# $A$ integral domain implies $A[x]$ integral domain (proof check)

I'm studying for my abstract algebra course and want to prove as an exercise that if $A$ is an integral domain then $A[x]$ is an integral domain. I realized later that there is a more direct proof, but my first attempt was this one, so I decided to complete the proof anyway to exercise my knowledge of the subject. I would like it if someone could verify that the following is correct:

Suppose that there exist non-zero $f,g\in A[x]$ such that $fg = 0$. This means that the homomorphisms $$\mathcal S_a: A[x]\to A \atop \small{f(x)\mapsto f(a)}$$

map $fg$ to $0$, for every $a\in A$. If $A$ is infinite, choose $\alpha$ that is root of neither $f$ nor $g$. Then, $f(\alpha )g(\alpha ) = 0$ and $A$ isn't an integral domain. If $A$ is finite and we can't choose such an $\alpha$, necessarily $f(x) = x^n-x$ (renaming $f$ and $g$ if necessary), where $|A| = n$. From the initial hypothesis this means that $$x^ng(x)-xg(x) = 0 \Rightarrow x^ng(x) = xg(x)$$ Since $A$ is integral (and in particular, a field) this means that $$\deg g + n = \deg g + 1$$

So either $g = 0$ when we supposed it wasn't, or $n=1$ which means $A$ is the trivial ring, which we can discard I guess? Or say it contradicts the integrality of $A$ because $1\cdot 1 = 0$, but also $1=0$ so I'm not sure how to formalize this final step.

If a polynomial $f$ has every element of $A$ as roots, then all you can say is that $(x^n-x)$ divides $f$.

This is not much of a problem, as you can use $f=(x^n-x)h(x)$ and argue on $h(x)g(x)$.

Then you get, from $fg=0$, that $x^nh(x)g(x)=xh(x)g(x)$, but without considerations on the degree you can't go much further. And the formula $$\operatorname{deg}(fg)=\operatorname{deg}(f)+\operatorname{deg}(g)$$ that you're using in the argument already proves what you want by itself.

The simplest way is as follows. If $f=a_0+a_1x+\dots+a^mx^m$ and $g=b_0+b_1x+\dots+b^nx^n$ are nonzero polynomials, with leading coefficients $a_m\ne0$ and $b_n\ne0$, prove that the coefficient of $x^{m+n}$ in $fg$ is $a_mb_n$.

Your proof can be salvaged by noting that the algebraic closure $K$ of a finite field $A$ is infinite and if $fg$ is the zero polynomial, it is also the zero polynomial when considered in $K[x]$. But this is using a sledgehammer.

• Actually that last remark helps, I wanted to skip the finitude if possible. Also I didn't deduce that $(x^n-x)g(x)$ from the $\forall a ...[ \ ]$, I got it from $fg=0$. – GPerez Dec 26 '14 at 15:33
• I should have noticed the "divides $f$" part though, thanks. – GPerez Dec 26 '14 at 15:36
• @GPerez OK, I'll modify the paragraph. – egreg Dec 26 '14 at 15:36

If $A$ is an integral domain, then the degree of the product of polynomials is the sum, this is enough to deduce the exercise.

• Yes I know, but got far into the proof before I realized it, and decided why not finish it. – GPerez Dec 26 '14 at 15:23