Skew Polynomial Rings: the Schreier Technique

Abstract

Schreier bases are introduced and used to show that skew polynomial rings are free ideal rings, i.e., rings whose one-sided ideals are free of unique rank, as well as to compute a rank of one-sided ideals together with a description of corresponding bases. The latter fact, a so-called Schreier-Lewin formula (Lewin Trans. Am. Math. Soc. 145, 455–465 1969), is a basic tool determining a module type of perfect localizations which reveal a close connection between classical Leavitt algebras, skew polynomial rings, and free associative algebras.

1 Introduction

In contrast to what one might expect, skew polynomial rings form a quite large, extremely diverse ring class. Elements of the (left) skew polynomial ring D[X;α, δ] over a division ring D with a nonzero endomorphism $\alpha :D\rightarrow D$ and an α-derivation $\delta :D\rightarrow D$, i.e., an additive homomorphism satisfying δ(cd) = δ(c)d + α(c)δ(d) for any two c, d ∈ D, are formal polynomials a₀ + a₁X + ⋯ + a_nXⁿ with the usual addition, but with multiplication twisted by putting Xa = α(a)X + δ(a), respectively. D[X;α,0] is denoted by D[X;α]. Typical examples for endomorphisms α are provided by Lüroth’s theorem about endomorphisms of rational function fields. D[X;α, δ] is a left but not right principal ideal domain unless α is an automorphism. Almost nothing is known about right ideals of skew polynomial rings except an obvious consequence of Cohn’s dependence relations [3, Theorem 1.1.1] that finitely generated right ideals are free of unique rank. Because of its importance we present a simple direct and elementary proof that one-sided principal ideal domains are semifirs. A ring is a semifir if every finitely generated one-sided ideal is free of unique rank. A ring is a fir, i.e., a free ideal ring if every one-sided ideal is a free module of unique rank. Consequently, a fir is projective-free, i.e., all projective modules are free. Following Rosenmann and Rosset [13], Schreier bases are introduced and used to show that skew polynomial rings are firs with a Schreier-Lewin formula [11] determining both rank and corresponding bases of one-sided ideals. The latter formula is a basic tool in the study of perfect localizations of skew polynomial rings which is inspired by the advanced theory of Leavitt algebras [6, 10] and [1]. In particular, we compute precisely both the module type and the Grothendick group of perfect localizations of skew polynomial rings. Furthermore, we describe also their maximal flat epimorphic ring of quotients as the ring of quotients with respect to the Gabriel topology defined by finite codimensional right ideals.

The title is suggested by Lewin’s nice article [11]. All rings have the identity 1 ≠ 0, and modules are unitary. A division ring is a ring whose nonzero elements are units. For undefined notions we refer to Stenström’s classic [14]. For a systematic investigation of skew polynomial rings, we refer to the classics [3, 5], or [7, 8] of either Cohn or Jacobson, respectively.

2 Skew Polynomial Rings are Free Ideal Rings

As a first step and for the sake of completeness, an obvious consequence of Cohn’s theory on dependence relations that one-sided principal ideal domains are semifirs is presented with a short elementary proof. However, it seems to be open whether one-sided principal ideal domains are firs.

Proposition 2.1

Every one-sided principal ideal domain is a semifir.

Proof

Let A be a left principal ideal domain. Then, A is a ring having IBN, whence every free A-module has a unique rank. Therefore, it suffices to verify by induction on the number of generators that every finitely generated right ideal is free. Since A is a domain, every cyclic right ideal is free. Assume that all right ideals generated by at most n − 1 (n > 1) generators are free and consider a nonzero right ideal $R={\sum }_{i=1}^{n} a_{i}A$. It is enough to assume that $a_{1}, \dots , a_{n}$ are not right linearly independent over A, that is, there are b_i ∈ A not all equal to 0 with ${\sum }_{i=1}^{n} a_{i}b_{i}=0$. Then, there is 0 ≠ b ∈ A with $b={\sum }_{i=1}^{n} c_{ni}b_{i}, b_{i}=d_{i}b$ for appropriate c_ni, d_i ∈ A. Hence, ${\sum }_{i=1}^{n} c_{ni}d_{i}=1$ and ${\sum }_{i=1}^{n} a_{i}d_{i}=0$ hold. Therefore, $f_{n}=c_{n1}e_{1}+{\dots } +c_{nn}e_{n} \in {{~}_{A}A}^{n}$ maps to 1 ∈ A under the left module homomorphism $\phi : A^{n}\rightarrow A: e_{i}\mapsto d_{i}=\phi (e_{i})\in A$, where $\{e_{i} | i=1, \dots , n\}$ is a basis of _AAⁿ. Consequently, $_{A}A^{n}=Af_{n}\oplus \ker \phi$ holds. This shows that $\ker \phi$ is a projective left A-module. Consequently, $\ker \phi$ is free of rank n − 1 because A is a left principal ideal domain. If $\{f_{i}={\sum }_{j=1}^{n} c_{ij}e_{j} | i=1, \dots , n-1\}$ is a basis of $\ker \phi$, then the square matrix (c_ij) is invertible with the inverse matrix (v_ij). For an arbitrary column vector $(r_{1}, \dots , r_{n})^{t}\in {A^{n}_{A}}$ the equality

$$ \begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{n} a_ir_i & = & (a_1, \cdots, a_n)(r_1, {\cdots}, r_n)^t=\{(a_1, \cdots, a_n)(v_{ij})\}\{(c_{ij})(r_1, \cdots, r_n)^t\}\\ &=& (a^{\prime}_1, \cdots, a^{\prime}_n)(r^{\prime}_1, \cdots, r^{\prime}_n)^t= \sum\limits_{i=1}^{n} a^{\prime}_ir^{\prime}_i \end{array} $$

with $a^{\prime }_{j}={\sum }_{i=1}^{n} a_{i}v_{ij}$ and $r^{\prime }_{j}={\sum }_{i=1}^{n} c_{ji}r_{i}$ for all $j=1, \dots , n$ ensures that R can be generated by $a^{\prime }_{1}, \dots , a^{\prime }_{n}$, i.e., $R={\sum }_{i=1}^{n} a_{i}A={\sum }_{i=1}^{n} a^{\prime }_{i}A$. In particular, one has

$$0=\sum\limits_{i=1}^{n} a_{i}d_{i}=\sum\limits_{i=1}^{n} a^{\prime}_{i}d^{\prime}_{i}$$

with $d^{\prime }_{j}={\sum }_{i=1}^{n} c_{ji}d_{i}$. The equality $d^{\prime }_{n}={\sum }_{i=1}^{n} c_{ni}d_{i}=1$ implies $a^{\prime }_{n}\in {\sum }_{i=1}^{n-1}a^{\prime }_{i}A$ and so R can be generated by n − 1 elements, whence R is free by the induction hypothesis. □

