob} ( \mathcal{C})} \overline{ \mathcal{C}(\sigma )}_{ij}\end{aligned}$$
$$\begin{aligned} M=eM=\bigoplus _{i\in \mathrm{Ob} ( \mathcal{C})}e_iM \end{aligned}$$
$$\begin{aligned} N\rightarrow \overline{ \mathcal{C}(\sigma )}N=\bigoplus _{k\in \mathrm{Ob} ( \mathcal{C})} \overline{ \mathcal{C}(\sigma )}_{ki}N \end{aligned}$$
Proposition 2.5
(Criterion for \( \mathcal{C}(\sigma )\) to be a right Noetherian algebra) The algebra \( \mathcal{C}(\sigma )\) is a right Noetherian algebra iff the following conditions hold
1.
the set \(\mathrm{Ob} ( \mathcal{C})\) is a finite set,
2.
for every object \(i\in \mathrm{Ob} ( \mathcal{C})\), the K-algebra \( \mathcal{C}(\sigma )_{ii}\) is a right Noetherian algebra, and
3.
for all objects \(i,j\in \mathrm{Ob} ( \mathcal{C})\) such that \(i\ne j\), the right \( \mathcal{C}(\sigma )_{jj}\)-module \( \mathcal{C}(\sigma )_{ij}\) is finitely generated.
Proof
The algebra \( \mathcal{C}(\sigma )= \bigoplus _{i\in \mathrm{Ob} ( \mathcal{C})} \mathcal{C}(\sigma )_{i*}\) is a direct sum of nonzero right ideals whereSo, the algebra \( \mathcal{C}(\sigma )\) is a right Noetherian algebra iff the set \(\mathrm{Ob} ( \mathcal{C})\) is a finite set and all right ideals \( \mathcal{C}(\sigma )_{i*}\) are Noetherian right \( \mathcal{C}(\sigma )\)-modules iff \(|\mathrm{Ob} ( \mathcal{C})|<\infty \) and the right \( \mathcal{C}(\sigma )_{jj}\)-module \( \mathcal{C}(\sigma )_{ij}\) is Noetherian for all \(i,j\in \mathrm{Ob} ( \mathcal{C})\) iff \(|\mathrm{Ob} ( \mathcal{C})|<\infty \), the rings \( \mathcal{C}(\sigma )_{ii}\) are right Noetherian and the right \( \mathcal{C}(\sigma )_{jj}\)-modules \( \mathcal{C}(\sigma )_{ij}\) are finitely generated for all \(i\ne j\). \(\square \)
$$\begin{aligned} \mathcal{C}(\sigma )_{i*}=\bigoplus _{j\in \mathrm{Ob} ( \mathcal{C})} \mathcal{C}(\sigma )_{ij}. \end{aligned}$$
Example 5
Let \( \mathcal{C}\): \(1{\mathop {\rightarrow }\limits ^{f}}2\) and the functor \(\sigma \) is as follows: \(\sigma (1) = \mathbb {Q}\), \(\sigma (2) =\mathbb {R}\), \(\sigma _{e_1}=\mathrm{id}_{\mathbb {Q}}\), \(\sigma _{e_2}=\mathrm{id}_{\mathbb {R}}\) and \(\sigma _f:\mathbb {Q}\rightarrow \mathbb {R}\), \(q\mapsto q\). Then the algebra \( \mathcal{C}(\sigma )\) is isomorphic to the lower triangular matrix algebra \(\begin{pmatrix} \mathbb {Q}&{} 0 \\ \mathbb {R}&{}\mathbb {R}\end{pmatrix}\). By Theorem 2.4, the algebra \( \mathcal{C}(\sigma )\) is left Noetherian but not right Noetherian, by Proposition 2.5 (since \(\mathbb {R}_\mathbb {Q}\) is not a finitely generated right \(\mathbb {Q}\)-module).
Example 6
Let \( \mathcal{C}\): \(1{\mathop {\rightarrow }\limits ^{f}}2\) and the functor \(\sigma \) is as follows: \(\sigma (1) = K[t]\) is a polynomial algebra in the variable t over K, \(\sigma (2) =K\), \(\sigma _{e_1}:K[t]\rightarrow K[t]\), \(t\mapsto 0\); \(\sigma _{e_2}=\mathrm{id}_{K}:K\rightarrow K\) and \(\sigma _f:K[t]\rightarrow K\), \(t\mapsto 0\). Then \(\mathfrak {a}= tK[t]e_1\) is not a finitely generated \(\mathbb {Z}\)-module. So, the algebra \( \mathcal{C}(\sigma )\) is not a left Noetherian algebra, by Theorem 2.4. Since the algebra \( \mathcal{C}(\sigma )_{11}=K[t]e_1\) is not a right Noetherian algebra, the ring \( \mathcal{C}(\sigma )\) is not a right Noetherian ring, by Proposition 2.5.
