# If $A \hookleftarrow B \to R$ each contain $R$, is $R\to A\otimes_B R$ injective?

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? –  Benja 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