Remark 2.2

The above argument can be used to simplify the demonstration of (b) ⇒ (c) ⇒ (a) in [5, Theorem 1.6.1] characterizing n-firs as algorithmically computable. Moreover, the proof of Proposition 2.1 shows also that one-sided Bezout domains are semifirs. Even more, it is suitable to verify with an obvious modification that a domain whose finitely generated left ideals are free of unique rank, is a semifir. In particular, the proof of Proposition 2.1 can be used to show in a direct and elementary way the symmetry of n-fir, i.e., if every one-sided ideal, say, every left ideal generated by at most n-elements is free of unique rank, then every right ideal generated by at most n elements is also a free right module of unique rank. Note the fact that a unique rank of right free modules generated by at most n elements becomes immediate by applying the duality between finitely generated free left and right modules induced by functors Hom_R(−;R)

From now on, D always denotes a division ring together with a non-zero, proper endomorphism $\alpha :D\rightarrow D$ and an α-derivation δ. Fix a basis B₁ = {b_i|i ∈ I}∋ 1 of the right vector space D over α(D) where I is not necessarily a finite set. Then αⁿ(B₁) = {αⁿ(b_i)|i ∈ I} = B_n+ 1 is a basis of the right vector space $\alpha ^{n}(D)_{\alpha ^{n+1}(D)}$. Since a skew polynomial ring A = D[X;α, δ] is obviously a right D-module, we will first construct a basis of A_D in terms of B₁ = {b_i|i ∈ I} of D_α(D). Put y_i = b_iX for all i ∈ I. For each $d={\sum }_{i} b_{i}\alpha (d_{i})\in D$ the multiplication rule

$$ \begin{array}{@{}rcl@{}} Xd=\alpha(d)X+\delta(d) \end{array} $$

(1)

implies

$$ \begin{array}{@{}rcl@{}} dX={\sum}_{i} b_{i}\alpha(d_{i})X\equiv {\sum}_{i} y_{i}d_{i} \quad \mod D \end{array} $$

(2)

whence {y_i|i ∈ I} is a right linearly independent set over D and {1, y_i|i ∈ I} is a basis of the right D-module D + DX. For the sake of simplicity, 1 is considered as the monomial of length 0, and a product $p=y_{i_{1}}{\cdots } y_{i_{n}} (i_{1}, \dots , i_{n} \in I)$ has a length |p| = n. A product $y_{i_{1}}{\cdots } y_{i_{m}}$ of length m ≤ n is called by definition an initial segment of length m of p, and is denoted by h_p(m)(h_p(0) = 1) while a remainder $y_{i_{m+1}}{\cdots } y_{i_{n}}$ is a tail t_p(m)(t_p(n) = t_p(|p|) = 1) of colength m of p. The obvious iteration using (2) shows that for each l > 0, monomials $y_{i_{j}}{\cdots } y_{i_{k}}$ of length at most l form a basis of D-D-bimodule D + DX + ⋯ + DX^l as a right D-module and

$$ \begin{array}{@{}rcl@{}} DX^{l}\equiv \sum\limits y_{i_{1}}{\cdots} y_{i_{l}}D=\sum\limits_{|p|=l} pD \quad \mod D+DX+{\cdots} +DX^{l-1}. \end{array} $$

(3)

Note an important fact that the 1-dimensional left D-module DX is not a right D-module unless δ = 0. In this case of δ = 0, all $DX^{n} (n\in \mathbb N)$ are also right D-modules. In any case, monomials in y_i form a basis B of A_D, called a standard Schreier basis with respect to a basisB₁ of D_α(D), shortly, a standard Schreier basis for A_D, and every elements of A can be written uniquely as a right linear combination of monomials in the y_i with coefficients from D on the right.

Schreier bases introduced by Rosenmann and Rosset [13] can be extended to skew polynomial rings as follows.

Definition 2.3

A Schreier basis for a right ideal R of a skew polynomial ring A = D[X;α, δ] with respect to a standard Schreier basis B of all monomials in y_i = b_iX(i ∈ I), where B₁ = {b_i|i ∈ I}∋ 1 is a basis of D_α(D), is a subset $B=B_{R}\subseteq \textbf {B}$ that spans a right vector space V = V_R over D that is complementary to R (that is, A = V + R, V ∩ R = 0), and is closed to taking initial segments. For each $n\in \mathbb N$, let V_n be a right vector space generated by elements of B having length at most n. A Schreier basis B is called a strong Schreier basis for R if every monomial in y_i of length n lies in V_n + R.

The argument of Rosenmann and Rosset [13, (3.2) Lemma] is used to show the existence of Schreier bases.

Proposition 2.4

There exists a Schreier basis B_R for any right ideal R of A = D[X;α, δ].

Proof

The case R = A is trivial because B_A is just the empty set. Therefore, one can assume without loss of generality that R is a proper right ideal, i.e., 1∉R. A Schreier basis B = B_R is constructed inductively by putting firstly ₀B = {1}. If D + R = A, let ₁B = ₀B. If D + R ≠ A, then DX is not a subset of D + R. Namely, $DX\subseteq D+R$ would imply $DX^{2}\subseteq DX+RX\subseteq D+R$ and hence all $DX^{n}\subseteq D+R$ hold whence A = D + R, a contradiction. Consequently, if D + R ≠ A, then let $_{1}B^{\prime }$ be a maximal subset of {y_i = 1y_i|i ∈ I} such that $_{1}B^{\prime }$ is right linearly independent modulo D + R over D. Put ${{~}_{1}B} = {{~}_{0}B} \cup {{~}_{1}B}^{\prime }$ and V₁ a right D-module spanned by D and ₁B. Then $D+DX\subseteq V_{1}+R$ holds. Therefore, one has the equality D + DX + R = V₁ + R. Assuming now that $_{n}B^{\prime }, _{n}B$ and V_n(n > 0) have been already constructed such that D + DX + ⋯ + DXⁿ + R = V_n + R. If V_n + R = A, let _n+ 1B = _nB. If V_n + R ≠ A, then DX^n+ 1 is not a subset of V_n + R because as above one can verify easily that $DX^{n+1}\subseteq V_{n}+R$ would imply V_n + R = A, a contradiction. Consequently, in case V_n + R ≠ A, let $_{n+1}B^{\prime }$ be a maximal subset

