A Question about the Proof of Eisenstein's Irreducibity Criterion Statement: Let $f(x) = a_n x^n + a_{n-1} x^{n-1}+ \cdots + a_0 \in \mathbb Z[x]$.
If there is a prime $p$ such that $p \nmid a_n, p \mid a_{n-1}, \dots,p \mid a_0$ and $p^2 \nmid a_0 $, then $f(x)$ is irreducible over $\mathbb Q$.
Proof: Let us assume that $f(x)$ is reducible over $\mathbb Q$.
Then from a previously known result $\exists\ g(x),h(x) \in \mathbb Z[x]$ such that $f(x)=g(x)h(x)$ and $1 \leq \deg g(x)<n ~~;~~ 1 \leq \deg h(x) < n$.
Let's suppose $g(x) = b_r x^r+ b_{r-1}x^{r-1}+ \cdots +b_0$ and $h(x)=c_sx^s+ c_{s-1}x^{s-1}+ \cdots + c_0$.
Then $g(x)h(x) = b_r c_s x^{r+s}+ \cdots +b_0c_0$
Now, since $p \mid a_0, p^2 \nmid a_0$ and $a_0=b_0c_0$

$p$ divides either of $b_0$ or $c_0$ . Let us assume that $p \mid b_0~~;~~p \nmid c_0~~~~.....(1)$

Since, $p \nmid a_n \implies p \nmid b_rc_s$
Let there be a least integer $t$ such that $p \nmid b_t$ and consider $a_t=b_tc_0+b_{t-1}c_1+\cdots+b_0c_t = $ Coffecient of $x^t$ in $g(x)h(x)$

Since, $p \mid a_t  : p \mid (b_tc_0) \implies p \mid c_0$ as $p \nmid b_t~~~~.....(2) $

But, From $(1) : p \nmid c_0$ which is a contradiction with the above result.
Hence, our assumption that $f(x)$ is reducible over $\mathbb Q$ is false.

My source of confusion is $(1)$. Why have we supposed that $p \mid b_0$ and $p \nmid c_0$? What happens if $t=0$ is actually the least integer which satisfies $p \nmid b_t$?
And if we instead assumed in $(1)$ that $p \nmid b_0 ;~ p \mid c_0$. Then everything falls in place?

Where am I making a mistake. Thank you for your help.

Since $b_0 c_0 = a_0$, the assumptions on $a_0$ ($p \mid a_0$ but $p^2 \nmid a_0$) implies that $p$ divides precisely one of $b_0$ or $c_0$ but not the other. Since the polynomials $g$ and $h$ are treated symmetrically up until this point, we simply make an arbitrary choice of which case to consider in the proof, with the understanding that the other case would be handled identically.
