Problem
Let R be a UFD, $f = a_0 + a_1 X + ... + a_n X^n \in R[X]$, where f is primitive. Suppose $p \in R$ is irreducible with $p \mid a_i$, $0 \le i \lt n$, $p^2 \nmid a_0$, $p \nmid a_n$.
To show f is irreducible, we argue by contradiction: Assume $f = gh$ where $g,h \in R[X]$, $g = b_0 + b_1 X + ... + b_r X^r$, $h = c_0 + c_1 X + ... + c_s X^s $ , where $r, s \ge 1$.
- Show that $p \mid b_0$ or $p \mid c_0$, but not both. Also show that $p \nmid b_r$ and $p \nmid c_s$.
- Suppose $p \mid b_0$. Let k be the smallest positive integer so that $p \nmid b_k$. Show that $k \lt n$ and $p \nmid a_k$.
Attempted Solution
We know that $p \mid a_0$, which means $p \mid b_0 c_0$, since $f=gh$. Because p is irreducible, and in a UFD, we know p is prime, so $p \mid b_0$ or $p \mid c_0$. If it divides both, $ p^2 \mid a_0$, which contradicts assumptions.
Since $p \nmid a_n$, $p \nmid c_s b_r$. This means $p \nmid c_s$ and $p \nmid b_r$, by negating $ p \mid ab \iff p \mid a \lor p \mid b $