Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm reading Lenstra's notes on the étale fundamental group, and I've got stuck on his exercise 3.9(c). He says that if $A$ and $B$ are separable algebras over a field $K$, and $f:A \to B$ a $K$-algebra homomorphism, then the image of $f$ is precisely $\{b \in B: b \otimes 1 = 1 \otimes b\text{ in }B \otimes_A B\}$.

By his theorem 2.7, every separable $K$-algebra is a finite product of separable extension fields $K_i$ of $K$. So the morphisms of two such algebras are fairly restricted: if I'm not mistaken, if $A = \prod K_i$ and $B = \prod L_j$, and $v_i$ and $w_j$ are the respective idempotents defining these splittings, then each $f(v_i)$ is a sum of distinct $w_j$, each generating an extension field of $K_i$.

I don't know how to get from this to the claim. More generally, for any map of commutative rings $A \to B$, we have this subring $\{b\in B:b \otimes 1 = 1 \otimes b \text{ in }B \otimes_A B\}$, which I feel like I've seen in other contexts as well, though I can't remember where. Is there a high-level way of thinking about this set, or more general conditions that tell you it's equal to $f(A)$? Thank you in advance.

share|improve this question
    
One approach: you have the natural inclusion of $A$ modules $f(A) \rightarrow \{b \in B: b \otimes 1 = 1 \otimes b\}$, to prove this is an isomorphism it suffices to do it locally, which reduces you to the case that $A$ is simply a field, which makes the tensor product $B \otimes_A B$ less mysterious. –  uncookedfalcon Dec 26 '12 at 4:41

1 Answer 1

up vote 2 down vote accepted

Let $A$ be a field, $f: A \rightarrow B$ a map of rings. I claim first that in this baby case $f(A) = \{b: b \otimes 1 = 1 \otimes b \in B \otimes_A B \}$. Indeed, we have the inclusion $\subseteq$, and in the other direction, if $b \notin f(A)$ we may extend $1,b$ to basis $b^i$ for $B$ over $A$, as $b^i \otimes b^j$ form a basis for $B \otimes_A B$, in particular we have $1 \otimes b \neq b \otimes 1$.

Let's reduce to this case for the question at hand: you have the natural inclusion of $A$ modules $f(A) \rightarrow \{b \in B: b \otimes 1 = 1 \otimes b\}$, to prove this is an isomorphism it suffices to do it locally (the latter is easily seen to be an $A$ module). Indeed, suppose $A = K_1 \oplus \ldots \oplus K_n$, the spectrum of $A$ is just $\mathfrak{p}_1 \ldots \mathfrak{p}_n$, where $\mathfrak{p}_i = \oplus_{j \neq i} K_j \subseteq A$.

Fix some $\mathfrak{p}_i =: p$. Now, $A_{p}$ is just $K_i$, and $f(A)_p = f_p(A_p)$, where $f_p$ is the induced map $A_p \rightarrow B_p$. We have directly that $$\{b \in B: b \otimes 1 = 1 \otimes b \in B \otimes_A B\}_p \subseteq \{b \in B_p: b \otimes 1 = 1 \otimes b \in B_p \otimes_{A_p} B_p\}$$where we use the canonical identification $(B \otimes_A B)_p \simeq B_p \otimes_{A_p} B_p$. By our earlier analysis (of $A$ just a field) then the result follows, as all of the RHS is $f_p(A_p)$.

share|improve this answer
    
Thanks a lot. So the only important thing is that $A$ is a product of fields. For the more general question -- for what maps $f:A \to B$ does $f(A) = \{b \in B: b \otimes 1 = 1 \otimes b \in B \otimes_A B\}$ -- your last paragraph makes it clear that we can work locally. Do you have any idea when this happens more generally (or good examples for when it doesn't happen), other than $A$ a field? –  Paul VanKoughnett Dec 26 '12 at 20:23
    
I really don't know man :(. This could be a great follow up question though! –  uncookedfalcon Dec 27 '12 at 0:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.