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