# Epimorphisms and free submodules

By inspecting the accepted answer to this question

Are epimorphisms from a division ring isomorphisms ?

one obtains the following necessary condition for epimorphisms:

Let $R \le S$ be rings with identity $1\neq 0$ such that $S$ is flat as right $R$-module. If the inclusion $R \hookrightarrow S$ is an epimorphism, then for each $s \in S$ there is $r\in R, r\neq 0$ such that $rs \in R$.

Proof: The statement says that $S/R$ has no left $R$-submodule isomorphic to $R$. If it is wrong, we have an embedding $R \hookrightarrow S/R$ and hence $0 \neq S = S \otimes_R R \hookrightarrow S\otimes_R S/R$ But by the quoted answer, $S\otimes_R S/R=0$ if $R \to S$ is epi. $\blacksquare$

However, without success I tried to drop the flatness condition. Whence my question:

Question: Is the statement above still valid, if $S$ is not supposed to be flat as right $R$-module ?

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Are you sure your proof is correct? I do not think that an (arbitrary) embedding of left $R$-modules $R^2 \hookrightarrow S$ induces $R \hookrightarrow S/R$. There are rings $R$ with $R \simeq R^2$ as left $R$-modules (see e.g. Lam, Lectures on Modules and Rings, Example (1.4)), and then $id: R = S$ is a counter-example, right? –  Torsten Schoeneberg Feb 11 '13 at 14:27
Torsten, thanks, you are right, I just need that $S/R$ has no $R$-submodule isomorphic to $R$. But this leads to an even nicer necessary condition. –  tj_ Feb 11 '13 at 17:17
This is indeed nicer. Certainly you want $r \neq 0$. Have you looked through the famous [Samuel seminar on epimorphisms][1]? In Mazet's exposé (p. 2-06), your assertion is proven for $R$ a commutative domain. [1]: numdam.org/numdam-bin/feuilleter?id=SAC_1967-1968__2_ –  Torsten Schoeneberg Feb 11 '13 at 23:29
I assume that you want $rs \neq 0$ as well? –  Manny Reyes Mar 4 '13 at 15:05
@Manny Reyes: I don't think so. The statement is strictly weaker than the similar-looking "$S$ is an essential extension of $R$ (as left $R$-modules)", which would read: For each $s \in S$ there is $r \in R$ such that $rs \in R \setminus \lbrace 0 \rbrace$. This is not true for all epimorphic extensions: Consider $\mathbb{Z} \hookrightarrow \mathbb{Q} \times \mathbb{Z}/2$). (It is true e.g. for localisations.) –  Torsten Schoeneberg Mar 4 '13 at 15:18

A simple counterexample can be constructed using free objects:

For a set $X$ let $\mathbb{Z}\langle X\rangle$ be the free ring on $X$ and let $\mathbb{Z}F(X)$ be the group ring of the free group $F(X)$ on $X$. The inclusion $X \hookrightarrow F(X)$ induces a ring homomorphism $i: \mathbb{Z}\langle X\rangle \to \mathbb{Z}F(X)$ which is an epimorphism. If $x\neq y$ are in $X$ then $xy^{-1} \in \mathbb{Z}F(X)$ and there is obviously no "polynomial" $f \in \mathbb{Z}\langle X\rangle$ such that $fxy^{-1} \in \mathbb{Z}\langle X\rangle$.

To see that $i$ is an epimorphism, note that $\mathbb{Z}F(X)$ is generated (as a ring) by $x,x^{-1}\;(x \in X)$ and a unitary ring homomorphism $\varphi: \mathbb{Z}F(X) \to T$ satisfies $\varphi(x^{-1})=\varphi(x)^{-1}$. Hence $\varphi$ is determined by its values on $X$.

Update: The property in question holds true for commutative rings.

Proof: At first assume this has already been proved for zero-dimensional local commutative rings. Let $R \le S$ be an epimorphic extensions of comm. rings and choose a minimal prime $\mathfrak{p}$ of $R$ (exists by Zorn's lemma). $R_\mathfrak{p}$ is a local ring of dimension $\text{ht}(\mathfrak{p})=0$. Since $R\setminus \mathfrak{p}$ is a multiplicative subset of $S$ we can localize and obtain a comm. diagramm $$\begin{array}{ccc} R & \xrightarrow[\scriptstyle\text{epi}]{i} & S\;\;\; \newline {\scriptstyle\text{epi}}\downarrow & & \downarrow\scriptstyle\text{epi} \newline R_\mathfrak{p} & \xrightarrow[i_\mathfrak{p}]{} & (R\setminus \mathfrak{p})^{-1}S \end{array}$$ Hence $i_\mathfrak{p}$ is epi and it's easy to see that $i_\mathfrak{p}$ is mono as well.

Let $s \in S$. By our assumption there are $r_i/t_i \in R_\mathfrak{p},\;r_1/t_1 \neq 0$ such that $\frac{r_1}{t_1}\frac{s}{1}=\frac{r_2}{t_2}$, i.e. there is $t \in R\setminus \mathfrak{p}$ with $(tt_2r_1)s=tt_1r_2 \in R$. Moreover $tt_2r_1 \neq 0$ (because $r_1/t_1 \neq 0$ in $R_\mathfrak{p}$ just says there is no $t \in R\setminus \mathfrak{p}$ such that $tr_1=0$ in $R)$ and we are done.

