Is $(p^{\frac{p^2+1}{p^5-1}}-x^{p^3})^{p^2}+p^{p^2+1}x-p^{p^2+\frac{p^2+1}{p^5-1}} $ eisenstein?

I could not decide whether the following polynomial is $$f(x)=(p^{\frac{p^2+1}{p^5-1}}-x^{p^3})^{p^2}+p^{p^2+1}x-p^{p^2+\frac{p^2+1}{p^5-1}}$$ an eisenstein polynomial?

If I write simply \begin{align} f(x)&=(p^{\frac{p^2+1}{p^5-1}}-x^{p^3})^{p^2}+p^{p^2+1}x-p^{p^2+\frac{p^2+1}{p^5-1}} \\ &=x^{p^{5}}-\binom{p^2}{p^2-1}p^{\frac{p^2+1}{p^5-1}}x^{p^3(p^2-1)}+\cdots+p^{p^2+1}x+p^{p^2 \cdot \frac{p^2+1}{p^5-1}}-p^{p^2+\frac{p^2+1}{p^5-1}} \end{align} To show $f(x)$ is Eisenstein, we need to show $p^2$ does not divide the constant term $\left( p^{p^2 \cdot \frac{p^2+1}{p^5-1}}-p^{p^2+\frac{p^2+1}{p^5-1}} \right)$. Clearly $p^2$ divides $p^{p^2+\frac{p^2+1}{p^5-1}}$.

So $ p^2$ must not divide $p^{p^2 \cdot \frac{p^2+1}{p^5-1}}=p^{\frac{p^4}{p^5-1}}\cdot p^{\frac{p^2}{p^5-1}}$.

I think this is true.

any comments please

Edit: As mentioned by @reuns, the polynomial $f(x)$ is not Eisenstein over $\mathbb{Q}_p$ but over some extension

Solution 1:

Your polynomial is not in $\Bbb{Z}_p[x]$.

To help your understanding: $x^2-2^{1/2}$ is not in $\Bbb{Q}_2[x]$, it is in $\Bbb{Q}_2(\sqrt2)[x]$ and in $K[x]$ for any field extension of $\Bbb{Q}_2$ containing $2^{1/2}$.

It is Eisenstein as a polynomial of $\Bbb{Q}_2(\sqrt2)[x]$ but not as a polynomial of $ \Bbb{Q}_2(2^{1/4})[x]$