How does Eisenstein's criterion use primeness?

Question

I'm not sure I understand this proof in Dummit-Foote.

When we reduce modulo $P$, we get $x^n=\overline{a(x)b(x)}=\overline{a(x)}\cdot\overline {b(x)}$. Let $$a(x)=x^k+\dots+a'_0\\b(x)=x^l+\dots+b_0$$

They claim $\overline a'_0=\overline b_0=\overline 0$ (or $a'_0,b_0\in P$). My explanation is this: $x^n$ divides the residues of $a(x)$ and $b(x)$ in $R/I[x]$. So $\overline a(x)=x^k, \overline b(x)=x^l$. Thus $a'_0,b_0\in P$. Is this a correct reasoning? But then the proof does not use that $P$ is prime. Or if it does use it, then how?

Also, I took a look at the proof in Artin (he only proves for $\mathbb Z$), and I also don't see how he uses that $p$ is a prime number! For reference, here is Artin's proof:

Answer 1

In Artin's proof, the primality of $p$ is used in the step:

$\overline{g}$ divides $\overline{a_n}x^n$, so $\overline{g}$ has the form $\overline{b_r}x^r$.

This step uses the fact that $\mathbb{Z}/p\mathbb{Z}$ is a field, so that $\mathbb{Z}/p\mathbb{Z}[x]$ is a unique factorization domain, which allows us to draw the above conclusion.