$$\{by_{i} | b\in {{~}_{n}B}^{\prime}\}$$

which is a right linearly independent set modulo V_n + R over D. Then, define $_{n+1}B={{~}_{n}B} \cup _{n+1}B^{\prime }$ and let V_n+ 1 be a right D-module spanned by _n+ 1B. Hence, $DX^{n+1}\subseteq V_{n+1}+R$ holds. If $_{n+1}B^{\prime }=\emptyset$, the process stops at this step and define B_R = _nB. If the process does not stop after finitely many steps define

$$ B = B_{R} = \bigcup\limits_{n=0}^{\infty} {{~}_{n}B} .$$

B is clearly a strong Schreier basis of R, completing the proof. □

Schreier’s technique [11] is now suitable for showing that skew polynomial rings are firs.

Let $\pi :A=V\oplus R\rightarrow V$ be the canonical projection of A along R onto the right D-module V spanned by the Schreier basis B of R constructed above. Then, for every element a ∈ A, one has a − π(a) ∈ R whence as − π(a)s ∈ R holds for every s ∈ A. In particular, the equality

$$ \begin{array}{@{}rcl@{}} \pi(as)=\pi(\pi(a)s), \quad \forall a, s \in A \end{array} $$

(4)

holds. Consequently, for every element b ∈ B and y_i (i ∈ I), the element by_i is either contained in B whence π(by_i) = by_i and so by_i − π(by_i) = 0, or not contained in B. In that case, by the construction of B, or equivalently, by the definition of a strong Schreier basis, 0 ≠ by_i − π(by_i) ∈ R holds. Therefore, for every monomial $y=y_{i_{1}}{\cdots } y_{i_{l}}$ of length l ≥ 0 in B, the associated element

$$ \begin{array}{@{}rcl@{}} u_{y,i}=yy_{i}-\pi(yy_{i}), \quad \forall y\in B ~ \& ~ i\in I \end{array} $$

(5)

is either 0 or a nonzero element of R. We are now ready to reach the main goal of this section.

Theorem 2.5

If δ is an α-derivation of a division ring with respect to a nonzero, proper endomorphism $\alpha :D\rightarrow D$, then A = D[X;α, δ] is a free ideal ring.

Proof

We have to show that every right ideal R of A is a free A-module. Since the case R = A or R = 0 is obvious, one can assume without loss of generality that R is a proper right ideal. For an arbitrary but fixed basis B₁ = {b_i|i ∈ I}∋ 1 of D_α(D) define elements (“variables”) y_i = b_iX (i ∈ I). Moreover, let B = B_R be a strong Schreier basis of R with respect to the standard Schreier basis B of A_D consisting of all monomials in y_i (i ∈ I). We show first that the nonzero elements u_{y, i} = yy_i − π(yy_i) (y ∈ B, i ∈ I) generate R freely. For an arbitrary monomial y and y_i = b_iX (i ∈ I), we have by (4)

$$\pi(y)y_{i}-\pi(yy_{i})=\pi(y)y_{i}-\pi(\pi(y)y_{i}).$$

Let $\pi (y)={\sum }_{j=1}^{l} z_{j}d_{j}$ for some monomials z_j ∈ B_R and d_j ∈ D. Write $d_{j}y_{i}=({\sum }_{k} y_{k}{{~}^{ij}c}_{k})+d_{ij}$ for appropriate ^ijc_k, d_ij ∈ D. Then we have, by putting $v={\sum }_{j=1}^{l} z_{j}d_{ij}\in V$ where V is a right D-module spanned by B_R, the following equality

$$\pi(y)y_{i}=\sum\limits_{j=1}^{l} z_{j}(d_{j}y_{i})=\sum\limits_{j=1}^{l} z_{j}\left( d_{ij}+\sum\limits_{k} y_{k}{{~}^{ij}c}_{k}\right)=\sum\limits_{j, k} z_{j}y_{k}{{~}^{ij}c}_{k}+v$$

whence

$$ \begin{array}{@{}rcl@{}} \pi(y)y_{i}-\pi(yy_{i})= \sum\limits_{j, k} (z_{j}y_{k}-\pi(z_{j}y_{k})){{~}^{ij}c}_{k}=\sum\limits_{j, k} u_{z_{j}, k}{{~}^{ij}c}_{k}\in \sum\limits_{y\in B \& i\in I} u_{y,i}A. \end{array} $$

(6)

The canonical projection 1 − π of A on R via the decomposition A = V ⊕ R sends every monomial $y=y_{i_{1}}{\cdots } y_{i_{n}} ~(n>0)$ to

$$ \begin{array}{@{}rcl@{}} (1-\pi)(y)=y-\pi(y)=\sum\limits_{j=0}^{n-1} \left\{ \pi(h_{y}(j))y_{i_{j+1}}-\pi(h_{y}(j+1))\right\} t_{y}(j+1) \end{array} $$

(7)

and so by formulas (6), (7) together with y − π(y), the image a − π(a) of an arbitrary element a ∈ A is contained in ${\sum }_{y\in B \& i\in I} u_{y,i}A$. Consequently, u_{y, i} (y ∈ B; i ∈ I) generate R.

