1. Definition, existence and uniqueness

This section is devoted to the proof of existence and uniqueness of free skew extensions. We also show that these rings satisfy a universal mapping property.

Let $R$ be a ring. An $R$-ring is an ordered pair $(A,\lambda_A)$, where $A$ is a ring and $\lambda_A\colon R\to A$ is a ring homomorphism. Given $R$-rings $(A,\lambda_A)$ and $(B,\lambda_B)$ an $R$-ring homomorphism from $(A,\lambda_A)$ into $(B,\lambda_B)$ is a ring homomorphism $\varphi\colon A\to B$ such that $\varphi\lambda_A=\lambda_B$. Clearly, if $(A,\lambda_A)$ is an $R$-ring and $I$ is an ideal of $A$, then the quotient ring $A/I$ has a natural $R$-ring structure which makes the canonical surjective ring homomorphism $A\to A/I$ an $R$-ring homomorphism.

Recall that an $R$-bimodule is an additive abelian group $M$ which is a left $R$-module, a right $R$-module and such that

\begin{equation*} (rx) s = r (x s) \end{equation*}

for all $r,s\in R$ and $x\in M$. So, if $(A,\lambda_A)$ is an $R$-ring, then $A$ has a natural $R$-bimodule structure given by

\begin{equation*} r a = \lambda_A(r)a \quad\text{and}\quad a r = a\lambda_A(r), \end{equation*}

for all $r\in R$ and $a\in A$. This $R$-bimodule structure on $A$ is such that

(2)\begin{equation} r (ab) = (r a)b, \quad (ar)b=a(rb), \quad (ab) r = a(b r), \end{equation}

for all $r\in R$ and $a,b\in A$. Conversely, given an $R$-bimodule structure on a ring $A$ satisfying (2), then $A$ becomes an $R$-ring via

\begin{equation*} \begin{array}{rcl} \lambda_A\colon R & \longrightarrow & A\\ r&\longmapsto &r 1_A \end{array}. \end{equation*}

We shall switch from these two viewpoints according to convenience.

If $A$ and $B$ are $R$-rings and $\varphi\colon A\to B$ is a ring homomorphism that is also a left (or right) $R$-module homomorphism, then $\varphi$ is an $R$-ring homomorphism.

Let $R$ be a ring, let $n$ be a positive integer and let $\sigma\colon R\to M_n(R)$ be a ring homomorphism. Let $R^n$ stand for the free left $R$-module of rank $n$, hereby identified with the set of row vectors with $n$ entries from $R$. An additive map $\delta\colon R\to R^n$ satisfying

(3)\begin{equation} \delta(rs) = \delta(r)\sigma(s)+r\delta(s), \end{equation}

for all $r,s\in R$, will be called a right $\sigma$-derivation. The concatenation $\delta(r)\sigma(s)$ on the right-hand side of (3) stands for the matrix product of the $1\times n$ matrix $\delta(r)$ by the $n\times n$ matrix $\sigma(s)$.

Writing $\sigma_{ij}(r)$ for the $(i,j)$-entry of the matrix $\sigma(r)$ and $\delta_j(r)$ for the $j$-component of the row vector $\delta(r)$, it follows that $\sigma_{ij}$ and $\delta_j$ are additive endomorphisms of $R$ satisfying

\begin{equation*} \sigma_{ij}(rs)=\sum_{k=1}^n\sigma_{ik}(r)\sigma_{kj}(s) \end{equation*}

and

\begin{equation*} \delta_j(rs) = \sum_{i=1}^n\delta_i(r)\sigma_{ij}(s)+r\delta_j(s), \end{equation*}

for all $r,s\in R$. Moreover,

\begin{equation*} \sigma_{ij}(1)=\begin{cases} 1, &\text{if $i=j$}\\ 0, &\text{otherwise}\end{cases} \quad\text{and}\quad\delta_j(1)=0. \end{equation*}

Given a ring $S$, we shall write $S^{\times}$ for the monoid structure on $S$ with operation given by the multiplication.

We note that since $1_S\in\langle X\rangle$, condition (ii) above implies that the ring homomorphism $R\to S$ affording the $R$-ring structure on a free skew extension $S$ of $R$ is injective and, therefore, $R$ can be regarded as a subring of $S$. For this reason, in condition (iii) we write simply $\delta_j(r)$ for the element $\delta_j(r)1_S$.

Our first result, following the ideas in [Reference Goodearl and jun4, Prop. 2.3] and [Reference McConnell and Robson14, 1.2.3], shows that free skew extensions always exist.

For inductive arguments used throughout this text, the following notation is introduced. For a free monoid $\langle X\rangle$ on a set $X=\{x_1,\dots, x_n\}$, given $w\in\langle X\rangle$ such that $w=x_{i_1}x_{i_2}\dotsm x_{i_r}$, with $i_k\in\{1,\dots,n\}$ for all $k=1,\dots, r$, we say that $w$ has length $r$ and write $\lvert w\rvert = r$. We also set $\lvert 1\rvert = 0$, where $1$ stands for the identity element of $\langle X\rangle$, that is, the empty word.

Proof. Let $\mathcal{F}$ be the free monoid on the $n$-element set $\{z_1,\dots,z_n\}$ and let $R^{\mathcal{F}}$ denote the set of functions $\mathcal{F}\to R$. Then $R^{\mathcal{F}}$ has a natural abelian group structure such that $(f+g)(m)=f(m)+g(m)$, for all $f,g\in R^{\mathcal{F}}$ and $m\in\mathcal{F}$. Let $E$ be the ring of group endomorphisms of $R^{\mathcal{F}}$, acting on the elements of $R^{\mathcal{F}}$ on the right. Each $r\in R$ defines $\lambda(r)\in E$ by right multiplication: $\bigl(f\lambda(r)\bigr)(m)=f(m)r$, for all $f\in R^{\mathcal{F}}$ and $m\in\mathcal{F}$. The map $r\mapsto\lambda(r)$ is easily seen to be an injective ring homomorphism $R\to E$, defining in $E$ an $R$-ring structure. Moreover, for each $j=1,\dots, n$, one has $x_j\in E$ defined by

\begin{equation*} (fx_j)(mz_i)= \sigma_{ij}\bigl(f(m)\bigr)+\delta_j\bigl(f(mz_i)\bigr), \quad\text{for $i=1,\dots,n$ and $m\in\mathcal{F}$} \end{equation*}

and

\begin{equation*} (fx_j)(\mathbf{1})= \delta_j\bigl(f(\mathbf{1})\bigr), \end{equation*}

where $\mathbf{1}$ denotes the unity element of $\mathcal{F}$.

For each $m\in\mathcal{F}$, let $\overline{m}\in R^{\mathcal{F}}$ be defined by

\begin{equation*} \overline{m}(l)=\begin{cases} 1, & \text{if $l=m$}\\ 0, &\text{otherwise.}\end{cases} \end{equation*}

We note that the function $m\mapsto\overline{m}$, from $\mathcal{F}$ into $R^{\mathcal{F}}$, is injective. By the definition of $x_j$, one has

\begin{equation*} \overline{m}x_j=\overline{mz_j}, \end{equation*}

for all $m\in\mathcal{F}$, and, so it follows by induction on $k$ that

(4)\begin{equation} \overline{\mathbf{1}}(x_{j_1}\dotsm x_{j_k})=\overline{z_{j_1}\dotsm z_{j_k}}. \end{equation}

Hence, the submonoid $\langle X\rangle$ of $E^{\times}$ generated by $X=\{x_1,\dots,x_n\}$ is free on $X$. In particular, there exists a monoid isomorphism $\phi\colon \langle X\rangle \to \mathcal{F}$ such that $\phi(x_j)=z_j$, and, hence, (4) gives $\overline{\mathbf{1}}(w)=\overline{\phi(w)}$, for all $w\in \langle X\rangle$.

We have seen that $\lambda\colon R\to E$ makes $E$ into an $R$-ring, in particular, we can regard $E$ as an $R$-bimodule via $rf=\lambda(r)f$ and $fr=f\lambda(r)$, for all $f\in E$ and $r\in R$. Our aim is to show that the right $R$-submodule of $E$ generated by $\langle X\rangle$ is a $(\sigma,\delta)$-free skew extension of $R$. We begin by showing that the subset $\langle X\rangle$ of $E$ is right linearly independent over $R$.

