# Classification of Simple Modules of the Ore Extension $$K[X][Y; f\frac{d}{dX}]$$

• V. V. Bavula
Open Access
Article

## Abstract

For the algebras $$\Lambda$$ in the title of the paper, a classification of simple modules is given, an explicit description of the prime and completely prime spectra is obtained, the global and the Krull dimensions of $$\Lambda$$ are computed.

## Keywords

A skew polynomial ring A prime ideal A completely prime ideal A simple module The global dimension The Krull dimension A normal element

## Mathematics Subject Classification

16D60 13N10 16S32

## 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 algebra
\begin{aligned} \Lambda =K[X]\left[ Y; \delta := f\frac{d}{dX} \right] =\bigoplus _{i\ge 0} K[X]Y^i \end{aligned}
is 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 algebra
\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}
In 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).

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, 6, 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. 1.

The Krull dimension of $$\Lambda$$ is $$\mathrm{Kdim } (\Lambda )=2$$.

2. 2.

The global dimension of $$\Lambda$$ is $$\mathrm{gldim}(\Lambda )=2$$.

3. 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. 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 $$\mathrm{Spec}(\Lambda )=\{ 0 , \Lambda (X-\lambda _i), (X-\lambda _i, Y-\mu )\, | \, i=1, \ldots , s;\; \mu \in K\}$$.

The proof of Theorem 1.1 is given in Sect. 3.

## 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 from
\begin{aligned} fY=Yf-f'f=(Y-f')f,\;\;\mathrm{where}\;\;f'=\frac{df}{dX}, \end{aligned}
that 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 $$\Lambda$$:
\begin{aligned} fu=\omega _f(u)f,\;\; u\in \Lambda ,\\ \omega _f:X\mapsto X,\;\;Y\mapsto Y-f'. \end{aligned}
The algebra $$\Lambda$$ can be identified with a subalgebra of the first Weyl algebra $$A_1$$ by the map
\begin{aligned} \Lambda \rightarrow A_1,\;\;X\mapsto X,\;\; Y\mapsto f\partial . \end{aligned}
(1)
The 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].
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 $$\alpha \in D$$,
\begin{aligned} X\alpha =\sigma (\alpha )X \;\; \mathrm{and}\;\; Y\alpha =\sigma ^{-1}(\alpha )Y,\;\; \ YX=a \;\; \mathrm{and}\;\; XY=\sigma (a). \end{aligned}
The algebra
\begin{aligned} A={\bigoplus }_{n\in \mathbb {Z}}\, A_n \end{aligned}
is a $$\mathbb {Z}$$-graded algebra, where $$A_n=Dv_n,$$$$v_n=X^n\,\, (n>0), \,\,v_n=Y^{-n}\,\, (n<0), \,\,v_0=1.$$
The Weyl algebra $$A_1$$ is a GWA,
\begin{aligned} A_1= D(\sigma ,a=H),\; X\leftrightarrow X,\; \partial \leftrightarrow Y,\; \partial X\leftrightarrow H,\;\;D=K[H], \end{aligned}
with coefficients from a polynomial ring K[H] where $$\sigma \in \mathrm{Aut}_K\,K[H]$$ and $$\sigma :H\rightarrow H-1$$.
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.
\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}
By (1), $$\Lambda _f$$ is a subalgebra of $$A_{1,f}$$ such that
\begin{aligned} \Lambda _f=A_{1,f}. \end{aligned}
(2)
The algebras $$\Lambda$$ and $$A_1$$ can be considered as subalgebras of $$A_{1,f}$$,
\begin{aligned} \Lambda \subseteq A_1\subseteq \Lambda _f=A_{1,f}. \end{aligned}
(3)
The algebra $$A_{1,f}$$ is a simple Noetherian domain with restricted minimum condition.
$$\hat{\Lambda }(f-\mathrm{torsion})$$. 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,
\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 is a classification of simple f-torsion $$\Lambda$$-modules.

### Lemma 2.1

Let $$\lambda _1, \ldots ,\lambda _s$$ be the roots of the polynomial f. Then
\begin{aligned} \hat{\Lambda }(f-\mathrm{torsion})= \{[ \Lambda /\Lambda (X-\lambda _i,Y-\mu )]\;\;|\;\; i=1,\ldots ,s; \; \mu \in K\}. \end{aligned}
All these $$\Lambda$$-modules are 1-dimensional and they are the only simple finite dimensional $$\Lambda$$-modules (by Theorem 2.10).

### 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 algebra
\begin{aligned} \Lambda /(f)=K[X, Y]/(f) \end{aligned}
which is isomorphic to the factor algebra of the polynomial ring K[XY] in two variables at the ideal (f) generated by f, and the equality in the lemma follows.
Let N be a simple $$\Lambda$$-module. Then the map
\begin{aligned} f_N:\; N\rightarrow N, \; n\mapsto fn \end{aligned}
is 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. $$[N]\in \hat{\Lambda }(f-\mathrm{torsion})$$. $$\square$$

$$\hat{\Lambda }(f-\mathrm{torsionfree})$$. 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. 1.
The canonical map
\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}
is a bijection with inverse $$[N]\rightarrow [\mathrm{soc}_\Lambda (N)]$$.

2. 2.
Each simple f-torsionfree $$\Lambda$$-module has the form
\begin{aligned} M_\mathfrak {m}:= \Lambda /\Lambda \cap \mathfrak {m}\end{aligned}
(6)
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. $$\square$$

### 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 $$y=\partial +A_1(X-\lambda _i)$$ and
\begin{aligned} \partial y^j=y^{j+1}, \;\; Xy^j=\lambda _iy^j+\cdots \;\; \mathrm{for}\;\; j\ge 0 \end{aligned}
where by three dots we denote the sum of elements of smaller degree in the variable y. Thus the linear operator
\begin{aligned} X-\mu : M_i\rightarrow M_i, \;\; m\mapsto (X-\mu )m, \;\;m\in M_i, \end{aligned}
is 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.