Therefore, it remains to show that the set {u_{y, i} ≠  0|y ∈ B; i ∈ I} is linearly independent over A. Instead of using the lengthy opaque argument of Lewin [11] we use the nice argument of [13, p. 363]. Consider finitely many nonzero elements $u_{z_{j}, i}$ defined by z_j ∈ B_R and y_i = b_iX, i ∈ I. Then, z_jy_i∉B and $u_{z_{j}, i}=z_{j}y_{i}-\pi (z_{j}y_{i})\neq 0$ hold. By definition

$$ \begin{array}{@{}rcl@{}} \pi(z_{j}y_{i})=\sum\limits_{k}{{~}^{ij}m_{k}}{{~}^{ij}d}_{k} \quad \left( 0\neq~ {{~}^{ij}d}_{k}\in D ~ \& ~^{ij}m_{k}\in B_{R},~ |^{ij}m_{k}|\leq |z_{j}|+1\right). \end{array} $$

(8)

Assume indirectly that these $u_{z_{j}, i}$’s are right linearly dependent over A, i.e., there are elements a_ij ∈ A not all 0 such that ${\sum }_{j} u_{z_{j}, i} a_{ij}=0$. If yd(d ∈ D) is a monomial of longest length among monomials appearing in the a_ij-s, say, it is a monomial of $a_{\underline {i} \underline { j}}$, then in view of the equality (8) the term $z_{\underline {j}}y_{\underline {i}}yd$, cannot cancel. This is an immediate result of the following observations. Because of its maximal length $z_{\underline {j}}y_{\underline {i}}$ is not an initial segment of any elements z_jy_i satisfying $(j, i)\neq (\underline {j}, \underline {i})$, nor of any of ^ijm_k, i.e., of monomials appearing in π(z_jy_i) (without any exception, even including the case $(j, i)=(\underline {j}, \underline {i}))$ in view of the fact that $z_{\underline {j}}y_{\underline {i}}$ is not contained in B_R. This contradiction finishes the proof. □

Theorem 2.5 implies, in view of the structure of projective modules over hereditary rings, that projective right modules over D[X;α, δ] are free. However, one can extend in an obvious manner the notion of (strong) Schreier basis to free right modules, and by the same proof one gets directly that projective right modules over D[X;α, δ] are free as

Corollary 2.6

Let M be a submodule of a free right module F on a set {t_j|j ∈ J} over D[X;α, δ] where $\textup {\textbf {B}}=\{y_{i_{1}}, \dots , y_{i_{n}} | i_{j}\in I, n\geq 0\}$ is a standard Schreier basis of F given by a basis B₁ = {b_i|i ∈ I} of D_α(D) and y_i = b_iX (i ∈ I). Then $\textup {\textbf {B}}_{F}=\bigcup _{j\in J} t_{j}\textup {\textbf {B}}$ is called astandard Schreier basis of F with respect to B. Let B_M be a strong Schreier basis of M constructed in the same manner as in Proposition 2.4, then F admits a decomposition F = V_M ⊕ M, where V_M is the right D-module spanned by B_M. Let π_M be the canonical projection of F onto V_M along M and put u_j = t_j − π_M(t_j),_ju_{b, i} = t_jby_i − π_M(t_jby_i) for j ∈ J, i ∈ I; b ∈B. Then, the nonzero elements $u_{j}, {{~}_{j}u}_{b, y_{i}}$ form a basis of M over D[X;α, δ].

We can now easily deduce the Schreier-Lewin formula [11] for the rank of submodules with a finite strong Schreier basis of finitely generated free modules over D[X;α, δ].

Corollary 2.7

Let F be a free right module of a finite rank l over D[x;α, δ] such that D has a finite dimension n over α(D), and M a submodule admitting a finite strong Schreier basis of $m\in \mathbb N$ elements, i.e., the factor A-module F/M has dimension m as a right D-module. Then nm − m + l is the rank of M.

Proof

For simplicity, we present a proof for the case l = 1 which works word-for-word for the arbitrary case. Consider a right ideal R of D[X;α, δ]. In the notation of preceding results there are exactly mn symbols u_{b, i} (b ∈ B_R, i ∈ I;|I| = n). Therefore, by Theorem 2.5 R is a free right A-module of rank mn − k where k is the number of u_{b, i} equal to 0. But this number is exactly m − 1 by the construction of B_R and the definition of π and u_{b, i}, completing the proof. □

Remark 2.8

In view of Lewin’s footnotes [11] that Cohn was able to obtain results in [11] from a general theory of the weak algorithm and the Hilbert series of filtered rings (cf. [3] and [4]), it seems that results of this section can be obtained by Cohn’s method, too. Our treatment is, however, elementary. On one hand, it provides evidence for the broad applicability of Cohn’s beautifully deep theory; and on the other hand, it supplies another particular, preliminary, introductory example different than the ordinary, classical example of free associative algebras for Cohn’s theory. Furthermore, it is worth keeping in mind that by its flexibility, Schreier-Lewin techniques can be applicable to not necessarily unital rings with zero divisors, for example, see [1].

3 Localizations of Skew Polynomial Rings

Inspired by the important work of Rosenmann and Rosset [13], we now turn to the study of perfect bimorphic localizations of skew polynomial rings because of their striking similarity to free associative algebras. In particular, Rosenmann and Rosset [13] proposed a nice method to compute module types of certain rings which do not have IBN. Roughly speaking, their idea is to find a subring A such that A has IBN, and then to reduce the problem to one on (unique) rank of free modules over A which is usually a projective-free ring, i.e., a ring that admits only free modules as projective modules. Recall that a module type of a ring A without IBN is a pair $(m, n)\in {\mathbb N}^{2} ~(m<n)$ such that m, n are the first two smallest integers satisfying ${A^{m}_{A}}\cong {A^{n}_{A}}$. In particular, it is easy to find rings of module type (1,2), i.e., rings over which all finitely generated free modules are isomorphic. But it is generally quite hard to show that certain free modules are not isomorphic. As an illustration, we include the following obvious folkloric result, with proof to aid the reader.

Proposition 3.1

If $\alpha :D\rightarrow D$ is a non-zero, non-surjective endomorphism, then the right maximal quotient ring of D[X;α, δ] has the module type (1,2). More generally, if a ring A has a dense right ideal R which is an infinite direct sum of isomorphic modules, then the module type of the maximal right quotient ring $Q_{\max \limits }(A)$ is (1,2).

