8
$\begingroup$

All rings I'll consider will be commutative with identity.

Given a homomorphism $f:R \to S$ we can give an $S$-module an $R$-module structure via restriction of scalars. In particular, $S$ can be thought of as an $R$-module with action $$r \circ s = f(r) \cdot s$$

I've long thought that, as an $R$-module, $S \otimes_R S \simeq S$, since it is the quotient of $S \otimes_S S$ by the ideal genreated by relations $(s_1 \circ r) \otimes s_2 = s_1 \otimes (r \circ s_2)$, or equivalently $f(r) \cdot s_1 \otimes s_2 = s_1 \otimes f(r) \cdot s_2$. Since $S \otimes_S S \simeq S$ via the map $s_1 \otimes s_2 \mapsto s_1 s_2$ it seemed like the ideal we were quotienting by was trivial.

Thinking about this though, $\mathbb{C} \otimes_R \mathbb{C} \simeq \mathbb{C} \oplus \mathbb{C}$

Are there any conditions on $R,S$ that would make $S \otimes_R S \simeq S$, as an $R$-module? For example $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q} \simeq \mathbb{Q}$ holds, as well as $\mathbb{Z}/p \otimes_{\mathbb{Z}} \mathbb{Z}/p \simeq \mathbb{Z}/p$

$\endgroup$
1
  • 2
    $\begingroup$ You have it backwards, $S \otimes_S S$ is a quotient of $S \otimes_R S$. In $S \otimes_S S$ there are more elements that you're allowed to swap between the factors so you're quotienting out by more relations than $S \otimes_R S$, hence you get a quotient of $S \otimes_R S$. In fact as $S \otimes_S S \simeq S$ the quotient map can be given as the multiplication map $S \otimes_R S \to S$ that you mentioned. $\endgroup$
    – Jim
    Jan 31, 2013 at 1:27

3 Answers 3

4
$\begingroup$

The generalizations of the two examples you gave are:

  1. If $R \to S$ is surjective.
  2. If $R \to T^{-1}R$ is a localization.

I'm sure other conditions are possible. I don't know of general conditions which classify when this does or does not hold.

$\endgroup$
2
  • $\begingroup$ Thanks Jim! This is a good start - I'll have to think about why these two cases work, and that might be enlightening $\endgroup$
    – Juan S
    Jan 31, 2013 at 1:52
  • $\begingroup$ No problem. p.s. Both these conditions make $S \otimes_R S \simeq S$ as $R$-algebras and not just as $R$-modules. $\endgroup$
    – Jim
    Jan 31, 2013 at 2:00
4
$\begingroup$

You have an exact sequence $$ 0\longrightarrow \mathrm{image}(f)\longrightarrow S \longrightarrow S/\mathrm{image}(f) \longrightarrow 0$$ which gives an exact sequence $$ \mathrm{image}(f)\otimes_RS \longrightarrow S\otimes_RS \longrightarrow \left(S/\mathrm{image}(f)\right)\otimes_RS \longrightarrow 0$$ and we know $ \mathrm{image}(f)\otimes_RS \cong S$ because in the category $\mathcal{Rings}$ of commutative rings with identity, $f(1_R)=1_S$ so the ideal $I=\ker(f)$ annihilates $S$ and w.l.o.g $R\cong R/I$ imbeds in $S$. By exactness, $\left(S/\mathrm{image}(f)\right)\otimes_RS= 0$ if and only if $S\otimes_R S\cong S$ as $S$-modules.

$\endgroup$
10
  • $\begingroup$ Is your Tor on the wrong side? Tensor is right exact. $\endgroup$
    – Jim
    Jan 31, 2013 at 4:17
  • 1
    $\begingroup$ @Jim :Of course you are right. Sorry about that complete blackout on my part. But that actually makes things easier, as $\mathrm{image}(f)\otimes_RS \cong S$ and the right exactness of $\otimes$ gives $S \rightarrow S\otimes_RS \rightarrow S/\mathrm{image}(f) \rightarrow 0$ is exact so $S\otimes_R S=S$ iff f is epic in the category of rings with 1. $\endgroup$ Jan 31, 2013 at 15:38
  • $\begingroup$ I see that we get an exact sequence $0 \to S \to S \otimes_R S \to S/\mathrm{im}(f) \otimes_R S \to 0$ (I think you missed the $\otimes_R S$ in the last term) but I don't see how the condition $S/\mathrm{im}(f) \otimes_R S = 0$ is equivalent to the condition that $f$ is epic in the category of rings. $\endgroup$
    – Jim
    Jan 31, 2013 at 16:27
  • $\begingroup$ Let me erase the original answer and replace it with a better, and hopefully more correct answer. $\endgroup$ Jan 31, 2013 at 21:50
  • $\begingroup$ The end of the answer should read "$S \otimes_R S \cong S$" (and the isomorphism should be via the arrow in the exact sequence). $\endgroup$
    – Ted
    Feb 1, 2013 at 4:27
2
$\begingroup$

Let $\varrho:R\to S$ be an homomorphism of commutative rings. Then the canonical mapping \begin{align*} &\eta:S\to S\otimes_R S& &x\mapsto(x\otimes 1) \end{align*} is surjective if and only if $\varrho:R\to S$ is a ring epimorphism.

Let \begin{align*} &\eta':S\to S\otimes_R S& &x\mapsto(1\otimes x) \end{align*} If $\varrho:R\to S$ is a ring epimorphism, then $\eta\circ\varrho=\eta'\circ\varrho$ implies $\eta=\eta'$. Consequently, $x\otimes 1=1\otimes x$ in $S\otimes_RS$ for all $x\in S$. Then for all $x,y\in S$ we have \begin{align} x\otimes y &=(x\otimes 1)(1\otimes y)\\ &=(x\otimes 1)(y\otimes 1)\\ &=(x\otimes 1)(y\otimes 1)\\ &=xy\otimes 1 \end{align} hence $\operatorname{Im}\eta=S\otimes_RS$.

Conversely, if $\operatorname{Im}\eta=S\otimes_RS$, then $\eta:S\to S\otimes_RS$ is bijective and it's inverse is \begin{align} &\mu:S\otimes_RS\to S& &x\otimes y\mapsto xy \end{align} which is thus injective. Consequently, $x\otimes 1=1\otimes x$ for all $x\in S$ thus proving $\eta=\eta'$. Now let $\xi,\xi':S\to T$ be two ring homomorphisms such that $\xi\circ\varrho=\xi'\circ\varrho$. Then there exists an $R$-algebra structure on $C$ making $\xi,\xi'$ both $R$-algebra homomorphisms. Consider the map $\xi\otimes\xi':S\otimes_RS\to C\otimes_R C$. Then \begin{align} \xi &=(\xi\otimes\xi')\circ\eta\\ &=(\xi\otimes\xi')\circ\eta'\\ &=\xi' \end{align} which proves that $\varrho:R\to S$ is a ring epimorphism.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .