Loading [MathJax]/jax/output/HTML-CSS/jax.js

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×(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)=¯¯sixiS[x1,...,xn]/IS[x1,...,xn] we can choose a preimage si¯xiS(R[x1,...,xn]/I) so that this homomorphism is surjective. Introduce the homomorphismS[x1,...,xn]SR(R[x1,...,xn]/I)sixisi¯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(ssi¯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)) 

No comments:

Post a Comment