Proof

The assumption on α implies that D[X;α, δ] has a dense right ideal which is free of infinite rank. In particular, if a right ideal R of a ring A is an infinite direct sum of isomorphic modules, then we have an isomorphism R_R≅Rⁿ for every $n\in \mathbb N$. Consequently, this direct decomposition ensures elements $a_{i}, a^{*}_{i}\in Q_{\max \limits }(A),~ i=1, \dots , n$ satisfying ${\sum }_{i=1}^{n} a_{i}a^{*}_{i}=1$ and $a^{*}_{i}a_{j}={\delta }^{i}_{j}$, where ${\delta }^{i}_{j}$ is the usual Kronecker symbol. Hence, all finitely generated free modules over $Q_{\max \limits }(A)$ are isomorphic, completing the proof. □

For the goals of this section recall that a ring homomorphism $\phi :A\rightarrow B$ is called an epimorphism if for any ring C and ring homomorphisms $\alpha , \beta : B\rightarrow C, \alpha \phi =\beta \phi$ implies α = β. Dually one gets a notion of a monomorphism. Epimorphisms are not necessarily surjective, for example $\mathbb Q$ is an epimorphic overring of $\mathbb Z$. However, monomorphisms are always injective. Namely, if there is 0 ≠ a ∈ A with ϕ(a) = 0, then $\alpha , \beta :\mathbb Z[X]\longrightarrow A$ defined by putting α(X) = 0 and β(X) = a, respectively, are two ring homomorphisms from Z[X] to A satisfying ϕα = ϕβ but α ≠ β. Two rings are bimorphic if there is a ring homomorphism between them which is both a monomorphism and an epimorphism. For example, the ring $\mathbb Z$ of integers and the field $\mathbb Q$ of rationals are bimorphic. An epimorphism $\phi :A\rightarrow B$ is flat if _AB is a left flat A-module. In this case B is called a perfect right localization or a flat epimorphic right ring of quotients of A. Flat epimorphisms form a non-trivial intersection of Cohn’s localization via epimorphisms and Gabriel’s theory of localization via both Gabriel topology and hereditary torsion theory. For a basic theory of perfect localizations we refer to classic books [14] and [12] by Stenström and Popescu, respectively.

If ϕ : A→Q is a flat bimorphism, then Q becomes a subring of the maximal ring $Q_{\max \limits }(A)$ of right quotients of A by identifying each a ∈ A with ϕ(a) ∈ Q. It is well-known that $Q_{\max \limits }(A)$ has the largest subring Q_tot(A), called the maximal flat epimorphic ring of right quotients of A which is a flat bimorphism of A and contains all flat bimorphisms of A. It is also well-known, and in fact, not hard to see that every element q of $Q_{\max \limits }(A)$ uniquely determines a largest right ideal dom(q) = {a ∈ A|qa ∈ A} of A, called a maximal right ideal of definition of q and q can be identified with an A-homomorphism q : dom(q)→A : a↦qa = q(a). This observation greatly simplifies notations when working inside maximal rings of quotients.

From now on, we assume that α is an endomorphism of a division ring D such that D_α(D) is of finite dimension n > 1. We shall use notations fixed in Section 2. Therefore, $B_{1}=\{b_{1}, \dots , b_{n}\}$ is a basis of D_α(D) and y_i = b_iX. Since D is a subring of A = D[X;α, δ] every right A-module is also a right vector space over D. We’ll start with some preparatory results. The first assertion is an adaption of [13, (3.3) Theorem].

Proposition 3.2

A right ideal R of A = D[X;α, δ] is of finite codimension over D, i.e., a factor module A/R is a finite dimensional D-module, if and only if R is essential and finitely generated. In particular, a nonzero two-sided ideal of A has a finite codimension over D if and only if it is finitely generated.

Proof

Since every cyclic right ideal is an infinite dimensional D-module, the necessity is obvious in view of Theorem 2.5. Conversely, assume indirectly that there is a finitely generated essential right ideal R which is of infinite (right) codimension over D. If B is a strong Schreier basis of R (constructed as in Proposition 2.4), then $\{u_{b, i}=by_{i}-\pi (by_{i})\neq 0 | b\in B, i=1, \dots , n\}$ is a basis for R as a free module over A. Since A has IBN, and by the assumption that R is finitely generated, then the above set must be finite. In particular, there exists an integer m > 0 such that for every b ∈ B = B_R satisfying |b|≥ m all monomials $by_{i}~ (i=1, \dots , n)$ are also in B. Consequently, all monomials of length > r whose initial segments of length r belong to B, are also contained in B. Hence, for a monomial b ∈ B of length r the right ideal bA meets R trivially, contradicting the essentiality of R. This shows that B = B_R is finite. The last claim follows trivially from the observation that a nonzero two-sided ideal of a domain is always essential. □

We reach now the first goal of this section as

Theorem 3.3

Let I be an ideal of right codimension 1 of A = D[X;α, δ] over D, where α is an endomorphism of a division ring D such that D_α(D) has a dimension n > 1. Then ideal powers $I^{l}~ (l\in \mathbb N)$ form a perfect Gabriel topology $\frak I$ and (1, n) is the module type of the associated ring $A_{\frak I}$ of quotients. Moreover, projective modules over $A_{\frak I}$ are free and the Grothendieck group $K_{0}(A_{\frak I})$ is cyclic of order n − 1.

Proof

By Theorem 2.5 I is a free right A-module of rank n. Consequently, I^l is a free right A-module of rank n^l. This implies by [14, Propositions XI.3.3 and XI.3.4 (d)] that the ideal topology defined by powers $I^{l} (l\in \mathbb N)$ is a perfect topology. Therefore, the canonical embedding $D[X; \alpha , \delta ]=A\longrightarrow A_{\frak I}$ is a flat bimorphism whence $A_{\frak I}$ is a subring of both the maximal flat epimorphic ring and the maximal ring of right quotients $Q_{\text {tot}}(A)\subseteq Q_{\max \limits }(A)$. Consequently, by [14, Corollary XI.3.8] an equality $(R\cap A)A_{\frak I}=R$ holds for every right ideal R of $A_{\frak I}$. Therefore, R is a free $A_{\frak I}$-module because R ∩ A is a free A-module by Theorem 2.5 and $_{A}A_{\frak I}$ is a flat left A-module. Hence, projective right modules over $A_{\frak I}$ are free.

