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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  • 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!🙂

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