Given a right linear combination

\begin{equation*} \sum_{w\in\langle X\rangle}wr_w, \end{equation*}

with $r_w\in R$, one has

\begin{equation*} \overline{\mathbf{1}}\left(\sum_{w\in\langle X\rangle}wr_w\right) = \sum_{w\in\langle X\rangle}\overline{\phi(w)}\lambda(r_w), \end{equation*}

and this function evaluated at $\phi(w)$ gives $r_w$. Thus, indeed, $\langle X\rangle$ is right linearly independent over $R$ in $E$.

We now show that, in $E$, for all $r\in R$ and $j\in\{1,\dots,n\}$, one has $rx_j=\sum_{k=1}^nx_k\sigma_{kj}(r) + \delta_j(r)$. Indeed, given $f\in R^{\mathcal{F}}$ and $m\in\mathcal{F}$,

\begin{equation*}\begin{split} \bigl(f(rx_j)\bigr)(mz_i) & = \bigl((fr)x_j\bigr)(mz_i)\\ & = \sigma_{ij}\bigl((fr)(m)\bigr) + \delta_j\bigl((fr)(mz_i)\bigr)\\ & = \sigma_{ij}\bigl(f(m)r\bigr)+\delta_j\bigl(f(mz_i)r\bigr)\\ & =\sum_{k=1}^n\sigma_{ik}\bigl(f(m)\bigr)\sigma_{kj}(r) + \sum_{k=1}^n\delta_k\bigl(f(mz_i)\bigr)\sigma_{kj}(r) + f(mz_i)\delta_j(r)\\ & = \sum_{k=1}^n\Bigl(\sigma_{ik}\bigl(f(m)\bigr)+\delta_k\bigl(f(mz_i)\bigr)\Bigr)\sigma_{kj}(r) + f(mz_i)\delta_j(r)\\ & = \sum_{k=1}^n(fx_k)(mz_i)\sigma_{kj}(r)+ f(mz_i)\delta_j(r)\\ & = \sum_{k=1}^n\bigl((fx_k)\sigma_{kj}(r)\bigr)(mz_i) + \bigl(f\delta_j(r)\bigr)(mz_i)\\ & = \sum_{k=1}^n \bigl(f(x_k\sigma_{kj})(r)+f\delta(r)\bigr)(mz_i)\\ & = \biggl(f\Bigl(\sum_{k=1}^nx_k\sigma_{kj}(r)+\delta_j(r)\Bigr)\biggr)(mz_i). \end{split}\end{equation*}

Also,

\begin{equation*}\begin{split} \bigl(f(rx_j)\bigr)(\mathbf{1}) & = \bigl((fr)x_j\bigr)(\mathbf{1}) = \delta_j\bigl((fr)(\mathbf{1})\bigr)= \delta_j\bigl(f(\mathbf{1})r\bigr)\\ & = \sum_{k=1}^n \delta_k\bigl(f(\mathbf{1})\bigr)\sigma_{kj}(r) + f(\mathbf{1})\delta_j(r)\\ & = \sum_{k=1}^n (fx_k)(\mathbf{1})\sigma_{kj}(r) + f(\mathbf{1})\delta_j(r)\\ & = \sum_{k=1}^n\bigl((fx_k)\sigma_{kj}(r)\bigr)(\mathbf{1})+ f(\mathbf{1})\delta_j(r)\\ & = \biggl(f\Bigr(\sum_{k=1}^n x_k\sigma_{kj}(r)+\delta_j(r)\Bigl)\biggr)(\mathbf{1}). \end{split}\end{equation*}

Hence, the operators $rx_j$ and $\sum_{k=1}^nx_k\sigma_{kj}(r) + \delta_j(r)$ do coincide. Besides, it follows that, in $E$,

\begin{equation*} Rx_1+\dots+Rx_n\subseteq x_1R+\dots+x_nR+R, \end{equation*}

and, as a consequence, an inductive argument on the length $\lvert w\rvert$ of $w\in\langle X\rangle$, gives

(5)\begin{equation} Rw\subseteq \sum_{\lvert v \rvert \leq \lvert w \rvert} vR. \end{equation}

So let $S$ be the (free) right $R$-submodule of $E$ generated by $\langle X\rangle$. Given $w_1,w_2\in\langle X \rangle$, it follows from (5) that $w_1Rw_2R\subseteq S$; this proves that $S$ is a sub- $R$-ring of $E$. By what has been shown above, $S$ is a $(\sigma,\delta)$-free skew extension of $R$ generated by $x_1,\dots,x_n$.

Having dealt with the existence of free skew extensions, we now turn to uniqueness, which will follow from the fact that these rings satisfy a universal property (stated, in a special case, without proof in [Reference Martínez-Peñas13, Lem. 2]) as the next result shows.

Proposition 1.2. Let $R$ be a ring, let $n$ be a positive integer, let $\sigma\colon R\to M_n(R)$ be a ring homomorphism, let $\delta \colon R\to R^n$ be a right $\sigma$-derivation and let $S$ be a $(\sigma,\delta)$-free skew extension of $R$ generated by $x_1,\dots,x_n$. Given an $R$-ring $A$ and $a_1,\dots,a_n$ elements of $A$ such that

(6)\begin{equation} ra_j=\sum_{i=1}^n a_i\sigma_{ij}(r) + \delta_j(r)1_A, \quad\text{for all $j=1,\dots,n$ and $r\in R$,} \end{equation}

there exists a unique $R$-ring homomorphism $\varphi\colon S\to A$ such that $\varphi(x_i)=a_i$, for all $i=1,\dots,n$.

Proof. Since the submonoid $\langle X\rangle$ of $S^{\times}$ is free on $X=\{x_1,\dots,x_n\}$, there is a monoid morphism $\rho\colon \langle X\rangle \to A^{\times}$ such that $\rho(x_i)=a_i$, for all $i=1,\dots,n$. Let $\varphi\colon S\to A$ be the right $R$-module homomorphism such that $\varphi(w)=\rho(w)$, for all $w\in\langle X\rangle$. Note that $\varphi(1)=1_A$. Let us show that $\varphi$ is an $R$-ring homomorphism. First, note that if $f=\sum_{w\in\langle X\rangle} wr_w$ is an arbitrary element of $S$, then, for $j=1,\dots,n$, one has

\begin{equation*}\begin{split} \varphi(fx_j) & = \varphi\biggl(\sum_{i=1}^n\sum_{w\in\langle X\rangle}wx_i\sigma_{ij}(r_w) + \sum_{w\in\langle X\rangle} w\delta_j(r_w)\biggr)\\ & = \sum_{i=1}^n\sum_{w\in\langle X\rangle}\rho(wx_i)\sigma_{ij}(r_w) + \sum_{w\in\langle X\rangle}\rho(w)\delta_j(r_w)\\ & = \sum_{w\in\langle X\rangle}\rho(w)\biggl(\sum_{i=1}^na_i\sigma_{ij}(r_w) + \delta_j(r_w) \biggr)\\ & = \sum_{w\in\langle X\rangle}\rho(w)r_wa_j\\ & = \varphi(f)a_j. \end{split}\end{equation*}

Hence, by an inductive argument on the length of a monomial $w$, one proves that

\begin{equation*} \varphi(fw)=\varphi(f)\rho(w), \end{equation*}

for all $w\in\langle X\rangle$. Now, given $f'=\sum_{w\in\langle X\rangle} wr_w'\in S$, one has

\begin{equation*} \varphi(ff')=\varphi\biggl(f\sum_{w\in\langle X\rangle}wr_w'\biggr) =\sum_{w\in\langle X\rangle} \varphi(fw)r_w' = \sum_{w\in\langle X\rangle} \varphi(f)\rho(w) r_w'=\varphi(f)\varphi(f'). \end{equation*}

It follows that $\varphi$ is an $R$-ring homomorphism such that $\varphi(x_i)=a_i$, for all $i=1,\dots,n$.

Uniqueness of $\varphi$ follows from the fact that $S$ is generated by $\{x_1,\dots,x_n\}$ as an $R$-ring.