Lemma 2.6
(Existence of 1 in \( \mathcal{C}(\sigma )\)) The algebra \( \mathcal{C}(\sigma )\) has 1 iff the set \(\mathrm{Ob} ( \mathcal{C})\) is a finite set and \(\sigma _{e_i}=\mathrm{id}_{D_i}\) for all \(i\in \mathrm{Ob} ( \mathcal{C})\). In this case, \(e=\sum _{i\in \mathrm{Ob} ( \mathcal{C})}e_i\) is the identity of the algebra \( \mathcal{C}(\sigma )\).
Proof
\((\Rightarrow )\) Suppose that 1 is an identity of \( \mathcal{C}(\sigma )\). Then necessarily the set \(\mathrm{Ob} ( \mathcal{C})\) is a finite set, otherwise \(1a=0\) for some nonzero element a of \( \mathcal{C}(\sigma )\). The \(1=\sum _{i,j} 1_{ij}\) where \(1_{ij}\in \mathcal{C}(\sigma )_{ij}\). The equalities \( 1e_j=e_j=e_j1\) for all \(j\in \mathrm{Ob} ( \mathcal{C})\) imply that \(1=\sum _{i\in \mathrm{Ob} ( \mathcal{C})} e_i=e\). Then, necessarily \(\sigma _{e_i}=\mathrm{id}_{D_i}\) for all \(i\in \mathrm{Ob} ( \mathcal{C})\).
\((\Leftarrow )\) Clearly, e is the identity of the algebra \( \mathcal{C}(\sigma )\). \(\square \)
Lemma 2.7
Suppose that \(n=|\mathrm{Ob}( \mathcal{C})|<\infty \). If I is an ideal of \( \mathcal{C}(\sigma )\) such that \(e_iIe_i=0\) for all \(i\in \mathrm{Ob}( \mathcal{C})\) then \(I^{n+1}=0\).
Proof
By (8), \( \mathcal{C}(\sigma )= \overline{ \mathcal{C}(\sigma )}\oplus \mathfrak {a}\). Hence, \(I\subseteq \overline{I}\oplus \mathfrak {a}\) where \(\overline{I}= (I+\mathfrak {a})/\mathfrak {a}= \sum _{i,j\in \mathrm{Ob}( \mathcal{C})} e_iIe_j\subseteq \overline{ \mathcal{C}(\sigma )}\). Notice that \(\overline{I}^n=0\) since \(e_iIe_i=0\) for all \(i\in \mathrm{Ob}( \mathcal{C})\). Now,since \(\mathfrak {a}^2=0\) and \( \mathcal{C}(\sigma )\mathfrak {a}=0\) (Lemma 2.2). \(\square \)
$$\begin{aligned} I^{n+1}\subseteq (\overline{I}+\mathfrak {a})^{n+1} \subseteq \overline{I}^{n+1}+\mathfrak {a}\overline{I}^n =0 \end{aligned}$$
Recall that a ring is a semiprime ring if the zero ideal is the only nilpotent ideal.
Theorem 2.8
(Criterion for \( \mathcal{C}(\sigma )\) to be a semiprime algebra) Suppose that \(n:=|\mathrm{Ob}( \mathcal{C})|<\infty \). Then the following statements are equivalent.
1.
The algebra \( \mathcal{C}(\sigma )\) is a semiprime algebra.
2.
The algebras \( \mathcal{C}(\sigma )_{ii}\) are semiprime where \(i\in \mathrm{Ob}( \mathcal{C})\) and, for all distinct \(i,j\in \mathrm{Ob} ( \mathcal{C})\), \(a_{ij} \mathcal{C}(\sigma )_{ji}\ne 0\) and \( \mathcal{C}(\sigma )_{ji}a_{ij}\ne 0\) for all nonzero elements \(a_{ij}\in \mathcal{C}(\sigma )_{ij}\).
3.
The algebras \( \mathcal{C}(\sigma )_{ii}\) are semiprime where \(i\in \mathrm{Ob}( \mathcal{C})\) and each ideal I of \( \mathcal{C}(\sigma )\) such that \(e_iIe_i=0\) for all \(i\in \mathrm{Ob}( \mathcal{C})\) is equal to zero.
