Homomorphism of Polynomial Domains induced by a homomorphism of the ring of coefficients (G7)

Question

This question originates from Pinter's Abstract Algebra, Chapter 24, G7.

Let $h:\mathbb{Z}\rightarrow\mathbb{Z}_n$ be a homomorphism with kernel $K$. Define $\bar{h}: \mathbb{Z}[x]\rightarrow\mathbb{Z}_n[x]$ by

$\quad\bar{h}(a_0 + a_1x +\cdots a_nx^n) = h(a_0) + h(a_1)x +\cdots +h(a_n)x^n$

Let $n$ be a prime. Prove that if $a(x)b(x)\in \operatorname{ker}\bar{h}$, then either $a(x)$ or $b(x)$ is in $\operatorname{ker}\bar{h}$.

Attempt:

1. Given $n$ is a prime, $\mathbb{Z}_n$ is an integral domain, for if $a,b\in\mathbb{Z}_n$ and $ab\equiv 0\,(\operatorname{mod}n)$, then $n$ necessarily divides either $a$ or $b$.

2. $a(x)b(x)\in\operatorname{ker}\bar{h}$ implies $\bar{h}(a(x)b(x))=0$.

3. $h$ being a ring homomorphism implies $\bar{h}$ is a ring homomorphism. Hence,$\bar{h}\left(a(x)b(x)\right)=\bar{h}\left(a(x)\right)\bar{h}\left(b(x)\right) = 0$.

4. $\mathbb{Z}_n$ is an integral domain implies $\mathbb{Z}_n[x]$ is an integral domain. Hence either $\bar{h}\left(a(x)\right)$ or $\bar{h}\left(b(x)\right)$ must be zero; and therefore either $a(x)$ or $b(x)$ is in $\operatorname{ker}\bar{h}$.

Correct?

Answer 1

Your proof works, as long as you prove that $\bar h$ is a homomorphism itself(you have not written the details, only said you have to use the fact that $h$ is a homomorphism : but if you are confident on this, then it is ok).

In more generality, you have proved one part of the statement : "Given $\phi :R \to S$, a ring homomorphism between $R$ and an integral domain $S$, the kernel of $\phi$ is always a prime ideal of $R$". Once you prove that it is closed under addition, which is a similar proof, the kernel is shown to be a prime ideal.