Solution Theorem 1 (The Fundamental Theorem of Ring Homomorphisms): Let (R,+1,∗1) and (S,+2,∗2) be homomorphic rings with ring homomorphism ϕ:R→S. Then R/ker(ϕ)≅ϕ(R)
.
Let R4=ker(ϕ) and let Φ:R2/R4→ϕ(R2) be defined for all (a+R4)∈R2 by ϕ(a+R4)=ϕ(a)
Then ψ is well defined, for if a+R4=b+R4⟹a=b+r for some r∈R4 and so:
ψ(a+R4)=ψ((b+r)+R4)=ψ((b+R4)+R4)=ψ(b+R4) To show that R2/R4 is an isomorphism to ϕ(R2), we need to show that ψ is bijective.
Let (a+R4),(b+R4)∈R2/R4 and suppose that
ψ(a+R4)=ψ(b+R4).
Then ϕ(a)=ϕ(b)
So,
ϕ(a−b)=0
So,
a−b∈K,
So, a+R4=b+R4
Hence ψ is injective
Furthermore, ∀a∈ϕ(R2) we have that (a+I)∈R2/R4 is such that ψ(a+R4)=a.
So ψ is surjective.
Hence ψ is bijective and so R2/R4 is proven to be an isomorphism, that is
R2/R4≅ϕ(R2)
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