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 map
\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}
is bijective.

### 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 ring
\begin{aligned} B=K(H)[X, X^{-1}; \sigma ], \;\; \sigma (H)=H-1, \end{aligned}
with coefficients from the field K(H) of rational functions. The algebra B is a right and left Euclidean domain with respect to the ‘length’ function
\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}
hence, it is a right and left principal ideal domain.
We have
\begin{aligned} \hat{A}_1=\hat{A}_1 (D-\mathrm{torsion})\coprod \hat{A}_1 (D-\mathrm{torsionfree}) \end{aligned}
where a simple $$A_1$$-module M belongs to the first (resp., second) set if $$S^{-1}M=0$$ (resp., $$S^{-1}M\ne 0$$).

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 map
\begin{aligned} K/\sim \;\rightarrow \;\hat{A}_1 (D-\mathrm{torsion}), \;\; [\Gamma ]\mapsto [L(\Gamma )], \end{aligned}
is a bijection, where
1. 1.

If $$\Gamma =\mathcal{O}(\lambda )\ne \mathbb {Z}$$, then $$L(\Gamma )= A_1/A_1(H-\lambda )$$.

2. 2.

If $$\Gamma =(-\infty ,0]$$, then $$L(\Gamma )= A_1/A_1X$$.

3. 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. 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. 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 $$\{ y^i=\partial ^i+A_1X, \; i\ge 0\}$$, and
\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}
for 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, $$M=M_f$$.
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.
\begin{aligned} M=L(\Gamma ), \;\; \Gamma =\mathcal{O}(\lambda )\ne \mathbb {Z}\;\; (\mathrm{resp}., \;\; \Gamma =[1, \infty )). \end{aligned}
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 $$\Lambda$$-submodules
\begin{aligned} M\supset fM \supset \cdots \supset f^nM\supset \cdots \;\;\mathrm{with}\;\; \bigcap _{n\ge 0}\; f^nM=0. \end{aligned}
Suppose 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. $$\square$$

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. 1.

The A-module $$M_\mathfrak {m}:=A/A\cap \mathfrak {m}$$ is simple.

2. 2.

The socle $$\mathrm{soc}_A(M_\mathfrak {m})\ne 0$$.

3. 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 to
\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}
where $${}^{\omega _s^{-1} }M_{\mathfrak {m}}$$ 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 $$M_{\mathfrak {m}}$$.

$$(3\Rightarrow 1)$$ 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. Then
\begin{aligned} \mathcal{M}_b:=A_1/A_1\cap Bb \end{aligned}
is 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 $$\mathcal{M}_b$$. $$\square$$
Set
\begin{aligned} B_f:=S_f^{-1}B=A_{1,f}\otimes _\Lambda \,B=\Lambda _f\otimes _\Lambda \,B \end{aligned}
for 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 $$b\in B$$, $$(Bb)_f=B_fb$$.

Theorem 2.10 is a classification of simple f-torsionfree $$\Lambda$$-modules.

### 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 that
1. 1.

$$\Lambda =\Lambda f+\Lambda \cap B_fb \; (= \Lambda f+\Lambda \cap Bb )$$, and

2. 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.

Then
\begin{aligned} \mathcal {M}_b:=\Lambda /\Lambda \cap Bb\; \, (=\Lambda /\Lambda \cap B_fb) \end{aligned}
is 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.

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,
\begin{aligned}{}[M]\in \hat{\Lambda }(f-\mathrm{torsionfree}) \Leftrightarrow [M_f]\in \hat{A}_{1,f}(\Lambda -\mathrm{socle}) \end{aligned}
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).
In the second case, by Theorems 2.4 and 2.9,
\begin{aligned} M_f\simeq (\mathcal{M}_b)_f=A_{1,f}/A_{1,f}\cap B_fb \end{aligned}
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,
\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)
By Theorem 2.8 and by the Remark after it,
\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}
In this case,
\begin{aligned} \mathrm{soc}_\Lambda (M_f)=\Lambda /\Lambda \cap (A_{1,f}\cap B_fb)=\Lambda /\Lambda \cap B_fb. \end{aligned}
Let us show that (in this case) the natural $$\Lambda$$-module epimorphism
\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}
is an isomorphism. Note that
\begin{aligned} \mathrm{ker}\, \varphi =\Lambda \cap B_fb/\Lambda \cap Bb. \end{aligned}
The $$A_1$$-module $$\mathcal{M}_b$$ is a submodule of $$(\mathcal{M}_b)_f\simeq M_f$$. So,
\begin{aligned} \mathrm{soc}_\Lambda (M_f)=\mathrm{soc}_\Lambda (\mathcal{M}_b)=\mathrm{soc}_\Lambda (\Lambda /\Lambda \cap Bb). \end{aligned}
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.
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. Since
\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}
by 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.

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 $$\Lambda$$ since
\begin{aligned} Yp_i=p_i(Y-p_i^{-1}f)\; \; \mathrm{and}\;\; Xp_i=p_iX. \end{aligned}
4. The algebra $$\Lambda$$ is a domain, hence $$0\in \mathrm{Spec}_c (\Lambda )$$.
Since
\begin{aligned} \Lambda / \Lambda p_i\simeq F_i[Y] \end{aligned}
(8)
is 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 $$\Lambda$$.

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 $$\Lambda$$,
\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}
by [11, Theorem 6. 5.9]. Therefore, $$\mathrm{Kdim } (\Lambda )=2$$.

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)],
\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}
Therefore, $$\mathrm{gldim}(\Lambda )=2$$. $$\square$$

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