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In this question, all rings and algebras are commutative with identity.

Let $R$ be a ring, and let $A$ be an $R$-algebra with an $R$-subalgebra $B$. Suppose that we have an $R$-algebra homomorphism $\phi: B\to R$; then we can form the tensor product $A\otimes_B R$. My question is:

If the structure maps $R\to B$ and $R\to A$ are injective, is the map $R\to A\otimes_B R$ injective as well?

My intuition says Yes: the tensor product $A\otimes_B R$ is a quotient $A / (b-\phi(b): b\in B)$, and since $\phi: B\to R$ is a ring homomorphism preserving elements of $R$, it's hard to see how this ideal could ever contain an element of $R$. But of course that's not enough to go on.

I'm particularly interested in the case where $A = R[x_1,\ldots,x_n]$ and $B = R[x_1,\ldots,x_n]^G$ for some subgroup $G\subseteq S_n$, so if it would help to use the fact that $A$ is a polynomial ring, then by all means please do.

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If $R \to A$ is injective then is it not automatic that $R \to B$ is also injective? – user38268 Nov 16 '12 at 15:54
Yes, and of course vice versa. I just didn't want to prejudice anyone by mentioning one and not the other. – Owen Biesel Nov 16 '12 at 16:11
up vote 3 down vote accepted

This is false. The ideal $(b - \phi(b))$ in $A$ can contain $1 \in R$ as soon as some $b - \phi(b)$ is a unit in $A$. Here is a concrete counterexample.

Let $R$ be a field and let $B$ be the power series ring $k[[x]]$. Let $A$ be the field of fractions of $B$, namely $k((x))$, and let $\phi : B \to R$ be the $k$-algebra homomorphism which sends $x$ to zero.

Then $A \otimes_R B = A \otimes_R B / xB = A / xA = 0$ since $x$ is invertible in $A$. Hence the map $R \to A \otimes_R B$ is not injective in this case.

In the positive direction, it is for example enough to know that $B$ has a $B$-module complement $C$ inside $A$, since then

$A = B \oplus C \Rightarrow R \hookrightarrow R \oplus (C \otimes_B R) = (B \otimes_B R) \oplus (C \otimes_B R) = A \otimes_B R$.

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I had also found the similar example $B=R[x]$, $A = B[x^{-1}]$, and $B\to R$ sending $x\mapsto 0$. Thanks for tying off this loose question! – Owen Biesel Jan 13 '14 at 8:42

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