restriction of scalars, reference or suggestion for proof

Let $f:R \rightarrow R'$ be a ring homomorphism that is epic in the category of rings. Let $M,N$ be $R'$-modules. Why is it that a homorphism $h:M \rightarrow N$ is $R'$-linear if and only if it is $R$-linear? I have used this result in specific cases in the past (and know why it's true),e.g., if $f$ is the usual map of $R$ to $S^{-1}R$,$S$ a multiplicatively closed subset of $R$, but never in the generality I stated. But now I need the general result and would like to know why it works. A reference or a sketch for a proof would be appreciated.

Solution 1:

Recall the characterization of epimorphisms in the category of rings (see for example this question on MathOverflow):

Silver-Mazet-Isbell Zigzag Lemma for rings. Let $f\colon R\to S$ be a ring homomorphism. Then $f$ is an epimorphism in the category of rings if and only if for every $s\in S$ there exist matrices $C$, $D$, and $E$, where $C$ is $1\times n$, $D$ is $n\times n$, $E$ is $n\times 1$; $C$ and $E$ have coefficients in $S$; $CD$, $D$, and $DE$ have coefficients in $f(R)$; and $s=CDE$. This is called a "zigzag in $S$ over $f(R)$ with value $s$."

For simplicity, suppose that $R\subseteq R'$ and that $f$ is the canonical inclusion. Now I'm going to be a bit sloppy below, by identifying $M$ with its image in obvious matrix rings, but I think the idea carries through:

Suppose $h\colon M\to N$ is $R$-linear, and let $m\in M$, $b\in R'$. We want to show that $h(bm) = bh(m)$. Since $R\hookrightarrow R'$ is an epimorphism, there is a zigzag in $R'$ over $R$ with value $b$, $b=CDE$, where $D$, $CD$, and $CE$ are matrices with coefficients in $R$. Since $h$ is $R$ linear, we have: \begin{align*} h(bm) &= h(CDEm)\\\ &= CDh(Em) &\qquad&\mbox{(since $CD$ has coefficients in $R$)}\\\ &= Ch(DEm) &&\mbox{(since $D$ has coefficients in $R$)}\\\ &= C(DEh(m)) &&\mbox{(since $DE$ has coefficients in $R$)}\\\ &= (CDE)h(m)\\\ &= bh(m). \end{align*} As I said, that's a bit sloppy, since in the middle I'm actually dealing with first with $\mathcal{M}_{1\times n}(M)$ as an $\mathcal{M}_{1\times n}(R')$ module, with the induced map $h$; and later I'm dealing with $\mathcal{M}_{n\times n}(M)$ and $\mathcal{M}_{n\times 1}(M)$ over suitable matrix rings. But essentially that is what is going on.

(You can also probably see why it's called a "zigzag": you zigzag between $CD$ and $DE$ to pull the entire scalar out.)

Isbell's Zigzag Lemma (characterizing elements that are in the dominion of a subalgebra in a large class of subalgebras) can be found in

• Isbell, John R. Epimorphisms and dominions. 1966 Proc. Conf. Categorical Algebra (La Jolla, Calif., 1965) pp. 232-246 Springer, New York, MR0209202 (35 #105a)

The statement of the Zigzag Lemma for rings is incorrect in that paper. A correct version appears in:

• Isbell, John R. Epimorphisms and dominions. IV. J. London Math. Soc. (2) 1 1969, pp. 265-273. MR0257120 (41 #1774)

More generally, the collection of all $s\in S$ for which there is a Zigzag in $S$ over $f(R)$ with value $s$ is called the dominion of $f(R)$ in $S$. It is precisely the subring of $S$ given by: $$\mathrm{dom}_S(f(R)) = \Bigl\{ s\in S\Bigm| \forall T,\ \forall g,h\colon S\to T( g|_{f(R)} = h|_{f(R)}\Rightarrow g(s)=h(s))\Bigr\}.$$ The argument above shows that any module map that is $R$ linear will necessarily be $\mathrm{dom}_{R'}(f(R))$-linear; in the special case when $\mathrm{dom}_{R'}(f(R)) = R'$, which is precisely the case where $f$ is an epimorphism, the conclusion you want follows.

Solution 2:

Here's an elementary approach, following this answer.

Claim. For any ring morphism $f:R\to S$, restriction of scalars is faithful.

Proof. We have a commutative triangle formed by restriction of scalars $f^\ast$ and each forgetful functor from modules to abelian groups. Each of the latter is faithful, and if $G\circ F$ is faithful so is $F$.

Proposition. If $f:R\twoheadrightarrow S$ is a commutative ring epimorphism then restriction of scalars $f^\ast$ is full.

Proof. Let $X,Y$ be two $S$-modules and write $f^\ast X,f^\ast Y$ for their pulled back $R$-module structures. We want to prove any $R$-linear map $\varphi:f^\ast X\to f^\ast Y$ satisfies $s\varphi(x)=\varphi (sx)$. We shall do this universally by considering all such morphisms at once.

Recall ${}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y)$ is an abelian group. As any abelian group, we may consider its endomorphism ring.

This endomorphism ring admits two ring morphisms from $S$, $$S\rightrightarrows \mathsf{Ab}({}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y),{}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y))$$ with one given by $s\mapsto (\varphi(x)\mapsto s \varphi(x))$ and the other by $s\mapsto (\varphi(x) \mapsto \varphi (sx))$. These are ring morphisms because $ X,Y$ are $S$-modules. We want to show these two ring morphisms coincide.

I claim these ring morphisms coincide upon precomposing with $f:R\to S$.

$$R\to S\rightrightarrows \mathsf{Ab}({}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y),{}_{R}\mathsf{Mod}(f^\ast X,f^\ast Y))$$ Indeed this amounts to the equation $f(r)\varphi(x)=\varphi(f(r)x)$ which holds because $X,Y$ are $S$-modules. We now apply the fact $f$ is epic to conclude the two of $S$-actions were already equal, as desired.

Corollary. If $f:R\twoheadrightarrow S$ is a commutative ring epimorphism then restriction of scalars $f^\ast$ is fully faithful (consequently also conservative).