Hilbert Basis Theorem applied to integral domains

Question

Was reading a solution to an exercise of the Atiyah-MacDonald "Introduction to commutative Algebra" and this passage catched my attention

"Let $A$ be an integral domain, so by H-B-T we can infer that $A[x]$ is an integral domain too"

I looked at the standard proof of H-B-T but I didn't manage to extrapolate a proof to this. Furthermore I didn't find over the net a reference to this.

Can someone provide some hints to help me prove this result?

It can be that this is a trivial corollary, my apologies in such case

Thanks in advance^^

You really don't need the Hilbert basis theorem for this. For one thing, $A$ may not be noetherian. — Zhen Lin, Nov 05 '13 at 22:29
yeah, i know the elementar proof, but the fact that according to this writer "it is a consequence of HBT shocked me :)" — Riccardo, Nov 05 '13 at 22:56
The name of the Atiyah-Mac$\boldsymbol{\mathrm d}$onald book is "Introduction to commutative algebra". — Matemáticos Chibchas, Nov 06 '13 at 02:31
Your ar right For the title, but wrong for the name :) at least my book has written MacDonald, and even google books. Anyway thanks:) — Riccardo, Nov 06 '13 at 13:47
@RicPed Now you are right! I was based on Wikipedia, not on the book cover. — Matemáticos Chibchas, Nov 07 '13 at 00:09

score 1 · Accepted Answer · answered Nov 06 '13 at 13:55

As Zhen Lin says in the comments, there's no need to apply the Hilbert basis theorem. In fact, I see no way of using it to prove the statement.

The statement is true for any integral domain $A$ and indeterminate $x$. For suppose that $A[x]$ is not an integral domain. Then there exists polynomials $f(x),g(x)$ such that $f(x)g(x)=0$. But by comparing coefficients, we deduce that that would imply the coefficients were zero-divisors, which they're not, since $A$ is an integral domain.

Hilbert Basis Theorem applied to integral domains

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