Question #135136
Using Fundamental theorem of homomorphism to prove that the rings R^2 and R^2/R^4 are isomorphic.
1
Expert's answer
2020-09-28T12:07:59-0400
SolutionSolution

Theorem 1 (The Fundamental Theorem of Ring Homomorphisms): Let (R,+1,1)(R, +_1, *_1) and (S,+2,2)(S, +_2, *_2) be homomorphic rings with ring homomorphism ϕ:RS.\phi: R \to S. Then R/ker(ϕ)ϕ(R)R/ker(\phi) \cong \phi (R)

.

Let R4=ker(ϕ)R_4 = ker(\phi) and let Φ:R2/R4ϕ(R2)\Phi :R_2/R_4 \to \phi(R_2) be defined for all (a+R4)R2(a + R_4) \in R_2 by ϕ(a+R4)=ϕ(a)\phi (a+R_4)=\phi(a)

Then ψ\psi is well defined, for if a+R4=b+R4    a=b+ra+R_4=b+R_4 \implies a=b+ r for some rR4r \in R_4 and so:

ψ(a+R4)=ψ((b+r)+R4)=ψ((b+R4)+R4)=ψ(b+R4)\psi (a+R_4)=\psi ((b+r)+R_4)=\psi((b+R_4)+R_4)=\psi(b+R_4)

To show that R2/R4R_2/R_4 is an isomorphism to ϕ(R2)\phi (R_2), we need to show that ψ\psi is bijective.


Let (a+R4),(b+R4)R2/R4(a+R_4), (b+R_4) \in R_2/R_4 and suppose that

ψ(a+R4)=ψ(b+R4).\psi(a+R_4)=\psi(b+R_4).

Then ϕ(a)=ϕ(b)\phi(a)=\phi(b)

So,

ϕ(ab)=0\phi(a-b)=0

So,

abK,a-b \in K,

So, a+R4=b+R4a+R_4=b+R_4

Hence ψ\psi is injective


Furthermore, aϕ(R2)\forall a \in \phi(R_2) we have that (a+I)R2/R4(a +I) \in R_2/R_4 is such that ψ(a+R4)=a\psi (a + R_4)=a.

So ψ\psi is surjective.


Hence ψ\psi is bijective and so R2/R4R_2/R_4 is proven to be an isomorphism, that is


R2/R4ϕ(R2)R_2/R_4\cong \phi(R_2)


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