Proof
Since \(|\mathrm{Ob}( \mathcal{C})|<\infty \), the direct product of algebras \(\mathcal{D}:= \prod _{i\in \mathrm{Ob}( \mathcal{C})} \mathcal{C}(\sigma )_{ii}\) is a semiprime algebra iff all the algebras \( \mathcal{C}(\sigma )_{ii}\) are semiprime.
\((1\Rightarrow 2)\) If \(\mathfrak {b}\) is a nonzero nilpotent ideal of the ring \(\mathcal{D}\) and \((\mathfrak {b}) = \mathcal{C}(\sigma )\mathfrak {b} \mathcal{C}(\sigma )\) is the ideal of \( \mathcal{C}(\sigma )\) generated by \(\mathfrak {b}\) thenwhere for a real number r, \(\lfloor r\rfloor :=\max \{ z\in \mathbb {Z}\, | \, z\le r\}\), and so the ideal \((\mathfrak {b})\) of the algebra \(\mathcal{D}\) is a nilpotent ideal. Therefore, the ring \( \mathcal{C}(\sigma )_{ii}\) must be semiprime for all \(i\in \mathrm{Ob} ( \mathcal{C})\).
$$\begin{aligned} (\mathfrak {b})^k \subseteq (\mathfrak {b}^{\lfloor \frac{k}{n^2}\rfloor })\;\; \mathrm{for\; all}\;\; k\ge 1 \end{aligned}$$
Suppose that there exists a nonzero element \(a_{ij}\in \mathcal{C}(\sigma )_{ij}\) for some distinct objects i and j such that either \(a_{ij} \mathcal{C}(\sigma )_{ji}=0\) or \( \mathcal{C}(\sigma )_{ji}a_{ij}=0\). Then \((a_{ij})^2= (a_{ij} \mathcal{C}(\sigma )_{ji}a_{ij})=0\), a contradiction.
\((2\Rightarrow 1)\) Since all rings \( \mathcal{C}(\sigma )_{ii}\) are semiprime, the ideal \(\mathfrak {a}\) is equal to zero, by Lemma 2.2. Therefore, if J is a nilpotent ideal of \( \mathcal{C}(\sigma )\) then necessarily \(J=\bigoplus _{i,j\in \mathrm{Ob} ( \mathcal{C})}J_{ij}\) where \(J_{ij}=e_iJe_j\). Furthermore, all \(J_{ii}=0\) since the rings \( \mathcal{C}(\sigma )_{ii}\) are semiprime (and \(J_{ii}^m\subseteq J^m\) for all \(m\ge 1\)). Suppose that \(J\ne 0\). We seek a contradiction. Then \(J_{ij}\ne 0\) for some \(i\ne j\). Then, by the assumption, either \( \mathcal{C}(\sigma )_{ji}J_{ij}\) is a nonzero nilpotent ideal of the algebra \( \mathcal{C}(\sigma )_{jj}\) or \(J_{ij} \mathcal{C}(\sigma )_{ji}\) is a nonzero nilpotent ideal of the algebra \( \mathcal{C}(\sigma )_{ii}\), a contradiction.
\((1\Rightarrow 3)\) The algebras \( \mathcal{C}(\sigma )_{ii}\) are semiprime for all \(i\in \mathrm{Ob}( \mathcal{C})\), by the implication \((1\Rightarrow 2)\). By Lemma 2.7, each ideal I of \( \mathcal{C}(\sigma )\) such that \(e_iIe_i=0\) for all \(i\in \mathrm{Ob}( \mathcal{C})\) is a nilpotent ideal, so it must be zero (since \( \mathcal{C}(\sigma )\) is a semiprime ring).
\((3\Rightarrow 1)\) If I is a nilpotent ideal of \( \mathcal{C}(\sigma )\) then for each \(i\in \mathrm{Ob}( \mathcal{C})\), \(I_{ii}\) is a nilpotent ideals of the semiprime ring \( \mathcal{C}(\sigma )_{ii}\), and so \(I_{ii}=0\). Then, we must have \(I=0\), by the second assumption of statement 3. \(\square \)
Theorem 2.9
(Simplicity criterion for \( \mathcal{C}(\sigma )\)) The algebra \( \mathcal{C}(\sigma )\) is a simple algebra iff the following conditions hold
1.
\(\mathfrak {a}=0\),
2.
for every \(i\in \mathrm{Ob} ( \mathcal{C})\), the ring \( \mathcal{C}(\sigma )_{ii}\) is simple,
3.
for all distinct \(i,j\in \mathrm{Ob} ( \mathcal{C})\), \( \mathcal{C}(\sigma )_{ij}\) is a simple \(( \mathcal{C}(\sigma )_{ii}, \mathcal{C}(\sigma )_{jj})\)-bimodule (in particular, \( \mathcal{C}(\sigma )_{ij}\ne 0\)), and
4.