To compute the module type of $A_{\frak I}$, write $I={\sum }_{i=1}^{n} z_{i}A=\oplus _{i=1}^{n}z_{i}A$ in view of Theorem 2.5. Let $z^{*}_{j}$ be the canonical j th projection $z^{*}_{j}:I\rightarrow A: {\sum }_{i=1}^{n} z_{i}a_{i}\mapsto a_{j}$ for each $j=1, \dots , n$. The obvious equalities

$$ \begin{array}{@{}rcl@{}} z^{*}_{j}z_{i}=\begin{cases} 1& \text{if } j=i,\\ 0& \text{if } j\neq i\end{cases}\quad \& \quad \sum\limits_{i=1}^{n} z_{i}z^{*}_{i}=1 \end{array} $$

(9)

in $A_{\frak I}$ imply that $A^{n}_{\frak I}\cong A_{\frak I}$ holds. Therefore, to verify that (1, n) is the module type of $A_{\frak I}$, it is enough to see m ≡ 1 mod n − 1 provided $A_{\frak I}\cong A^{m}_{\frak I}$. In fact, assume $A_{\frak I}={\sum }_{i=1}^{m} c_{i}A_{\frak I}=\oplus _{i=1}^{i=m} c_{i}A_{\frak I}$ such that the annihilator of each $c_{i}\in A_{\frak I}$ is trivial. Consequently, multiplication by c_i on the left is injective on $A_{\frak I}$. Therefore, the right ideals $\text {dom}(c_{i})\cong c_{i}\text {dom}(c_{i})~ (i=1, \dots , m)$ of A are free A-modules of rank ≡ 1 mod (n − 1) by Schreier-Lewin formula established in Corollary 2.7 and the finite codimensionality of dom(c_i). This shows that $R=\bigoplus _{i} c_{i}\text {dom }(c_{i})$ is a free A-module of rank ≡ m mod (n − 1). On the other hand, R is an essential right ideal of A. Namely, for any $0\neq a={\sum }_{i=1}^{m} c_{i}q_{i}\in A (q_{i}\in A_{\frak I})$ with at least one of the q_i ≠  0, say q₁, there are nonzero elements $r_{1}\in q_{1}\text {dom}(q_{1})\cap (\text {dom}(c_{1})\cap (\cap _{i=1}^{m} \text {dom}(q_{i}))); r_{2}, \dots , r_{m} \in A$ such that q_ir₁r₂⋯r_i ∈dom(c_i) for all $i=2, \dots , m$ in view of the fact that all dom(c_i),dom(q_i) are essential right ideals of A and q₁dom(q₁) ≠  0 by q₁ ≠  0. By the choice of r₁, one has 0 ≠ q₁(r₁) = q₁r₁ ∈dom(c₁) whence 0 ≠ (c₁q₁(r₁))r₂⋯r_m = c₁q₁(r₁r₂⋯r_m) = c₁q₁r₁⋯r_m ∈ c₁dom(c₁) because A is a domain. Similarly the product ar₁⋯r_m ≠  0 holds, and all c_iq_ir₁⋯r_m are not necessarily nonzero elements of c_idom(c_i), respectively, for all i > 1. Consequently, $0\neq arr_{1}{\cdots } r_{m}={\sum }_{i=1}^{m} c_{i}q_{i}(r_{1}{\cdots } r_{m})\in R$ holds. Hence, R is a right ideal of A of finite codimension in view of Proposition 3.2. Hence, again by the Schreier-Lewin formula in Corollary 2.7, R is a free right A-module of free rank ≡ 1 mod (n − 1). Therefore, m − 1 ≡ 0 mod (n − 1) holds, i.e., n is the smallest integer satisfying $A_{\frak I}\cong A_{\frak I}^{m}$, whence the module type of $A_{\frak I}$ is (1, n). Consequently, the Grothendieck group $K_{0}(A_{\frak I})$ is cyclic of order n − 1 because projective modules over $A_{\frak I}$ are free. □

Remark 3.4

There are essential differences between free associative algebras and skew polynomial rings. For example, the annihilator ideals of finite codimensional right ideals are not necessarily finite codimensional. Consequently, the topology defined by finite codimensional right ideals are, in general, not an ideal topology. Furthermore, it is unclear whether there always exist two-sided ideals of codimension 1, even of finite codimension. However, there are simple skew polynomial rings when α is an automorphism and even the identity automorphism. Namely, let F = K(y) be a field of rational functions over a field K of characteristic 0 with the usual derivation $^{\prime }$, then the skew polynomial ring $F[X;1,^{\prime }]$ is a simple ring, whence it does not admit maximal two-sided ideal of codimension 1 over F. The simplicity is obvious in view of the facts that $F[X;1,^{\prime }]$ is a localization of the first Weyl algebra A₁(K) generated by x, y over K subject to yx − xy = 1 at the set of all monic polynomials on y over K and Weyl algebras over fields of characteristic 0 are simple. It is, in fact, more interesting to study the ring $A_{\frak I}$ of quotients with respect to an ideal topology defined by an ideal I of D[X;α, δ] under the extra assumption that I has codimension 1. It is then quite important to look for (quasi-)normal forms for elements of $A_{\frak I}$ and good D-bases for A. If μ^∗ denotes a product $z^{*}_{i_{l}}{\cdots } z^{*}_{i_{1}}=\mu ^{*}$ for every monomial $\mu =z_{i_{1}}{\cdots } z_{i_{l}}$ of $z_{i}~ (i=1, \dots , n)$ where $z_{i}, z^{*}_{i}$ are defined as in the proof of Theorem 3.3, then it is unclear whether elements of $A_{\frak I}$ can be written in the form ${\sum }_{i=1}^{m} \mu _{i} \nu ^{*}_{i} d_{i}$ for monomials μ_i, ν_i of “variables” z_j and coefficients d_i ∈ D. Same questions arise when I_D is assumed more generally to be of finite codimension over D.

It seems to be an important and difficult problem whether the associated ring $A_{\frak I}$ of quotients considered in Theorem 3.3 is simple. In the particular case of trivial derivation and I = Ax we have a positive answer.

Proposition 3.5

If α is an endomorphism of a division ring D such that D_α(D) has a finite dimension n > 1, then the ring $A_{\frak I}$ of quotients of A = D[X;α] with respect to the Gabriel topology induced by ideals $AX^{l} ~ (l\in \mathbb N)$ is simple.

