Ramification and inertia degree for $\mathbb{Q}_{p}[a]$ where $0=g(a)=a^{3}+25a^{2}+a-9$

Question

The problem is to find e and f for p-adic rationals for p=2,3,5,7. Because g is not Eisenstein for each p, the extension will not be tottaly ramified and thus $3=ef\Rightarrow e=1$ and $f=3$. I feel I am missing sth. Thanks

Any suggestions on how to tackle this problem?

Answer 1

True: If $f(x)\in{\cal O}_K[x]$ is Eisenstein and $\alpha$ a root then $K(\alpha)/K$ is totally ramified.

True: If $L/K$ is totally ramified then it is generated by the root of an Eisenstein polynomial.

False: If $f(x)\in{\cal O}_K[x]$ isn't Eisenstein, $\alpha$ a root, then $K(\alpha)/K$ isn't totally ramified.

Answer 2

The extension can still be totally ramified even if $g(a)$ is not Eisenstein. Take for example the cyclotomic polynomial $\Phi_p(x) = x^{p-1} + x^{p-2} + ... + 1$ which is not Eisenstein but generates a totally ramified extension of $\mathbb{Q}_p$.

The converse is true, i.e. if $g(a)$ is Eisenstein, then the extension would have to be totally ramified.

Answer 3

One more hint or partial answer: Your polynomial clearly has a root in $\mathbb Z_3$, and after some machine-aided computation, I seem to see that the quadratic factor is irreducible with $f=2$. You should do this yourself, though.