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Thursday, June 20, 2013

Tensor Products and Extensions of Polynomial Ring Quotients (10.4.26)

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

MathJax TeX Test Page Let S and RS be commutative rings with 1R=1S, and let IR[x1,...,xn] be an ideal. ProveSR(R[x1,...,xn]/I)S[x1,...,xn]/IS[x1,...,xn]as S-algebras.

Proof: We must prove there is an isomorphism of rings and of S-modules between the two. Define a mappingφ : 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 homomorphismS[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 demonstrates\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

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