Tensor Products and Extensions of Polynomial Ring Quotients (10.4.26)

Dummit and Foote Abstract Algebra, section 10.4, exercise 26:

Let $S$ and $R \subseteq S$ be commutative rings with $1_R = 1_S$, and let $I \subseteq R[x_1,...,x_n]$ be an ideal. Prove$$S \otimes_R (R[x_1,...,x_n]/I) ≅ S[x_1,...,x_n]/IS[x_1,...,x_n]$$as $S$-algebras.

Proof: We must prove there is an isomorphism of rings and of $S$-modules between the two. Define a mapping$$φ : S \times (R[x_1,...,x_n]/I) → S[x_1,...,x_n]/IS[x_1,...,x_n]$$$$(s,\overline{r(x)}) \mapsto \overline{\overline{sr(x)}}$$This map is well defined since$$(s,\overline{r(x)}) = (s,\overline{r(x)+i}) ⇒$$$$φ(s,\overline{r(x)}) = φ(s,\overline{r(x)+i}) - is⇒$$$$\overline{\overline{φ(s,\overline{r(x)})}} = \overline{\overline{φ(s,\overline{r(x)+i})}}$$This map is evidently $R$-balanced, and thus induces a homomorphism $\Phi$ between the targets. For each $s(x)= \overline{\overline{\sum s_ix^i}}∈S[x_1,...,x_n]/IS[x_1,...,x_n]$ we can choose a preimage $\sum s_i \otimes \overline{x^i}∈S \otimes (R[x_1,...,x_n]/I)$ so that this homomorphism is surjective. Introduce the homomorphism$$S[x_1,...,x_n] → S \otimes_R (R[x_1,...,x_n]/I)$$$$\sum s_ix^i \mapsto \sum s_i \otimes \overline{x^i}$$which is seen to factor through $IS[x_1,...,x_n]$ so that there is an inverse to $\Phi$ making it injective, as well.

Now we may demonstrate$$s\Phi(\sum s_i \otimes \overline{r_i(x)})=\sum \overline{\overline{ss_ir_i(x)}} = \Phi(s\sum s_i \otimes \overline{r_i(x)})$$so that $\Phi$ is an $S$-module isomorphism. Finally, to show the ring isomorphism, we have$$\Phi(\sum s_i \otimes \overline{r_i(x)})\Phi(\sum s_j \otimes \overline{r_j(x)})=\sum \sum \overline{\overline{s_is_jr_i(x)r_j(x)}}=$$$$\Phi(\sum s_i \otimes \overline{r_i(x)} \cdot \sum s_j \otimes \overline{r_j(x)})~\square$$