Proof

We use Cohn’s trick presented in the proof of [2, Proposition 8.1]. AX is obviously an ideal of codimension 1. Let $B_{1}=\{b_{1}, \dots , b_{n}\}$ be a basis of the right vector space D over α(D). Then Ax is a free right A-module with the free generators y_i = b_iX, and monomials in y_i form a basis of A_D. Let $y^{*}_{i}~ (i=1, \dots , n)$ be the elements of $A_{\frak I}$ sending y_i to 1 and y_j to 0 for every index j ≠ i. If J is a nonzero ideal of $A_{\frak I}$, then J ∩ A is a non-zero ideal of A. Let a be a non-zero element of J ∩ A such that it can be written as a linear combination $a={\sum }_{i=1}^{m} \mu _{i}d_{i}$ of monomials μ_i in the y_j with coefficients d_i with possibly smallest m and among them with possibly smallest degree. By considering $\mu ^{*}_{l}a$ where μ_l has a minimal length, i.e., of minimal degree; if it is necessary, one sees clearly that some μ_i is a constant, i.e., of length 0. If m > 1, by considering $\mu ^{*}_{k}a\mu _{k}$, where μ_k has a positive length, one obtain a contradiction. Thus, m = 1 and a is a non-zero constant whence the statement follows. □

It is worth noting that even in the situation of Proposition 3.5, it is not clear whether elements of $A_{\frak I}$ can be written in the form ${\sum }_{i=1}^{m} \mu _{i} \nu ^{*}_{i} d_{i}$ for monomials μ_i, ν_i of “variables” y_j and coefficients d_i ∈ D.

We will now construct the maximal flat epimorphic ring of skew polynomial rings. To reach this aim we follow the idea of Rosenmann and Rosset [13]. First it is routine to verify that finite codimensional right ideals of A = D[X;α, δ] define a linear right topology on A, named the fc-topology by [13]. It is a little bit harder to show that the fc-topology is a Gabriel topology. It is instructive to modify the argument of Rosenmann and Rosset [13] (that right A-modules which can be generated by A-submodules of finite dimension over D, form a hereditary torsion theory in the category of right A-modules because of the fact that D is neither commutative nor commuting with the y_i’s) to obtain this claim. For undefined necessary notions and results on torsion theory, Gabriel topologies and localizations, we refer to Stenström classic [14]. However, to help the reader and to get a better grasp of the notorious special axiom T4 (see [14, Chapter VI.5]) in the definition of Gabriel topology, we present here a direct proof. To show that the fc-topology is a Gabriel topology, one has to show that it satisfies axiom T4, that is, a right ideal R is open, i.e., of finite codimension if there is a finite codimensional right ideal J such that R : b = {a ∈ A|ba ∈ R} has a finite codimension for all b ∈ J. By Proposition 3.2 and Theorem 2.5 there is a finite basis $\{b_{1}, \dots , b_{l}\}$ for $J=\sum b_{i}A=\oplus b_{i}A$. Moreover, there are l right ideals $R_{1}, \dots , R_{l}$ of finite codimension such that $b_{i}R_{i}\subseteq R$. Again by Proposition 3.2 and Theorem 2.5 each right ideal R_i(0 < i < l + 1) is an essential right ideal of A which is also a free right A-module of finite rank. Consequently, $\bar R={\sum }_{i=1}^{l} b_{i}R_{i}\subseteq R$ is clearly a finitely generated right ideal of A. On the other hand, it is obvious that b_iR_i is an essential submodule of b_iA for every index i. Therefore, by [9, Folgerung 5.1.8], $\bar R=\sum b_{i}R_{i}=\oplus b_{i}R_{i}$ is an essential submodule of ⊕ b_iA = J. Since J is an essential right ideal of A, $\bar R$ is an essential right ideal of A. Consequently, being both finitely generated and an essential right ideal of A, $\bar R$ is finite codimensional by Proposition 3.2. This implies, in view of the inclusion $\bar R\subseteq R$, that R is a finite codimensional right ideal of A, that is, it is open with respect to the fc-topology. Thus, we have shown that the fc-topology is a Gabriel topology and a right ideal of A is open in the fc-topology if and only if it is essential and free of finite rank in view of Proposition 3.2. By [14, Proposition XI.3.4 (d)] every open right ideal defined by a flat epimorphism $\phi : A=D[X; \alpha , \delta ] \rightarrow Q$ contains a finitely generated essential right ideal and so has a finite codimension by Proposition 3.2. This shows that the fc-topology defines the maximal flatepimorphic localization Q^fc(D[X;α, δ]) = Q_tot(D[X;α, δ]) of D[X;α, δ] which is a subring of the maximal ring $Q_{\max \limits }(D[X; \alpha , \delta ])$. By the same proof as for Theorem 3.3 we obtain another goal of this section as

Theorem 3.6

Right finite codimensional ideals of A = D[X;α, δ] where the dimension of D_α(D) is finite, form a perfect Gabriel topology, called the fc-topology. The ring Q^fc(A) of quotients with respect to this fc-topology is the maximal flat epimorphic localization of A. Its Grothendieck group K₀(Q^fc(A)) is cyclic of order n − 1 and their projective modules are free.

It seems that the maximal flat epimorphic localization of D[X;α, δ] is not necessarily simple but I am not able to construct a counterexample. However, if δ = 0, then we have a positive result.

Corollary 3.7

If α is an endomorphism of a division ring D such that D is a right vector space of dimension n > 1 over α(D), then the maximal flat epimorphic localization Q^fc(D[X;α]) is simple.

Proof

By assumption, D[X;α]X is a two-sided ideal of codimension 1 whence every element of D[X;α] can be written uniquely as a polynomial in y_i with coefficients from D on the right where y_i = b_iX with respect to a right basis $B_{1}=\{b_{1}, \dots , b_{n}\}$ of D over α(D). The proof of Proposition 3.5 can be used to complete the verification. □

In contrast to the case of free unital associative algebras we do not have the description of all flat bimorphisms of A = D[X;α, δ] because largest ideals contained in finite-codimensional right ideals of A are not necessarily finite-codimensional right ideals over A! However, thanks to [14, Proposition I.6.10], products of ideals from any (fixed) set Λ of maximal two-sided ideals of A which are finite-codimensional right ideals of A, still form a perfect Gabriel topology on A. The proof for [1, Theorem 3.10] yields the following result.