Now suppose $R$ is zero-dimensional local comm. with max. ideal $\mathfrak{m}$. Again from the comm. diagramm $$\begin{array}{ccc} R & \xrightarrow[\scriptstyle\text{epi}]{i} & S\;\;\; \newline {\scriptstyle\text{epi}}\downarrow\;\; & & \downarrow\scriptstyle\text{epi} \newline R/\mathfrak{m} & \xrightarrow[i_\mathfrak{m}]{} & S/\mathfrak{m}S \end{array}$$ we conclude that $i_\mathfrak{m}$ is epi and since $R/\mathfrak{m}$ is a field, $i_\mathfrak{m}$ is an isomorphism (see the link in the OP's question). Hence each $s \in S$ can be written as $$s=r + \sum_{i=1}^l m_is_i\qquad (r\in R, s_i \in S, m_i \in \mathfrak{m}, m_i \neq 0)$$ We want to show that there is $r_0 \in R,\;r_0 \neq 0$ such that $r_0s\in R$. If $l=0$ then $s=r\in R$ and we are done. Otherwise, since $\mathfrak{m}=\sqrt{0}$ there is $n_1> 0$ maximal such that $m_1^{n_1} \neq 0$ and multiplying yields $$m_1^{n_1}s=m_1^{n_1}r + \sum_{i=2}^l (m_1^{n_1}m_i)s_i.$$ If $m_1^{n_1}m_2=0$ ignore the corresponding summand. Otherwise, there is $n_2>0$ maximal such that $m_1^{n_1}m_2^{n_2}\neq 0$. Multiplying again yields $$m_1^{n_1}m_2^{n_2}s=m_1^{n_1}m_2^{n_2}r + \sum_{i=3}^l(m_1^{n_1}m_2^{n_2}m_i)s_i.$$ Proceeding this way, we obtain the required $r_0$ in the form $r_0 = \prod_{j=1}^k m_{i_j}^{n_{i_j}}.\;\;$ qed.

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Very nice. Just a comment that the free ring on $X$ can also be seen as the monoid ring of the free monoid on $X$; the ring epimorphism comes about by applying the monoid ring functor (which, being a left adjoint, preserves epis) to the epic inclusion of the free monoid into the free group. Of course, this is clear enough from what you wrote. –  Todd Trimble Mar 5 '13 at 0:30
Todd, thanks for clarifying the conceptual background why $i$ is epi (actually I wasn't very satisfied with my reasoning above, because I felt there should be a less computational argument). –  Ralph Mar 5 '13 at 1:13
Something I had found puzzling about the OP was the asymmetry of the condition: why does the factor $r$ appear just on the left? I.e., why not consider the condition that for all $s \in S$, there exist non-zero $r$, $r'$ such that $rsr' \in R$? But of course, with a slight adjustment, your example shoots that down as well (consider $xy^{-1}x$ instead of $xy^{-1}$). –  Todd Trimble Mar 5 '13 at 4:44
Just to comment that Manny and Ralph's examples are very similar in a sense, in that both are examples of the natural homomorphism from a ring to its universal localization with respect to a set of maps between finitely generated projective modules, which is always a ring epimorphism. –  Jeremy Rickard Mar 5 '13 at 13:04
@Todd: You're right. I could also have said: If $S$ is flat as $R$-module (either from the left or the right) then for each $s \in S$ there are $r,r' \neq 0$ in $R$ such that $rsr' \in R$. But I think which version one more likes is a matter of taste. –  tj_ Mar 5 '13 at 16:55

Let $k$ be a field, let $V = k \oplus k \cdot x$ be the subspace of $k[x]$ spanned by $1$ and $x$, and consider the following inclusion of rings: $$R := \begin{pmatrix} k & V \newline 0 & k \end{pmatrix} \subseteq \ \mathbb{M}_2(k[x]) =: S.$$ I learned from George Bergman that the inclusion $R \hookrightarrow S$ is an epimorphism (proof below), and this flavor of example has served to help me understand just how strangely behaved noncommutative ring epimorphisms can be.

Assuming for the moment that this is indeed an epimorphism, let's see that the answer to your question is no. If we simply take $s = x^2 I_2 = \left(\begin{smallmatrix}x^2 & 0 \newline 0 & x^2\end{smallmatrix}\right)$, then it's quite clear that for all $r \in R$ (and indeed, even all $r \in S$) we have $rs \in R \iff r = 0$.

So why is $R \hookrightarrow S$ an epimorphism? As you seem well aware, this is the case if and only $1 \otimes s = s \otimes 1$ in $S \otimes_R S$ for all $s \in S$. It suffices to check that this holds for a generating set of $S$ over $R$. One such generating set consists of $E_{21}$, $xE_{11}$, and $xE_{22}$.

Let's compute with the first of those three generators, recalling that this is in $S \otimes_R S$: \begin{align*} E_{21} \otimes 1 &= E_{21} \otimes E_{11} + E_{21} \otimes E_{22} \newline &= E_{21} \otimes E_{11} + E_{21} \cdot E_{22} \otimes 1 \newline &= E_{21} \otimes E_{11} \newline &= E_{21} \otimes E_{12} E_{21} \newline &= E_{21} E_{12} \otimes E_{21} \newline &= E_{22} \otimes E_{21} \newline &= 1 \otimes E_{22} E_{21} \newline &= 1 \otimes E_{21}, \end{align*} as desired. (It still feels like magic to me!)

Now let's try the second generator: \begin{align} xE_{11} \otimes 1 &= (xE_{12})E_{21} \otimes 1 \newline &=xE_{12} \otimes E_{21} \qquad \mbox{(as computed above)}\newline &= 1 \otimes (xE_{12}) E_{21} \qquad \mbox{(since xE_{12} \in R)}\newline &= 1 \otimes xE_{11}. \end{align} The computation for $xE_{22}$ is similar.

Now, if you're looking to restrict to commutative rings, then I have no idea what might change...

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Nice counter-example. Thanks. –  tj_ Mar 9 '13 at 12:21