Corollary 1.3. Let $R$ be a ring, let $n$ be a positive integer, let $\sigma\colon R\to M_n(R)$ be a ring homomorphism, let $\delta \colon R\to R^n$ be a right $\sigma$-derivation and let $S_1$ be a $(\sigma,\delta)$-free skew extension of $R$ generated by $x_1,\dots,x_n$. Let $S_2$ be an $R$-ring containing elements $y_1,\dots,y_n$ such that

\begin{equation*} ry_j=\sum_{i=1}^n y_i\sigma_{ij}(r) + \delta_j(r)1_{S_2}, \quad\text{for all $j=1,\dots,n$ and $r\in R$,} \end{equation*}

and such that for any $R$-ring $A$ and elements $a_1,\dots,a_n$ satisfying (6) there exists a unique $R$-ring homomorphism $\psi\colon S_2\to A$ such that $\psi(y_i)=a_i$, for all $i=1,\dots,n$. Then there is an $R$-ring isomorphism $S_1\to S_2$, sending $x_i$ to $y_i$, for all $i=1,\dots,n$.

In particular if $S_1$ and $S_2$ are $(\sigma,\delta)$-free skew extensions of $R$ generated by $x_1,\dots,x_n$ and $y_1,\dots,y_n$, respectively, then there exists a unique $R$-ring isomorphism $S_1\to S_2$ sending $x_i$ to $y_i$ for all $i=1,\dots,n$.

It follows that the universal property in Proposition 1.2 completely characterizes free skew extensions. This fact will be used in the proof of Corollary 2.4.

In view of Theorem 1.1 and Corollary 1.3, given a ring $R$, a positive integer $n$, a ring homomorphism $\sigma\colon R\to M_n(R)$ and a right $\sigma$-derivation $\delta\colon R\to R^n$, we can speak of the $(\sigma,\delta)$-free skew extension of $R$ generated by $x_1,\dots,x_n$. This ring will, henceforth, be denoted by $R\langle x_1,\dots,x_n;\sigma,\delta\rangle$. When the right $\sigma$-derivation $\delta$ is the zero map, we shall simply write $R\langle x_1,\dots,x_n;\sigma\rangle$.

It is often the case that, as is the case of ordinary skew polynomial rings (cf. [Reference Cohn2, 2.1.3]), a suitable change of variables allows one to reduce the right $\sigma$-derivation $\delta$ to the zero map or the ring homomorphism $\sigma$ to the scalar map, as the next result shows.

Proof. (i) Consider the $n$-element set $Y=\{y_1,\dotsc,y_n\}$ where $y_j=x_jc-cx_j$, for all $j=1,\dots, n$. Note that

\begin{equation*} [y_1 \ y_2 \ \dots \ y_n] = [x_1 \ x_2 \ \dots \ x_n]\bigl(cI_n-\sigma(c)\bigr)-\delta(c). \end{equation*}

Then,

\begin{equation*} [x_1 \ x_2 \ \dots \ x_n] = \bigl([y_1 \ y_2 \ \dots \ y_n] + \delta(c)\bigr)\bigl(cI_n-\sigma(c)\bigr)^{-1}. \end{equation*}

It follows that that the submonoid $\langle Y\rangle$ of $S^{\times}$ generated by $Y$ is free on $Y$ and that $S$ is the free right $R$-module with basis $\langle Y\rangle$. Moreover, for each $j=1,\dots,n$ and $r\in R$,

\begin{equation*}\begin{split} ry_j & = r(x_jc-cx_j) \\ & = rx_jc-rcx_j \\ & = \sum_{i=1}^nx_i\sigma_{ij}(r)c +\delta_j(r)c - c\sum_{i=1}^nx_i\sigma_{ij}(r)-c\delta_j(r) \\ & =\sum_{i=1}^n(x_ic-cx_i)\sigma_{ij}(r) \\ & = \sum_{i=1}^n y_i\sigma_{ij}(r). \end{split}\end{equation*}

This shows that $S=R\langle y_1,\ldots,y_n; \sigma\rangle$.

(ii) Since the elements $c_j$ are in the centre of $R$,

\begin{equation*} \delta_j(r)c_j+r\delta_j(c_j)=\delta_j(rc_j)=\delta_j(c_jr)=\delta_j(c_j)\sigma_{jj}(r)+c_j\delta_j(r), \end{equation*}

for all $r\in R$ and $j=1,\dots,n$. Hence,

\begin{equation*} \sigma_{jj}(r)\delta_j(c_j)^{-1}=\delta_j(c_j)^{-1}r, \end{equation*}

for all $r\in R$ and $j=1,\dots,n$.

For each $j=1,\dots, n$, set $y_j=x_j\delta_j(c_j)^{-1}$. For all $r\in R$ and $j=1,\dots,n$,

\begin{equation*}\begin{split} ry_j & = rx_j\delta_j(c_j)^{-1}\\ & = \bigl(x_j\sigma_{jj}(r)+\delta_j(r)\bigr)\delta_j(c_j)^{-1} \\ & = x_j\delta_j(c_j)^{-1}r+\delta_j(r)\delta_j(c_j)^{-1}\\ & = y_jr+\delta_j(r)\delta_j(c_j)^{-1}. \end{split}\end{equation*}

It remains to show that $\delta_j'\colon R\rightarrow R$, defined as $\delta_j'(r)=\delta_j(r)\delta_j(c_j)^{-1}$ for all $r\in R$, is an ordinary derivation. Indeed, if $r,s\in R$, then

\begin{equation*}\begin{split} \delta_j'(rs) & = \delta_j(rs)\delta_j(c_j)^{-1} \\ & = \bigl(\delta_j(r)\sigma_{jj}(s)+r\delta_j(s)\bigr)\delta_j(c_j)^{-1}\\ & = \delta_j(r)\delta_j(c_j)^{-1}s + r\delta_j(s)\delta_j(c_j)^{-1} \\ & = \delta_j'(r)s+r\delta_j'(s). \end{split}\end{equation*}

This proves that $S=R\langle y_1,\dots,y_n; \tau, \delta'\rangle$.

At this point, it should be noted that when $R$ is a division ring and $n\geq 2$, the centre $Z(S)$ of the free skew extension $S=R\langle x_1,\dots,x_n; \sigma,\delta\rangle$, as described in [Reference Leroy and Merdach9, Prop. 3.1], is given by

\begin{equation*} Z(S) = \{r\in Z(R) \mid \sigma(r)=rI_n \text{ and }\delta(r)=0\}. \end{equation*}

Remark 1. One can consider free skew extensions with coefficients on the left. For that, given a ring $R$, a positive integer $n$ and a ring homomorphism $\sigma\colon R\to M_n(R)$, a left $\sigma$-derivation is an additive map $\delta\colon R\to {}^nR$, where ${}^nR$ stands for the (free) right $R$-module formed by column vectors, such that

\begin{equation*} \delta(rs) = \sigma(r)\delta(s) + \delta(r)s, \end{equation*}

for all $r,s\in R$. The $(\sigma,\delta)$-free skew extension of $R$ generated by $x_1,\dots,x_n$ is then defined to be the $R$-ring $S$ which is free as left $R$-module on the free monoid $\langle X\rangle$ freely generated by $X=\{x_1,\dots,x_n\}$ satisfying $x_ir = \sum_{j=1}^n\sigma_{ij}(r)x_j + \delta_i(r)$, for all $i=1,\dots,n$ and $r\in R$.

2. Free skew extensions as universal rings on bimodules

A free skew extension can also be obtained as a universal $R$-ring on an appropriate bimodule, as we show in Corollary 2.4, which closes this section. The advantage of this approach is that it can be easily generalized.

