Generalization of Eisenstein's Criterion

Question

Let $f(X)=a_{2n+1}X^{2n+1}+\ldots+a_0\in \mathbb{Z}[X]$ with $$\begin{align*} a_{2n+1}&\not \equiv 0 \pmod p\\ a_{2n},\ldots,a_{n+1} &\equiv 0 \pmod p\\ a_n,\ldots,a_0&\equiv 0 \pmod{p^2} \\ \text{but } a_0&\not\equiv 0 \pmod{p^3} \end{align*}$$ Prove $f$ is irreducible over $\mathbb{Q}[X]$

Because the power is odd, we must have $f(X)=g(X)h(X)$ with different degrees. Suppouse without loss of generality that $\deg h=m>\deg g=r$. Let us fix the following notation: $g=\sum b_jX^j$ and $h=\sum u_j X^j$. Clearly $p\nmid b_r$ and $p\nmid u_m$.

If $p^2 $ divides the entire first coefficient of either $g$ or $h$ we get a contradiction identical to that of Eisenstein's original criterion.

Therefore, we must have $p|u_0$ and $p|b_0$. In this case let $k$ be the first such that $p\nmid u_k$ or $p\nmid b_k$. We have that the degree inequality implies $k\leq r\leq n<n+1\leq m$. Therefore $p^2|a_k$. On the other hand:

$$a_k=u_kb_0+...+u_0b_k\Rightarrow p^2| u_kb_0+u_0b_k$$

This implies $p|u_k (b_0/p)+(u_0/p)b_k$. The only way for this to happen is that $p\nmid u_k$ and $p\nmid b_k$.Therefore the first coefficient not divible by $p$ has the same accompanying degree in both $g$ and $h$. I am not sure how to draw a contradiction out of this.

EDIT: The point I didn't quite understand in the proof of the answer made me finish things off:

Taking the congruence modulo $p$ we have

$$\bar{a}_{2n+1}X^{2n+1}=\left(\sum_{j=k}^r\bar{b}_jX^j\right)\left(\sum_{j=k}^m \bar{u}_jX^j\right)$$

But that would imply $\bar{b}_k \bar{u}_k$ is the non zero term on the lefthandside. Therefore, $2k=2n+1$. But this is absurd!

Answer 1

Suppose $f$ is reducible over $\mathbb Q[X]$; since it's a polynomial in $\mathbb Z[X]$, it's reducible over $\mathbb Z[X]$. Then we can write (as in the proof of Eisenstein's condition) $$f(X)=(uX^r+pg(X))(vX^s+ph(X))$$ for integers $u$ and $v$ not divisible by $p$, polynomials $g,h\in\mathbb Z[X]$, and $r+s=2n+1$. Let $g_0$ and $h_0$ be the unit coefficients of $g$ and $h$. Without loss of generality, suppose $r\leq s$; then $r\leq n$. We have \begin{align*} a_r=[X^r]f(X) &=[X^r]\big(uvX^{r+s}+puX^rh(X)+pvX^sg(X)+p^2g(X)h(X)\big)\\ &\equiv puh_0\pmod{p^2}. \end{align*} Since $p^2\mid a_r$ as $r\leq n$, we conclude $p\mid h_0$. However, $$a_0=[X^0]f(X)=p^2g_0h_0,$$ and so $p^3\mid a_0$, a contradiction.