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Let $A$ be a commutative ring with $1$, and $B\subseteq A$ be a subring.

Is there a simple condition on $B$ and $A$ guaranteeing that $B\to A\rightrightarrows A\otimes_B A$ is an equalizer?
In other words, when does $a\otimes_B 1 = 1\otimes_B a$ imply that $a\in B$?

This always holds, for example, if $B=K$ is a field and $A=L$ is a finite separable extension of $K$. For then, if $\ell\otimes 1 = 1\otimes \ell$ in $L\otimes_K L$, the same clearly holds in $\bar K\otimes_K L$, where $\bar K$ is a separable closure of $K$ containing $L$. But the $\bar K$-algebra homomorphism $\bar K\otimes_K L\to \bar K^{\mathrm{Hom}_K(L,\bar K)}$ sending $\alpha\otimes\ell\mapsto (\alpha\cdot s(\ell))_{s:L\to\bar K}$ is an isomorphism, and the equation becomes $\ell = s(\ell)$ for all $s:L\to \bar K$. But this means $\ell$ is fixed by the action of the absolute Galois group of $K$ (which acts transitively on the $s$), so $\ell\in K$.

Does the result hold for general $B$ and $A$? If not, is there a simple condition describing when it does hold? Or, failing that, is there a simpler proof in the case of a separable field extension, that doesn't need Galois theory?

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If $B=\mathbb Z$ and $A=\mathbb Q$ then the equalizer is the whole $\mathbb Q$. This is a simple exercise, which an of course be generalized to localizations of arbitrary commutative domains. –  Fernando Muro Sep 1 '12 at 23:51
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I think this question is equivalent to "what are the regular monos in the category of commutative rings?". Not that that makes the answer any more obvious to me. –  Mike Shulman Sep 2 '12 at 0:11
    
Well, adding to what Fernando and Mike said, a regular mono which is an epimorphism (localizations being examples of epis in the category of commutative rings) is automatically an isomorphism. @Owen: I seem to have seen this topic discussed by Joyal and Tierney in the context of descent theory, in their An Extension of the Galois Theory of Grothendieck. (But I don't have that handy to follow up on for an answer.) –  Todd Trimble Sep 2 '12 at 0:18
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This holds when $A$ is fully faithful over $B$. This is a well known lemma in descent theory, and always holds when $B$ is a field. –  Angelo Sep 2 '12 at 4:24

3 Answers 3

up vote 8 down vote accepted

I think what you want is exactly that the dominion (in the sense of Isbell) of $B$ in $A$ be equal to $B$.

Recall that if you are in a category of algebras (in the sense of Universal Algebra), and $B\subseteq A$ is a subalgebra of $A$, then the dominion of $B$ in $A$ (relative to the category of context) is the set $$\{a\in A\mid \forall C\forall f,g\colon A\to C (f|_B=g|_B\implies f(a)=g(a))\}.$$

If $x$ lies in the dominion of $B$ in $A$, then the fact that the two embeddings of $A$ into $A\otimes_B A$ agree on $B$ implies that they agree on $x$; that is, $x\otimes 1=1\otimes x$. Conversely, suppose that the two embeddings of $A$ into $A\otimes_B A$ agree on $x$; if $C$ is any commutative ring and $f,g\colon A\to C$ are two maps that agree on $B$, then the universal property of $A\otimes_B A$ guarantees a homomorphism $\Phi\colon A\otimes_B A\to C$ such that $f = \Phi\circ\lambda$ and $g=\Phi\circ\rho$, where $\lambda$ and $\rho$ are the left and right embeddings; then since $\lambda(x)=\rho(x)$, we conclude that $f(x)=g(x)$. Hence any element of the equalizer is in th dominion.

A characterization of dominions in the category of commutative rings is given in the Isbell-Mazet-Silver Zigzag Theorem: the dominion of $B$ in $A$ consists precisely of the elements of $A$ that can be written in the form $XYZ$, where $X$ is a row matrix, $Z$ is a column matrix, $Y$ is a square matrix of the appropriate size, $X$, $Y$, and $Z$ have entries in $A$, and $XY$ and $YZ$ have entries in $B$.

(See also this previous post.)

So the inclusion of $B$ is the equalizer of the left and right embeddings if and only if $B$ is equal to its own dominion in $A$.

(Essentially, when the category is right-closed, the dominion of $B$ in $A$ is always equal to the equalizer of the two embeddings $A\to A\amalg_{B}A$; in the category of commutative rings the tensor product functions as a binary coproduct, hence the dominion of $B$ in $A$ is the equalizer of the two embeddings $A\to A\otimes_B A$; but this description does not actually answer your question, it just asks it in a different language; so the actual answer you want is contained in the Isbell-Mazet-Silver Zigzag Theorem, when all is said and done.)

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Yes, yes ... the dominion. I had forgotten about that. Thanks for writing this up! –  Todd Trimble Sep 2 '12 at 4:31

For an analysis for which rings $R$ the sequence $R \to A \rightrightarrows A \otimes_R A$ is exact for every $R \subseteq A$, see the paper "Epimorphismen von kommutativen Ringen" by Hans Storrer. There these rings are called "dominant".

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Commentarii Mathematici Helvetici Volume 43, Number 1 (1968), 378-401, DOI: 10.1007/BF02564404 –  Todd Trimble Sep 8 '12 at 14:52
    
I'll just note that in the language of dominions, we would say that $R$ is "absolutely closed", meaning that it is dominion-closed in any ring that contains it. –  Arturo Magidin Sep 8 '12 at 20:38

I've not ground this out, but:

For $(x,y)\in A\times A$, let [x,y] be the corresponding generator in the free $B$-module over the set $A\times A$.

Then if $x\otimes1-1\otimes x=0$, there must be a relation in the free module of the form

$$ [x,1]-[1,x]= \sum_i [a_i,r_ib_i]+\sum_i [a_i,r_ic_i]-\sum_i[r_ia_i,b_i+c_i] +...$$

where the ... consists of the other generators of the multilinearity condition (e.g. terms analogous to the above with the $b$'s and $c$'s moved to the left and the $a$'s to the right). (The $r$'s are in $B$ and the other variables are in $A$).

Now if you move terms across the equal sign to make all the signs positive, then each individual term on the left must be identical to some corresponding term on the right (because these are generators of a free module). Thus, for example, $[x,1]$ might equal $[ a_1, r_1b_1 ]$ .

There are a finite (but messy) number of ways this can happen; if you work through all of them you should get a complete catalogue of all the ways you can get $x\otimes 1-1\otimes x=0$.

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