Given a ring $R$ and an $R$-bimodule $M$, the tensor ring on $M$ is defined to be the (external) direct sum

\begin{equation*} R\langle M\rangle = \bigoplus_{n\geq 0} M^{\otimes n}, \end{equation*}

where $M^{\otimes 0} = R$, $M^{\otimes 1} = M$, and, for $n\geq 2$, $M^{\otimes n}=M\otimes_R\dots\otimes_R M$ ( $n$ factors). The $R$-bimodule $R\langle M\rangle$ becomes a ring with multiplication given by

\begin{equation*} (a_0,a_1,a_2,\dots)(b_0,b_1,b_2,\dots) =(c_0, c_1, c_2, \dots), \end{equation*}

with $c_n= \sum_{i=0}^n \rho_{i,n-i}(a_i\otimes b_{n-i})$, where $\rho_{i,n-i}\colon M^{\otimes i}\otimes_R M^{\otimes (n-i)}\to M^{\otimes n}$ are the natural $R$-bimodule maps. In fact, $R\langle M\rangle$ is an $R$-ring with structure map given by

\begin{equation*} \begin{array}{rcl} \lambda\colon R & \longrightarrow & R\langle M\rangle\\ r&\longmapsto &(r,0,0,\dots) \end{array}. \end{equation*}

So $(R\langle M\rangle,\lambda)$ is an $R$-ring with an $R$-bimodule map

(7)\begin{equation} \begin{array}{rcl} \mu\colon M & \longrightarrow & R\langle M\rangle\\ x&\longmapsto &(0,x,0,\dots) \end{array}, \end{equation}

which is universal with regard to this property, as shown in the next result.

Proof. Define $\psi_0=\lambda_A$, $\psi_1=f$ and, for $n\geq 2$, $\psi_n\colon M^{\otimes n}\to A$ to be the $R$-bimodule map induced by the $n$-balanced map

\begin{equation*} \begin{array}{rcl} M\times\dots\times M & \longrightarrow & A\\ (x_1,\dots,x_n)&\longmapsto &f(x_1)\dotsm f(x_n) \end{array}. \end{equation*}

Then $\psi = \bigoplus_{n\geq 0} \psi_n$ is an $R$-ring homomorphism with the desired property. Uniqueness follows from the fact that $R\langle M\rangle$ is generated by $\mu(M)$ as an $R$-ring.

We shall need a related concept.

The $R$-ring $U=U(M,\xi)$ constructed above, whose uniqueness up to an $R$-ring isomorphism is guaranteed by the last statement of Theorem 2.2, will be called the universal $R$-ring on the bimodule $M$ along $\xi$.

A particular case is of interest. Given a right $R$-module $V$, we shall denote the ring of all $R$-module endomorphisms of $V$ by $\operatorname{End}(V_R)$, with multiplication given by composition of functions. The set $\operatorname{Hom}(V_R,R_R)$ of all $R$-module maps $f\colon V\to R$ has a natural left $R$-module structure such that $(rf)(v)=rf(v)$, for all $r\in R$ and $v\in V$. Given a ring homomorphism $\sigma\colon R\to \operatorname{End}(V_R)$, we say that a map $\delta\colon R\to \operatorname{Hom}(V_R,R_R)$ is a $\sigma$-derivation if it is additive and $\delta(rs) =\delta(r)\sigma(s)+r\delta(s)$, for all $r,s\in R$.

Suppose, now, that we are given a right $R$-module $M$ having $R$ as a complemented $R$-submodule, that is, suppose that $R$ is a right $R$-submodule of $M$ and that there exists a right $R$-submodule $V$ of $M$ such that $M$ is the (internal) direct sum $M=R\oplus V$. Suppose, further, that $M$ has a left $R$-module structure making it into an $R$-bimodule in such a way that $R$ is a sub-bimodule. This implies that there exist functions $\sigma\colon R\to \operatorname{End}(V_R)$ and $\delta\colon R\to \operatorname{Hom}(V_R,R_R)$ such that

(8)\begin{equation} r(s+v) = rs + \delta(r)(v) + \sigma(r)(v), \quad\text{for all $r,s\in R$ and $v\in V$.} \end{equation}

It is then a straightforward verification that $\sigma$ must be a ring homomorphism and $\delta$ a $\sigma$-derivation. Conversely, given a right $R$-module $V$, a ring homomorphism $\sigma\colon R\to\operatorname{End}(V_R)$ and a $\sigma$-derivation $\delta \colon R\to \operatorname{Hom}(V_R,R_R)$, (8) defines an $R$-bimodule structure on the right $R$-module $M=R\oplus V$ in which $R$ is a sub-bimodule. We shall call $M$ the rank $1$ bimodule extension of $V$ defined by $\sigma$ and $\delta$.

Note that when $\delta$ is the zero derivation, $V$ is a sub-bimodule of $M$ and $R$ is, then, complemented as a sub-bimodule. In this case, the universal $R$-ring on $M$ coincides with the tensor ring on $V$, as the following result shows.

The universal $R$-ring $U$ on a rank $1$ bimodule extension $M=R\oplus V$ has been considered before by Cohn in [Reference Cohn2, Sec. 5.4], where the term pointed $R$-bimodule is used for $M$. There, an alternative construction of the universal $R$-ring on $M$ is given as a direct limit of a system of pointed $R$-bimodules; the term filtered tensor ring on $M$ is used for $U$ and it is denoted by $R_k\langle M;R\rangle$, where $k$ stands for a central commutative subfield of $R$, but this requirement (that is, of $R$ being an algebra over a field) is not necessary for the general construction. For this reason, we shall denote Cohn’s filtered tensor ring on a pointed $R$-bimodule $M$ by $R\langle M;R\rangle$. Further, a universal property for $R\langle M;R\rangle$ is provided in [Reference Cohn2, Thm. 5.4.6], which can be obtained from Theorem 2.2, when $M$ is taken to be a rank $1$ bimodule extension. We shall come back to Cohn’s results on the filtered tensor ring on a pointed $R$-bimodule in Section 4.

Now, we return to free skew extensions, as defined in Section 1. Given a ring $R$, a positive integer $n$, a ring homomorphism $\sigma\colon R\to M_n(R)$ and a right $\sigma$-derivation $\delta\colon R\to R^n$, let $S=R\langle x_1,\dots,x_n;\sigma,\delta\rangle$ be the $(\sigma,\delta)$-free skew extension of $R$ generated by $x_1,\dots,x_n$. We shall show that $S$ can be regarded as a universal $R$-ring on a rank $1$ bimodule extension of a free right $R$-module of rank $n$.

Let $V$ be the free right $R$-module of rank $n$ with basis $v_1,\dots,v_n$. Let $\overline{\sigma} \colon R\to \operatorname{End}(V_R)$ be the ring homomorphism obtained by composing $\sigma$ with the ring isomorphism $M_n(R)\to\operatorname{End}(V_R)$ which sends the matrix $(r_{ij})$ to the endomorphism of $V$ such that $v_j\mapsto \sum_{i=1}^{n}v_ir_{ij}$. And let $\overline{\delta}\colon R\to \operatorname{Hom}(V_R,R_R)$ be obtained by composing $\delta$ with the left $R$-module isomorphism $R^n\to \operatorname{Hom}(V_R,R_R)$ which sends $[r_1 \ r_2 \ \dots \ r_n]$ to the right $R$-module map from $V$ to $R$ such that $v_i\mapsto r_i$. It is straightforward to verify that $\overline{\delta}$ is a $\overline{\sigma}$-derivation. Let $M=R\oplus V$ be the rank $1$ bimodule extension of $V$ defined by $\overline{\sigma}$ and $\overline{\delta}$, and let $U(M,\iota)$ be the universal $R$-ring on $M$ along the inclusion map $\iota\colon R\to M$.

Our task is to show that $U(M,\iota)$ and $S$ are isomorphic $R$-rings. We accomplish this by showing that $U(M,\iota)$ satisfies the universal property of Proposition 1.2 that characterizes $S$.

First, note that in $U(M,\iota)$ one has

\begin{equation*} \lambda_U(r)\overline{v_j}=\sum_{i=1}^n\overline{v_i}\lambda_U(\sigma_{ij}(r)) + \lambda_U(\delta_j(r)), \end{equation*}

for all $j=1,\dots, n$, where $\overline{v_j}=\pi\mu(v_j)$, in the notation of the proof of Theorem 2.2. Now, let $A$ be an $R$-ring and let $a_1,\dots,a_n$ be elements of $A$ satisfying (6). The right $R$-module map $f\colon M\to A$ such that $1\mapsto 1_A$ and $v_i\mapsto a_i$, for all $i=1,\dots, n$ is, in fact, an $R$-bimodule map and, clearly, $f\iota=\lambda_A$, the ring homomorphism $R\to A$ affording $A$ its $R$-ring structure. The universal property of $U(M,\iota)$ guarantees that there exists a unique $R$-ring homomorphism $\varphi\colon U(M,\iota)\to A$ such that $\varphi(\overline{v_i})=f(v_i)=a_i$, for all $i=1,\dots,n$. It follows that $U(M,\iota)\cong S$ as $R$-rings. We have proved

We remark that this perspective allows one to consider free skew extensions on a not necessarily finite set of generators, by taking universal $R$-rings of the form $U(M,\iota)$, where $M$ is a rank $1$ bimodule extension of an arbitrary free right $R$-module. Ring homomorphisms into column-finite infinite matrices would be an alternative approach.

