Proof: We must prove there is an isomorphism of rings and of S-modules between the two. Define a mappingφ:S×(R[x1,...,xn]/I)→S[x1,...,xn]/IS[x1,...,xn](s,¯r(x))↦¯¯sr(x)This map is well defined since(s,¯r(x))=(s,¯r(x)+i)⇒φ(s,¯r(x))=φ(s,¯r(x)+i)−is⇒¯¯φ(s,¯r(x))=¯¯φ(s,¯r(x)+i)This map is evidently R-balanced, and thus induces a homomorphism Φ between the targets. For each s(x)=¯¯∑sixi∈S[x1,...,xn]/IS[x1,...,xn] we can choose a preimage ∑si⊗¯xi∈S⊗(R[x1,...,xn]/I) so that this homomorphism is surjective. Introduce the homomorphismS[x1,...,xn]→S⊗R(R[x1,...,xn]/I)∑sixi↦∑si⊗¯xiwhich is seen to factor through IS[x1,...,xn] so that there is an inverse to Φ making it injective, as well.
Now we may demonstratesΦ(∑si⊗¯ri(x))=∑¯¯ssiri(x)=Φ(s∑si⊗¯ri(x))so that Φ is an S-module isomorphism. Finally, to show the ring isomorphism, we haveΦ(∑si⊗¯ri(x))Φ(∑sj⊗¯rj(x))=∑∑¯¯sisjri(x)rj(x)=Φ(∑si⊗¯ri(x)⋅∑sj⊗¯rj(x)) ◻
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