Theorem 3.8

Let α : D→D be an endomorphism of a divison ring D such that the right dimension of D over α(D) is n ≥ 2, and A = D[X;α, δ] be a skew polynomial ring with respect to some α-derivation δ of D. Furthermore, let Λ be an arbitrary set of maximal two-sided ideals which are also finite codimensional right ideals of A, then finite products of elements from Λ form a perfect Gabriel topology ${\frak T}_{{{\varLambda }}}$ of A with the associated ring Q of right quotients. For each ideal I_λ ∈Λ the factor ring A/I_λ is a matrix ring $_{\lambda }D_{m_{\lambda }}$ over a division ring _λD of right dimension l_λ over D. Put d_λ = lλm_λ and let d be the greatest common divisor of d_λ’s. Then, the module type of Q is (1, d(n − 1) + 1). In addition, if the ideals of Λ are commutable, then the completion of A with respect to ${\frak T}_{\lambda }$ is the direct products ${\prod }_{I} \underleftarrow {\lim } A/I^{l}$ of inverse limits $\underleftarrow {\lim } A/I^{l}$ of the canonical inverse systems $\{A/I^{l} | l\in \mathbb N\}$, where I runs over all elements of Λ.

Since the work [1] is under submission, I sketch here the main idea of the proof of Theorem 3.8. First we show that $A_{\frak I}$ has module type (1, d(n − 1)) if I_D is a two-sided ideal of finite codimension over D where d is a dimension of a minimal right ideal of A/I. Here $\frak I$ denotes the perfect Gabriel topology defined by powers of I. The proof can be carried out in the same manner as that for Theorem 3.3 after the observation that for an open right ideal R of A, there is a power I^l contained in R, so that the factor module A/R is an A/I^l-module whence the dimension of A/R over D is md if m is a length of a right A module A/R. Consequently, R is a free module of rank dm(n − 1) + 1 by Corollary 2.7. The assertion for a set Λ of maximal ideals of finite right codimension over D is then a trivial consequence of elementary number theory! We now complete the proof by using the following equalities on commutable two-sided maximal ideals I, J of A

$$A=(I+J)(I+J)=I+J^{2}=I^{2}+J=I^{2}+J^{2}=\cdots=I^{l}+J^{m}, \quad \forall l, m\in \mathbb N,$$

and

$$I\cap J=(I\cap J)(I+J)=IJ+JI=IJ=JI.$$

By iterating these equalities one obtains the following more general equalities for finitely many pairwise coprime, commutable ideals $I_{1}, \dots , I_{m}$

$$I^{n_{i}}_{i} +\cap_{j\neq i} I^{n_{j}}_{j}=A \qquad \& \qquad \cap I^{n_{i}}_{i}=\prod\limits_{i} I^{n_{i}}_{i}, \qquad \forall n_{i}\in \mathbb N.$$

Now the claim on completion becomes obvious.

As a corollary of Theorem 3.3 we reobtain one more direct but “unusual” proof of the classical result [10] of Leavitt in the following

Corollary 3.9

The classical Leavitt algebra L_K(1, n) (n > 1) over a field K has module type (1, n).

Proof

Consider the rational function field D = K(t) in one variable t over a field K and let α be the endomorphism of D sending t to tⁿ. By Lüroth’s theorem D is a finite field extension of D₁ = α(D) with basis $\{b_{i}=t^{i-1} | i=1, \dots , n\}$. Moreover, K is exactly the set of all elements in D fixed by α. By Theorem 3.3, the subring of $A_{\frak I}=D[X, \alpha ]_{\frak I}$, where I is the two-sided ideal AX of A = D[X, α], generated by $y_{i}=t_{i}X, y^{*}_{i}$ where $y^{*}_{i}$ is an A-homomorphism from AX to A sending y_i to 1, y_j to 0 for all j ≠ i, is isomorphic to the classical Leavitt algebra L_K(1, n) and its module type is clearly (1, n). □

We end with some remarks pointing out the structural diversity of a particular skew polynomial ring A = D[X;α]. Since A is a left principal ideal domain, $\{X^{n} | n\in \mathbb N\}$ is a left Ore set. Hence, in the localized ring A_X, the equality α(d) = X^− 1Xα(d) = X^− 1dX implies that $D\cong D_{n}=X^{n}DX^{-n}~ (n\in \mathbb Z, D_{0}=D)$ form an ascending chain of rings and α can be extended naturally to an automorphism, denoted again by α of $E=\cup _{n=-\infty }^{\infty } {D_{n}}$ by putting α(XⁿdX⁻ⁿ) = Xⁿα(d)X⁻ⁿ = XⁿX^− 1Xα(d)X⁻ⁿ = X^n− 1dX^{−(n− 1)}. This shows that A_X is isomorphic to a localization of a two-sided skew polynomial ring E[X, α] at the two-sided Ore set $\{X^{n} | n\in \mathbb N\}$ and hence A_X is a simple two-sided principal ideal domain! On the other hand, substituting Y = X + c (0 ≠ c ∈ D) for X does not, in general, induce an automorphism of A. However, A can be considered as a skew polynomial ring with respect to Y. The multiplication is modified, in view of the equality Y d = (X + c)d = α(d)X + cd = α(d)(X + c) + (cd − α(d)c), by an inner α-derivation δ_c(d) = cd − α(d)c. Hence, A is now the skew polynomial ring D[Y, α, δ_c]. Conversely, if A = D[X, α, δ_c] is a skew polynomial ring defined by an endomorphism a of a division ring D together with an inner derivation δ_c given by an element c ∈ D, then A becomes a skew polynomial ring D[Z;α] by putting Z = X − c. The situation changes completely when we consider the localized ring $A_{\frak I}=D[X, \alpha ]_{\frak I}~ (I=AX)$ even in the case when D is commutative and α is an endomorphism of D such that D is finite dimensional over α(D) with a basis $B_{1}=\{b_{1}, \dots , b_{n}\}$. For y_i = b_iX, let $y^{*}_{i}: I\longrightarrow A$ be an A-module homomorphism induced by the rule y_i↦1;y_j↦0;j ≠ i. It is unclear whether elements of $A_{\frak I}$ can be written in a normal form ${\sum }_{i=1}^{m} \mu _{i} \nu ^{*}_{i} d_{i}$ for monomials μ_i, ν_i of “variables” y_j and coefficients d_i ∈ D.