3. Filtered-graded reductions

In this section, we consider a natural filtration on a free skew extension and point to the fact that the properties that will be dealt with in the next two sections can be reduced to the associated graded ring, which is again a free skew extension.

We recall, from [Reference Cohn2, Sec. 2.6], that given a ring $S$, a filtration of $S$ is a family $\{S_n\}_{n\in\mathbb{N}\cup \{-\infty\}}$ of additive subgroups of $S$ that satisfy the following conditions

1. (i) $\{0\}=S_{-\infty}\subseteq S_0\subseteq S_1\subseteq \dots $,

2. (ii) $\bigcup_{n\in\mathbb{N}\cup \{-\infty\}} S_n=S$,

3. (iii) $S_nS_m\subseteq S_{n+m}$, for all $m,n\in\mathbb{N}\cup \{-\infty\}$,

4. (iv) $1\in S_0$.

The function $\deg\colon S \to \mathbb{N}\cup \{-\infty\}$ defined by

(9)\begin{equation} \deg(f) = \min\{m \mid f\in S_m\}, \end{equation}

is a pseudo-valuation, that is, one has

1. (I) $\deg(f)\geq 0$, for all $f\in S\setminus\{0\}$, and $\deg(0)=-\infty$,

2. (II) $\deg(f-g)\leq\max\{\deg(f),\deg(g)\}$, for all $f,g\in S$,

3. (III) $\deg(fg)\leq\deg(f)+\deg(g)$, for all $f,g\in S$,

4. (IV) $\deg(1)=0$.

When equality holds in III, we say that $\deg$ is a degree-function. By a filtered ring we understand a ring with a given filtration.

Conversely, a ring $S$ with a pseudo-valuation can be filtered by setting

\begin{equation*} S_n = \{f\in S \mid \deg(f)\leq n\}, \end{equation*}

for all $n\in\mathbb{N}\cup \{-\infty\}$.

A ring $T$ is said to be graded if there is a family $\{T_n\}_{n\in\mathbb{N}}$ of additive subgroups of $T$ such that

\begin{equation*} T = \bigoplus_{n\in \mathbb{N}} T_n \quad\text{and}\quad T_m T_n\subseteq T_{m+n}\text{, for all $m,n\in\mathbb{N}$.} \end{equation*}

If $S$ is a filtered ring with filtration $\{S_n\}_{n\in\mathbb{N}\cup \{-\infty\}}$, the associated graded ring $\operatorname{gr}(S)$ is the ring whose underlying additive group is given by

\begin{equation*} \operatorname{gr}(S)=\bigoplus_{n\in\mathbb{N}} \frac{S_n}{S_{n-1}}, \end{equation*}

where it is understood that $S_{-1}=S_{-\infty}=\{0\}$. For the multiplication on $\operatorname{gr}(S)$, given $a\in S_m$ and $b\in S_m$, define

\begin{equation*} (a+S_{m-1})(b+S_{n-1})=ab+S_{m+n-1}\in \frac{S_{m+n}}{S_{m+n-1}} \end{equation*}

and extend by distributivity.

We recall the following known facts on the relation of a filtered ring with its associated graded ring.

Let $R$ be a ring, let $n$ be a positive integer, let $\sigma\colon R\to M_n(R)$ be a ring homomorphism and let $\delta \colon R\to R^n$ be a right $\sigma$-derivation. Consider the $(\sigma,\delta)$-free skew extension $S=R\langle x_1,\dots,x_n;\sigma,\delta \rangle$ of $R$. Set $S_{-\infty}=\{0\}$ and, for a non-negative integer $m$, set $S_m$ to be the right $R$-submodule of $S$ generated by all the monomials of $\langle X\rangle$ of length $\leq m$. Then $\{S_n\}_{n\in\mathbb{N}\cup \{-\infty\}}$ is a filtration of $S$, called its natural filtration. The next result characterizes the associated graded ring of $S$ as a free skew extension with trivial derivation.

Proof. First note that $\operatorname{gr}(S)$ is an $R$-ring with $R=\frac{S_0}{S_{-1}}$. For $i=1,\dots,n$, set $\overline{x}_i=x_i+S_0\in \frac{S_1}{S_0}$ and let $\overline{X}=\{\overline{x}_1,\dots,\overline{x}_n\}$. For each $m\geq 1$, $\frac{S_m}{S_{m-1}}$ is a free left $R$-module with basis formed by all the words of length exactly $m$ in the letters $\overline{x}_1, \dots,\overline{x}_n$. Therefore, $\operatorname{gr}(S)$ is the free right $R$-module with basis $\langle \overline{X}\rangle$. Moreover, for each $r\in R$ and $j=1,\dots, n$, we have

\begin{equation*}\begin{split} r\overline{x}_j & = \left(\sum_{i=1}^nx_i\sigma_{ij}(r) + \delta_j(r)\right) + S_0 \\ &= \left(\sum_{i=1}^nx_i\sigma_{ij}(r)\right) + S_0 \\ &= \sum_{i=1}^n\overline{x}_i\sigma_{ij}(r). \end{split}\end{equation*}

Hence, $\operatorname{gr}(S) \cong R\langle x_1,\dots,x_n;\sigma \rangle$ as $R$-rings.

4. Regularity

In this section, we shall investigate conditions for a free skew extension to be a domain.

As is well-known, given a ring $R$, a ring endomorphism $\sigma\colon R\to R$ and a right $\sigma$-derivation $\delta\colon R\to R$, the skew polynomial ring $R[x;\sigma,\delta]$ is a domain provided that $R$ is a domain and $\sigma$ is injective (cf., e.g., [Reference McConnell and Robson14, 1.2.9]). Once more than one variable is present, these conditions are no longer sufficient, as the following simple example shows.

Example 4.1. Let $k$ be a field, let $\sigma\colon k[t]\to M_2(k[t])$ be the $k$-algebra homomorphism such that

\begin{equation*} \sigma(t)=\begin{pmatrix} t&0\\0&0\end{pmatrix}, \end{equation*}

that is, if $f=\lambda_0+\lambda_1t+\dots+\lambda_nt^n\in k[t]$, with $\lambda_i\in k$, for all $i=0,\dots,n$, then

\begin{equation*} \sigma(f) =\begin{pmatrix} f&0\\0&\lambda_0\end{pmatrix}, \end{equation*}

and let $S=k[t]\langle x_1,x_2;\sigma\rangle$. Although $k[t]$ is a domain and $\sigma$ is injective, in $S$ one has

\begin{equation*} tx_2 = x_1\sigma_{12}(t) + x_2\sigma_{22}(t) = 0, \end{equation*}

but both $x_2$ and $t$ are nonzero.

Cohn, in [Reference Cohn2, Cor. 5.4.9] shows that the filtered tensor ring $R\langle M;R\rangle$ of a pointed $R$-bimodule $M$ over an integral domain $R$ is again an integral domain, provided that the quotient $R$-bimodule $\overline{M}=M/R$ is left and right flat. In the proof of this result, right flatness of $\overline{M}$ is used in order to regard the associated graded ring of $R\langle M;R\rangle$ as the tensor ring $R\langle \overline{M}\rangle$ and, then, right and left flatness of $\overline{M}$ implies that the tensor powers $\overline{M}^{\otimes r}$ of $\overline{M}$ are all right flat and left torsion-free. And these imply that $R\langle \overline{M}\rangle$ and, a fortiori, $R\langle M;R\rangle$ are domains.

