Help with Showing isomorphism in rings (Modern Algebra)?

Question

I have a problem dealing with ring isomorphism.

My question can be seen in better detail at the link below at Cramster.com:

http://answerboard.cramster.com/math-topic-5-146838-0.aspx

Answer 1

Let α:R → S be an isomorphism. Extend it to a function β:R[x] → S[x] by letting β([k=0, n]∑c_kx^k) = [k=0, n]∑α(c_k)x^k. This function has an inverse, β⁻¹([k=0, n]∑c_kx^k) = [k=0, n]∑α⁻¹(c_k)x^k, so is a bijection. By definition, we have that:

[k=0, n]∑c_kx^k + [k=0, n]∑d_kx^k = [k=0, n]∑(c_k + d_k)x^k

So from this and the fact that α is an isomorphism:

β([k=0, n]∑c_kx^k) + β([k=0, n]∑d_kx^k)
= [k=0, n]∑α(c_k)x^k + [k=0, n]∑α(d_k)x^k
= [k=0, n]∑(α(c_k) + α(d_k))x^k
= [k=0, n]∑(α(c_k + d_k))x^k
= β([k=0, n]∑(c_k + d_k)x^k)
= β([k=0, n]∑c_kx^k + [k=0, n]∑d_kx^k)

And also we have, by definition:

([i=0, n]∑c_ix^i) * ([j=0, m]∑d_jx^j) = [k=0, n+m]∑([i+j=k]∑c_id_j)x^k

So from that and the fact that α is an isomorphism:

β([i=0, n]∑c_ix^i) * β([j=0, m]∑d_jx^j)
([i=0, n]∑α(c_i)x^i) * ([j=0, m]∑α(d_j)x^j)
[k=0, n+m]∑([i+j=k]∑α(c_i)α(d_j))x^k
[k=0, n+m]∑([i+j=k]∑α(c_id_j))x^k
β([k=0, n+m]∑([i+j=k]∑c_id_j)x^k)
β(([i=0, n]∑c_ix^i) * ([j=0, m]∑d_jx^j))

Therefore, β is a homomorphism, and since we already know it is bijective, it is an isomorphism as well.

Answer 2

Sorry, I'm not sure how to do this, but as for your second question,

"I will need to show one-to-one, onto, and operation preserving for both addition and multiplication to show that R[x] and S[x] are isomorphic, correct?"

the answer is yes.