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.