As we have seen in Corollary 2.4, given a ring homomorphism $\sigma\colon R\to M_n(R)$ and a right $\sigma$-derivation $\delta\colon R\to R^n$, the free skew extension $S=R\langle x_1,\dots, x_n;\sigma,\delta\rangle$ is just the filtered tensor ring of the pointed $R$-module $M=R\oplus V$, where $V$ is the free right $R$-module with basis $\{x_1,\dots,x_n\}$. Since, as right $R$-modules $\overline{M}$ and $V$ are isomorphic, this guarantees right flatness of $\overline{M}$. The left $R$-module structure of $\overline{M}$ induces a left $R$-module structure on $V$ such that

(10)\begin{equation} rx_j=\sum_{i=1}^nx_i\sigma_{ij}(r)\text{, for all $j=1,\dots, n$ and $r\in R$.} \end{equation}

This way, $\overline{M}$ and $V$ become isomorphic $R$-bimodules. And, as we saw in the preceding paragraph, a sufficient condition for $S$ to be a domain is that the tensor powers $V^{\otimes r}$ are left torsion-free, since, as right $R$-modules they are free, and, therefore, flat. Now, the tensor powers $V^{\otimes r}$ are the homogeneous components of the tensor ring $R\langle V\rangle$, which, in turn, is isomorphic to the free skew extension $S_0= R\langle x_1,\dots,x_n;\sigma\rangle$ with zero derivation. For each positive integer $r$, we can identify $V^{\otimes r}$ with the free right $R$-submodule of $S_0$ with basis formed by the words of length $r$ in the letters of the set $X=\{x_1,\dots,x_n\}$. We shall investigate necessary and sufficient conditions on $\sigma$ for the tensor powers of $V$ to be left torsion-free.

For a positive integer $r$, denote the subset of $\langle X\rangle$ consisting of all monomials of length $r$ by $\langle X\rangle_r$. Order the free generators in the natural way, that is, declare $x_i \lt x_j$ if, and only if, $i \lt j$, and then order $\langle X\rangle_r$ lexicographically: $x_{i_1}x_{i_2}\dotsm x_{i_r} \lt x_{j_1}x_{j_2}\dotsm x_{j_r}$ if, and only if, $x_{i_1} \lt x_{j_1}$ or there exists $s\in\{2,\dots,r\}$ such that $x_{i_1}=x_{j_1}, x_{i_2}=x_{j_2},\dots, x_{i_{s-1}}=x_{j_{s-1}}$ and $x_{i_s} \lt x_{j_s}$. This defines a total ordering on the set $\langle X\rangle_r$. We shall list the $n^r$ elements of $\langle X\rangle_r$ ascendantly according to this order:

(11)\begin{equation} w_1 \lt w_2 \lt \dots \lt w_{n^r}, \end{equation}

and, in case $r\geq 2$, do the same with the $n^{r-1}$ elements of $\langle X\rangle_{r-1}$ lexicographically ordered:

(12)\begin{equation} v_1 \lt v_2 \lt \dots \lt v_{n^{r-1}}. \end{equation}

Given $k,l\in\{1,\dots,n^{r-1}\}$ and $p,q\in\{1,\dots,n\}$, it easy to see that

\begin{equation*} v_kx_p \lt v_lx_q\quad\text{if, and only if, $kl$ or ($k=l$ and $pq$).} \end{equation*}

Therefore, there are exactly $n(l-1)+q-1$ elements in $\langle X\rangle_r$ that are smaller than $v_lx_q$. It follows that $v_lx_q=w_{n(l-1)+q}$. Or, equivalently, when $r\geq 2$, given $j\in\{1,2,\dots,n^r\}$, if we let $l$ and $q$ be the integers such that $j=n(l-1)+q$ with $1\leq q \leq n$, then $w_j=v_lx_q$.

For the next lemma, we need a piece of notation. The homomorphism $\sigma\colon R\to M_n(R)$ defines inductively a sequence of ring homomorphisms $\sigma^{(r)}\colon R\to M_{n^r}(R)$ by $\sigma^{(1)}=\sigma$ and for $a\in R$, if

\begin{equation*} \sigma^{(r)}(a) = \begin{pmatrix} a_{11} &\dots& a_{1n^r} \\ \vdots & \ddots & \vdots \\ a_{n^r 1} & \dots & a_{n^rn^r} \end{pmatrix}\in M_{n^r}(R), \end{equation*}

then

\begin{equation*} \sigma^{(r+1)}(a) = \begin{pmatrix} \sigma(a_{11}) &\dots& \sigma(a_{1n^r}) \\ \vdots & \ddots & \vdots \\ \sigma(a_{n^r 1}) & \dots & \sigma(a_{n^rn^r}) \end{pmatrix}\in M_{n^{r+1}}(R). \end{equation*}

Lemma 4.2. Let $r$ be a positive integer, let $j\in\{1,2,\dots, n^r\}$, let $w_j\in\langle X\rangle_r$ be as in (11) and let $a\in R$. Then, in $S_0$,

\begin{equation*} aw_j = \sum_{i=1}^{n^r} w_i\sigma^{(r)}_{ij}(a), \end{equation*}

where, for all $i=1,\dots, n^r$, $w_i\in \langle X\rangle_r$ are listed according to (11).

Proof. We proceed by induction on $r$, with the base case $r=1$ following from (10).

Suppose that $r\geq 2$ and that the result holds for $r-1$. By what we have seen above, we can write $w_j=v_lx_q$, for some $q\in\{1,\dots,n\}$, $l\in\{1,\dots,n^{r-1}\}$ and $v_l\in \langle X\rangle_{r-1}$ as in (12). Using the induction hypothesis, we can write

\begin{equation*} aw_j = av_lx_q =\biggl(\sum_{k=1}^{n^{r-1}}v_k\sigma^{(r-1)}_{kl}(a)\biggr)x_q =\sum_{k=1}^{n^{r-1}}\sum_{i=1}^nv_kx_i\sigma_{iq}\bigl(\sigma^{(r-1)}_{kl}(a)\bigr). \end{equation*}

Since $\sigma_{iq}\bigl(\sigma^{(r-1)}_{kl}(a)\bigr)$ are precisely the elements in column $n(l-1)+q$ of $\sigma^{(r)}(a)$, the result follows.

Given $b_1,b_2,\dots,b_{n^r}\in R$, it follows from Lemma 4.2, that, in $S_0$,

\begin{equation*} a\sum_{j=1}^{n^r}w_jb_j = \sum_{i=1}^{n^r}w_i\biggl(\sum_{j=1}^{n^r}\sigma^{(r)}_{ij}(a)b_j\biggr), \end{equation*}

for all $a\in R$. This provides the following criterion:

This lemma motivates our next definition.

Equivalently, $\sigma$ is megainjective if for all $a\in R\setminus\{0\}$ and $r\geq 1$, $\sigma^{(r)}(a)$ is neither the zero matrix nor a left zero divisor in the ring $M_{n^r}(R)$. Clearly, if $\sigma$ is megainjective, then $\sigma$ is injective and $R$ is a domain. In fact, for $n=1$, a ring homomorphism $\sigma\colon R\to R$ is megainjective if, and only if, $R$ is a domain and $\sigma$ is injective.

From our discussion of the proof of [Reference Cohn2, Cor. 5.4.9] (see the paragraphs following Example 4.1), in view of Lemma 4.3, we obtain the following sufficient condition for a free skew extension to be a domain.

Megainjectivity of $\sigma$ is a sufficient condition for $S$ to be a domain, because it is a sufficient condition for its associated graded ring $S_0=R\langle x_1,\dots,x_n; \sigma\rangle$ to be a domain, but this, in view of Proposition 3.1 and Lemma 3.3, is equivalent with the pseudo-valuation $\deg$ on $S$ being a degree function. As it happens, the condition is also necessary.

