A proof of Eisenstein's Criterion

Question

The book I am using provides a nice proof of Eisenstein's Criterion, I'm lost on the last couple lines. The particular questions follow the proof:

Suppose the usual divisibility conditions on the coefficients in Eisenstein's Criterion. Call the prime in the supposition $p$. We want to show $f(x) \in \mathbb{Z}[x]$, a polynomial of degree $n$, is irreducible over $\mathbb{Q}$. Using a previous result we know $f(x)$ reducible over $\mathbb{Q}$ means there exist $g(x),h(x) \in \mathbb{Z}[x]$ such that $f(x)=g(x)h(x)$. Further, we know from previous results, the degree of $g$ and $h$ are less than $n$. Consider $\bar{f},\bar{g},\bar{h} \in \mathbb{Z}_p[x]$ obtained from $f(x),g(x),h(x) \in \mathbb{Z}[x]$ by reducing all coefficients modulo $p$. Then, since $p$ divides all the coefficients of $f(x)$ except $a_n$ (the coefficient of highest order), we have $a_nx^n=\bar{f}=\bar{g}\bar{h}$ Since $\mathbb{Z}_p$ is a field, $\mathbb{Z}_p[x]$ is a unique factorization domain. Thus, $x | \bar{g}(x)$ and $x | \bar{h}(x)$. So, $\bar{g}(0)=\bar{h}(0)=0$ and, therefore, $p | g(0)$ and $p | h(0)$. But then $p^2|g(0)h(0)=f(0)=a_0$ (i.e. the constant coefficient in $f$). This contradicts the assumptions we made on the coefficients.

1. Why does $x | \bar{g}(x)$ and $x | \bar{h}(x)$? Can't all the $x$'s divide one of the polynomials, for instance, $x^n|\bar{h}(x)$? I think this has to do with $\bar{f}$ being UFD, but I'm still a little unclear on this.
2. Why does $\bar{g}(0)=\bar{h}(0)=0$ imply $p | g(0)$ and $p | h(0)$?
3. I understand the contradictory statement, but I didn't know we were doing a proof by contradiction! Usually, these go by assuming the opposite of something. Where was that claim made?

Edit: Okay the answer to #3 is assuming a factorization existed.

Answer 1

The proof is by contradiction, because, in order to prove that $f$ is an irreducible polynomial, we are assuming that $f$ is not irreducible, i.e., it has a non-trivial factorization, in order to obtain a contradiction with the assumptions on $f$.

Once this is clear, we can clarify the two tricky points of the proof:

1. Notice that $\mathbb{Z}_p[x]$ is a UFD. The argument shows that $\overline{f}=a_nx^n$, therefore we have $\overline{g}=bx^r$ and $\overline{h}=cx^{n-r}$ for some $b,c,r$ such that $bc=a_n$ and $r\leq n$ (using the uniqueness of factorizations). The reason that $r$ can be neither $0$ nor $n$, i.e., $x$ divides both $\overline{g}$ and $\overline{h}$, is because $\overline{g}$ and $\overline{h}$ are polynomials of degree $<n$ (because they are obtained as reductions of polynomials of degree $<n$) and so $r<n$ and $n-r<n$. Hence $x$ divides both $\overline{g}$ and $\overline{h}$.
2. The main idea behind this is that reducing modulo $p$ and evaluating commute. In other words, $\overline{f}(0)=\overline{f(0)}$ and $\overline{g}(0)=\overline{g(0)}$. Thus $\overline{f}(0)$ implies that $\overline{f(0)}=0$ which means that $f(0)$ is divisible by $p$.