## Establish an isomorphism between two R-algebras

Let $S$ be a commutative ring with 1 and let $R \subseteq S$ be a unital subring. Prove that as $S$-algebras, $S[x]$ and $S \otimes_R R[x]$ are isomorphic.

Define $\varphi : S \times R[x] \rightarrow S[x]$ by $\varphi(s,r) = sr$. This mapping is certainly $R$-bilinear, and so (since $S$ is commutative) induces an additive group homomorphism $\Phi : S \otimes_R R[x] \rightarrow S[x]$ such that $s \otimes r \mapsto sr$. Note that $\Phi((s_1 \otimes r_1)(s_2 \otimes r_2)) = \Phi(s_1s_2 \otimes r_1r_2)$ $= s_1s_2r_1r_2$ $= s_1r_1s_2r_2$ $= \Phi(s_1 \otimes r_1) \Phi(s_2 \otimes r_2)$, so that $\Phi$ is a ring homomorphism. Moreover, we have $\Phi(1 \otimes 1) = 1$ and $\Phi(a (s \otimes r)) = \Phi(as \otimes r)$ $= asr$ $= a\Phi(s \otimes r)$, so that $\Phi$ is an $S$-algebra homomorphism.

Since $\Phi(s \otimes x^k) = sx^k$, $\Phi$ is surjective. Note that every simple tensor (hence every element) of $S \otimes_R R[x]$ can be written in the form $\sum s_i \otimes x^i$. If $\Phi(\sum s_i \otimes x^i) = \sum s_ix^i = 0$, then we have $s_i = 0$ for all $i$, and thus $\sum s_i \otimes x^i = 0$. In particular, $\mathsf{ker}\ \Phi = 0$, so that $\Phi$ is injective.

Thus $S[x]$ and $S \otimes_R R[x]$ are isomorphic as $S$-algebras.

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### Comments

• Emily  On November 21, 2011 at 5:22 pm

I thought that S-bilinear -> S-module homomophism by Cor. 12 in 10.4. But then why did you actually have to write out g(ax) = ag(x)? Are we not allowed to use Cor. 12?

• Emily  On November 21, 2011 at 5:33 pm

Sorry, nevrrmind, I get it now! 🙂