Proof. What is left to prove is that if $\deg$ is a degree-function, then $\sigma$ is megainjective. Fix $a\in R$, let $r$ be a positive integer, let $L_1, \dots, L_{n^r}$ be the columns of $\sigma^{(r)}(a)$ and let $w_1 \lt w_2 \lt \dots \lt w_{n^r}$ be the monomials of $\langle X\rangle_r$. Given $b_1,b_2,\dots,b_{n^r}\in R$, by a refined version of Lemma 4.2 with a similar proof,

\begin{equation*} a\sum_{j=1}^{n^r}w_jb_j \equiv \sum_{i=1}^{n^r}w_i\sum_{j=1}^{n^r}\sigma^{(r)}_{ij}(a)b_j \ {\pmod{S_{r-1}}}, \end{equation*}

where $S_{r-1}$ denotes the right $R$-submodule of $S$ generated by all the monomials of $\langle X\rangle$ of length $\leq r-1$. Therefore,

(13)\begin{equation} \deg\biggl(a\sum_{j=1}^{n^r}w_jb_j\biggl) \lt r, \quad\text{if, and only if,}\quad L_1b_1+L_2b_2+\dots+L_{n^r}b_{n^r}=0, \end{equation}

since the elements of $\langle X\rangle_r$ are right linearly independent over $R$.

Suppose that $\deg$ is a degree function. If $a\neq 0$ and $b_1,b_2,\dots,b_{n^r}\in R$ are such that $L_1b_1+L_2b_2+\dots+ L_{n^r}b_{n^r}=0$, then, by (13),

\begin{equation*} r \gt \deg\biggl(a\sum_{j=1}^{n^r}w_jb_j\biggl) = \deg(a)+\deg\biggl(\sum_{j=1}^{n^r}w_jb_j\biggr)= \deg\biggl(\sum_{j=1}^{n^r}w_jb_j\biggr), \end{equation*}

which is only possible if $b_1=b_2=\dots=b_{n^r}=0$. It follows that $\sigma$ is megainjective.

The following are immediate consequences of the theorem.

In the next result, the group of invertible elements of a ring $T$ is denoted by $\mathcal{U}(T)$.

We end this section with examples of megainjective homomorphisms.

In view of this example and Corollary 4.7, any free skew extension of a division ring is a domain. As stated in [Reference Leroy and Merdach9, Thm. 2.8], more is true: any free skew extension of a division ring is, in fact, a free ideal ring (because it satisfies Cohn’s weak algorithm with respect to the degree-function $\deg$) and, as so, has a universal division ring of fractions.

We say that a ring homomorphism $\sigma\colon R\to M_n(R)$ is upper (respectively, lower) triangularizable if there exists an invertible matrix $P\in M_n(R)$ such that $P\sigma(a)P^{-1}$ is upper (respectively, lower) triangular for all $a\in R$. Therefore, if $R$ is a domain and $\sigma\colon R\to M_n(R)$ is upper triangularizable under conjugation by $P$ with all diagonal entries of $P\sigma(a)P^{-1}$ nonzero for all $a\in R\setminus\{0\}$, then $\sigma$ is megainjective. This is because the ring homomorphism $\tau\colon R\to M_n(R)$ defined by $\tau(a)=P\sigma(a)P^{-1}$, for $a\in R$, satisfies the condition in Example 4.9 for being megainjective and, for all $a\in R$, the matrices $\tau^{(r)}(a)$ and $\sigma^{(r)}(a)$ are conjugate over $R$.

Example 4.10. If $R$ is a right Ore domain and $\sigma\colon R\to M_n(R)$ is a ring homomorphism such that, for all $a\in R\setminus\{0\}$, the columns of $\sigma(a)$ are right $R$-linearly independent, then $\sigma$ is megainjective. This fact can be proved as follows. Suppose the Ore division ring of fractions is $\varphi\colon R\hookrightarrow Q$. Given a positive integer $s$, and a ring homomorphism $\psi\colon S\to S'$, we shall denote by $M_s(\psi)$ the homomorphism from $M_s(S)$ into $M_s(S')$ defined by $M_s(\psi)\bigl((s_{ij})\bigr)=\bigl(\psi(s_{ij})\bigr)$. If $A\in M_n(R)$ is a matrix whose columns are right $R$-linearly independent, then, clearly, the columns of $M_n(\varphi)(A)$ are right $Q$-linearly independent, and, therefore, $M_n(\varphi)(A)$ is invertible in $M_n(Q)$. Thus, there exists a homomorphism $\overline{\sigma}\colon Q\rightarrow M_n(Q)$ such that the diagram

commutes. And, therefore, we obtain the commutativity of the square on the right-hand side of

by applying the functor $M_n$ to the square on the left-hand side.

If $a\in R\setminus\{0\}$ is such that the columns of $\sigma(a)$ are right $R$-linearly independent, then, as we have seen above, $\overline{\sigma}\varphi(a)$ is invertible in $M_n(Q)$ and, hence, $M_{n^2}(\varphi)M_n(\sigma)\sigma(a)= M_{n}(\overline{\sigma})\bigl(\overline{\sigma}\varphi(a)\bigr)$ is an invertible element of $M_{n^2}(Q)$. Since $M_n(\sigma)\sigma=\sigma^{(2)}$, this implies that the columns of $\sigma^{(2)}(a)$ must be right $R$-linearly independent. An inductive argument proves that the columns of $\sigma^{(r)}(a)$ are right $R$-linearly independent for all $r\geq 1$.

5. Primeness

Here we shall show that a condition analogous to the one described in [Reference Lam, Leroy and Matczuk8, Theorem 4.4] for a skew polynomial ring to be prime holds in the case of ‘triangular’ free skew extensions.

Let $R$ be a ring and let $n$ be a positive integer. Throughout this section we will focus exclusively on upper triangular ring homomorphisms $\sigma\colon R\to M_n(R)$, that is, homomorphisms for which $\sigma_{ij}\colon R\to R$ are zero maps for all $i \gt j$. In this case, it easy to see that $\sigma_{ii}$ are ring endomorphisms of $R$, for all $i=1,\dots,n$. Also, if $\delta\colon R\to R^n$ is a right $\sigma$-derivation, then in $S=R\langle x_1,\dots,x_n;\sigma,\delta\rangle$, one has

\begin{equation*} ax_j = \sum_{i=1}^jx_i\sigma_{ij}(a) + \delta_j(a), \end{equation*}

for all $j=1,\dots,n$ and $a\in R$.

We start by defining a total ordering on the set $\langle X\rangle$ of all monomials on $X=\{x_1,\dots,x_n\}$ that extends the ordering on $\langle X\rangle_r$ used in Section 4. Once fixed that $x_1 \lt x_2 \lt \dots \lt x_n$, let $u,v\in \langle X\rangle\setminus\{1\}$ and define $u \lt v$ if $\lvert u\rvert \lt \lvert v\rvert$ or $\lvert u\rvert =\lvert v\rvert$ and $u$ precedes $v$ lexicographically, that is, $u=x_{i_1}\dotsm x_{i_r}, v=x_{j_1}\dotsm x_{j_r}$ and $x_{i_1} \lt x_{j_1}$ or there exists $s\in\{2,\dots,r\}$ such that $x_{i_1}=x_{j_1}, x_{i_2}=x_{j_2},\dots, x_{i_{s-1}}=x_{j_{s-1}}$ and $x_{i_s} \lt x_{j_s}$. Note that since $\lvert 1\rvert = 0$, we have $1 \lt w$, for all $w\in\langle X\rangle$, $w\neq 1$. Because for every positive integer $n$ there are only finitely many monomials of length $n$, the set $\langle X\rangle$ is, in fact, well-ordered by $ \lt $: the minimal element of a nonempty subset $U$ of $\langle X\rangle$ is the smallest monomial with minimal length in $U$. Moreover, $ \lt $ is a monomial ordering in the sense that if $u \lt v$, then $w_1uw_2 \lt w_1vw_2$ for all $w_1,w_2\in\langle X\rangle$.

For every $w\in\langle X\rangle$, let $S_{ \lt w}$ be the right $R$-submodule of $S$ generated by $\{v\in\langle X\rangle \mid v \lt w\}$. It is clear that $uS_{ \lt w}\subseteq S_{ \lt uw}$ for all $u,w\in\langle X\rangle$. When $\sigma$ is upper triangular, we get a symmetric inclusion, as we shall see in Lemma 5.2.

Given $0\neq f\in S$, say $f=\sum_{w\in\langle X\rangle} wa_w$, with $a_w\in R$, the monomial $\max\{w\in\langle X\rangle \mid a_w\neq 0\}$ is called the leading monomial of $f$. And, of course, if $w$ is the leading monomial of $f$, then we can write $f=wa+g$, with $a\in R\setminus\{0\}$ and $g\in S_{ \lt w}$.