\( \mathcal{C}(\sigma )_{ij} \mathcal{C}(\sigma )_{jk}\ne 0\) for all \(i,j,k\in \mathrm{Ob} ( \mathcal{C})\).
Proof
\((\Rightarrow )\) Let \( \mathcal{C}_{ij}= \mathcal{C}(\sigma )_{ij}\).\((\Leftarrow )\) Suppose that conditions 1–4 hold. By conditions 1–3, condition 4 can be replaced by condition \(4'\): \( \mathcal{C}_{ij} \mathcal{C}_{jk}= \mathcal{C}_{ik}\) for all \(i,j,k\in \mathrm{Ob} ( \mathcal{C})\). Let J be a nonzero ideal of \( \mathcal{C}(\sigma )\). We have to show that \(J= \mathcal{C}(\sigma )\). By condition 1, \(e_iJe_j\ne 0\) for some i and j. By conditions 2 and 3, \(J_{ij}=J\cap \mathcal{C}_{ij}= \mathcal{C}_{ij}\). By condition \(4'\), \( \mathcal{C}_{st}= \mathcal{C}_{si} \mathcal{C}_{ij} \mathcal{C}_{jt}\subseteq J\) for all s, t. This means that \(J= \mathcal{C}(\sigma )\), as required. \(\square \)
(ii)
For every\(i\in \mathrm{Ob} ( \mathcal{C})\), \( \mathcal{C}_{ii}\)is a simple ring: Suppose that \(\mathfrak {b}\) is a proper ideal of the ring \( \mathcal{C}_{ii}\) then \((\mathfrak {b})\) is a proper ideal of \( \mathcal{C}(\sigma )\) since \((\mathfrak {b}) \cap \mathcal{C}_{ii}= \mathfrak {b}\), a contradiction.
(iii)
For all distinct objects\(i,j\in \mathrm{Ob} ( \mathcal{C})\), \( \mathcal{C}_{ij}\ne 0\): Suppose that \( \mathcal{C}_{ij}=0\) for some distinct objects i and j. Then the ideal \(( \mathcal{C}_{ii})\) of \( \mathcal{C}(\sigma )\) is a proper ideal since \(( \mathcal{C}_{ii})\cap \mathcal{C}_{jj}= \mathcal{C}_{ji} \mathcal{C}_{ii} \mathcal{C}_{ij}=0\), a contradiction.
(iv)
For all distinct objects\(i,j\in \mathrm{Ob} ( \mathcal{C})\), \( \mathcal{C}_{ij}\)is a simple\(( \mathcal{C}_{ii}, \mathcal{C}_{jj})\)-bimodule: Suppose that \(\mathfrak {b}\) is a proper \(( \mathcal{C}_{ii}, \mathcal{C}_{jj})\)-sub-bimodule of \( \mathcal{C}_{ij}\) then \((\mathfrak {b})\) is a proper ideal of the algebra \( \mathcal{C}(\sigma )\) since \((\mathfrak {b}) \cap \mathcal{C}_{ij}=\mathfrak {b}\), a contradiction.
(v)
\( \mathcal{C}_{ij} \mathcal{C}_{jk}\ne 0\)for all objects\(i,j,k\in \mathrm{Ob} ( \mathcal{C})\): The statement (v) holds in the following cases \(i=j=k\) (by (ii)), \(i=j\) or \(j=k\) (by (iii)). Suppose that \(i=k\) and \( \mathcal{C}_{ij} \mathcal{C}_{ji}=0\), we seek a contradiction. Then the ideal \(( \mathcal{C}_{ij})\) of \( \mathcal{C}(\sigma )\) is a proper ideal since \(( \mathcal{C}_{ij})\cap \mathcal{C}_{ii}= \mathcal{C}_{ij} \mathcal{C}_{ji}=0\), a contradiction. Suppose that \( \mathcal{C}_{ij} \mathcal{C}_{jk}=0\) for some distinct i, j and k. Then the ideal \(( \mathcal{C}_{ij})\) of \( \mathcal{C}(\sigma )\) is a proper ideal since \( ( \mathcal{C}_{ij})\cap \mathcal{C}_{kk}= \mathcal{C}_{ki} \mathcal{C}_{ij} \mathcal{C}_{jk}=0\), a contradiction.
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