每个不可约多项式都能用Eisenstein判定吗？

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Irreducibility check for polynomials not satisfying Eisenstein Criterion.
Is there an irreducible polynomial which cannot be judged by Eisenstein criterion

guests.mpim-bonn.mpg.de/spc/teaching/algebra2 … ein_discriminant.pdf
Proposition. If Eisenstein works for the polynomial $f(x)=a_n x^n+\cdots+a_1 x+a_0 \in$ $\mathbb{Z}[x]$, with the prime $p$, then $p \mid \Delta(f)$.

Proof. Since Eisenstein works, we have that $p \mid a_i$, for $0 \leq i \leq n-1$, and $p \nmid a_n$. Reducing modulo $p$ gives that
\[
\bar{f}(x)=\overline{a_n} x^n \quad \text { in } \mathbb{Z} / p[x] .
\]
Therefore $\bar{f}$ has a repeated root $x=\overline{0}$, modulo $p$. So we find, $\overline{\Delta(f)}=\Delta(\bar{f})=0$, in $\mathbb{Z} / p$. This means $\Delta(f)$ is divisible by $p$, as claimed.