Lemma 5.1. Suppose that $\sigma$ is upper triangular and let $a\in R$, $w\in\langle X\rangle$. Then,

\begin{equation*} aw \equiv w\sigma_w(a) \ {\pmod{S_{ \lt w}}}, \end{equation*}

where if $w=x_{j_1}\dotsm x_{j_r}$ then $\sigma_w = \sigma_{j_rj_r}\dotsm \sigma_{j_2j_2}\sigma_{j_1j_1}$, and $\sigma_1=\operatorname{Id}_R$.

Proof. The result is clearly valid for $w=1$. Let $w=x_{j_1}\dotsm x_{j_r}$, with $r\geq 1$. We proceed by induction on $r$. For $r=1$, since $\sigma$ is upper triangular, we have

\begin{equation*} ax_{j_1} = \sum_{i=1}^{j_1}x_i\sigma_{ij_1}(a) + \delta_{j_1}(a) = x_{j_1}\sigma_{j_1j_1}(a) + \sum_{i=1}^{j_1-1}x_i\sigma_{ij_1}(a) + \delta_{j_1}(a). \end{equation*}

So, $ax_{j_1}-x_{j_1}\sigma_{x_{j_1}}(a)\in S_{ \lt x_{j_1}}$. Suppose the result valid for $r\geq 1$ and let $w=x_{j_1}\dotsm x_{j_r}$. Then, there exists $\sum_{v \lt w} vb_v\in S_{ \lt w}$, with $b_v\in R$, such that

\begin{multline*} awx_{j_{r+1}} = wx_{j_{r+1}} \sigma_{wx_{j_{r+1}}}(a) + \sum_{i=1}^{j_{r+1}-1} wx_i\sigma_{ij_{r+1}}(\sigma_w(a)) \\ + w\delta_{j_{r+1}}(\sigma_w(a)) + \sum_{v \lt w}\sum_{i=1}^{j_{r+1}} vx_i\sigma_{ij_{r+1}}(b_v) + \sum_{v \lt w}v\delta_{j_{r+1}}(b_v). \end{multline*}

So, indeed, $awx_{j_{r+1}} - wx_{j_{r+1}} \sigma_{wx_{j_{r+1}}}(a) \in S_{ \lt wx_{j_{r+1}}}$.

Proof. For (i), let $a\in R$ and $f=\sum_{u \lt w}ub_u\in S_{ \lt w}$, with $b_u\in R$. Then, for each $u\in\langle X\rangle$, by Lemma 5.1, there exists $f_u\in S_{ \lt u}$ such that $au=u\sigma_u(a)+f_u$. Hence,

\begin{equation*} af = \sum_{u \lt w}u\sigma_u(a)b_u + \sum_{u \lt w}f_ub_u\in S_{ \lt w}. \end{equation*}

For (ii), let $g=\sum_{u \lt v}ud_u\in S_{ \lt v}$, with $d_u\in R$. Then, for each $u\in\langle X\rangle$ with $u \lt v$, by Lemma 5.1, there exists $h_u\in S_{ \lt w}$ such that $d_uw = w\sigma_w(d_u)+h_u$. Then,

\begin{equation*} gw = \sum_{u \lt v} uw\sigma_w(d_u) + \sum_{u \lt v} uh_u. \end{equation*}

Since $u \lt v$ implies $uw \lt vw$, we have $uh_u\in uS_{ \lt w}\subseteq S_{ \lt uw}\subseteq S_{ \lt vw}$. So, $gw\in S_{ \lt vw}$.

Inclusion (iii) follows from (ii).

We now show that a result analogous to [Reference Lam, Leroy and Matczuk8, Lemma 4.2] holds in this more general setting. A word on notation, given a ring $R$ and a subset $U$ of $R$, we denote the right (respectively, left) annihilator of $U$ in $R$ by $r-ann_R(U) = \{z\in R \mid Uz=0\}$ (respectively, $l-ann_R(U)=\{z\in R\mid zU=0\}$). When $U$ is a singleton, say $U=\{u\}$, we shall write $r-ann_R(u)$ for $r-ann_R(U)$ (and, similarly, for $l-ann_R(u)$).

Proof. We prove only (ii), the proof of (i) being similar. Suppose that $l-ann_S(p)\not\subseteq l-ann_S\bigl(\sigma_w^{-1}(a)\bigr)$ and take $h\in l-ann_S(p)\setminus l-ann_S\bigl(\sigma_w^{-1}(a)\bigr)$ of the form $h=uc+g$, with $u\in\langle X\rangle$, $c\in R\setminus\{0\}$ and $g\in S_{ \lt u}$ such that $u$ is minimal among the leading monomials of the elements of $l-ann_S(p)\setminus l-ann_S\bigl(\sigma_w^{-1}(a)\bigr)$. Then

\begin{equation*} 0=hp = (uc+g)(wa+f) = ucwa + ucf + gwa + gf. \end{equation*}

By Lemma 5.2, $ucf + gwa + gf\in S_{ \lt uw}$ and, by Lemma 5.1, $ucwa \equiv uw\sigma_w(c)a \pmod{S_{ \lt uw}}$. It follows that $\sigma_w(c)a=0$ and, hence, $c\sigma_w^{-1}(a)=0$. By the hypothesis, $cp=0$. But this implies that $h-uc\in l-ann_S(p)\setminus l-ann_S\bigl(\sigma_w^{-1}(a)\bigr)$, contradicting the minimality of $u$.

We remark that Theorem 5.5 holds more generally for triangularizable homomorphisms $\sigma\colon R\to M_n(R)$. More precisely, suppose that there exists an invertible $P\in M_n(R)$ such that, the map

\begin{equation*} \begin{array}{rcl} \tau\colon R & \longrightarrow & M_n(R)\\ a & \longmapsto & P\sigma(a)P^{-1} \end{array} \end{equation*}

is a ring homomorphism such that $\tau_{ij}$ is the zero map for all $i \gt j$ and $\tau_{ii}\in\operatorname{Aut}(R)$, for all $i=1,\dots,n$. Given a right $\sigma$-derivation $\delta \colon R\to R^n$, the map $\rho\colon R\to R^n$ defined by $\rho(a) = \delta(a)P^{-1}$, for all $a\in R$, is a right $\tau$-derivation. Let $S=R\langle x_1,\dots,x_n;\sigma,\delta\rangle$ and $S'=R\langle y_1,\dots, y_n;\tau,\rho\rangle$. By [Reference Martínez-Peñas12, Corollary 21], there exists an $R$-ring isomorphism $\varphi\colon S\to S'$. Thus, if $aSb\neq 0$ for all $a,b\in R\setminus\{0\}$, then, applying $\varphi$, one obtains that $a S'b\neq 0$ for all $a,b\in R\setminus\{0\}$. By Theorem 5.5, $S'$ is prime and, so, $S$ is prime.

We end by presenting sufficient conditions for a triangular free extension to be a prime ring. Because of Lemma 3.3 and Proposition 3.2, we can restrict to free skew extensions with zero derivation; more precisely, the following holds.

Let $S_0=R\langle x_1,\dots,x_n;\sigma\rangle$ be the free skew extension of a ring $R$ defined by an upper triangular ring homomorphism $\sigma\colon R\to M_n(R)$ with zero right $\sigma$-derivation. So, in $S_0$, one has

\begin{equation*} ax_j = \sum_{i=1}^jx_i\sigma_{ij}(a), \end{equation*}

for all $j=1,\dots,n$ and $a\in R$. Let $r$ be a positive integer and let $w=x_{j_1}\dotsm x_{j_r}\in\langle X\rangle_r$. Define the following subset of $\langle X \rangle$:

\begin{equation*} \langle X \rangle^w = \{v= x_{i_1}\dotsm x_{i_r}\in\langle X\rangle _r \mid x_{i_k}\leq x_{j_k}\text{, for all $k=1,\ldots,r$}\}. \end{equation*}

Given $v\in \langle X\rangle^w$, let $\sigma_v^w = \sigma_{i_rj_r}\dotsm \sigma_{i_2j_2}\sigma_{i_1j_1}$. We also denote $\sigma_1^1=\operatorname{Id}_R$ and $\langle X\rangle^1=\{1\}$.