Let $f = X^3 - 135X - 270 \in \mathbb{Q}_5[X]$ and $L$ the splitting field of $f$ over $\mathbb{Q}_5$.
Let $e$ be the ramification index of $L/\mathbb{Q}_5$. If I am not mistaken, the degree of $L/\mathbb{Q}_5$ is $6$, so $e \in \{1,2,3,6\}$.
Argument 1
Since $f \equiv X^3$ (mod $5$), we do not need to extend the residue field $\mathbb{F}_5$ of $\mathbb{Q}_5$, so $L/\mathbb{Q}_5$ is totally ramified. Or equivalently, $e=6$.
Argument 2
Let $\alpha \in L$ be a root of $f$. In the discussion of this post, we see that $L$ is also the splitting field of the polynomial $$g = \frac 1 {\alpha^3} f(\alpha x) = x^3 - \frac {135}{\alpha^2} x - \frac{270}{\alpha^3}$$.
From the same post I mentioned, the reduction of $g$ modulo the uniformizer $\alpha$ of the intermediate field $\mathbb{Q}_5(\alpha)$ is $$\bar{g} = g \mod \alpha = x^3 - 1 = (x-1)(x^2 + x + 1).$$
The second factor has no roots over $\mathbb{F}_5$ (the residue field of $\mathbb{Q}_5(\alpha)$), so in order for $\bar{g}$ to split over $\mathbb{F}_5$, we must extend the residue field, so $e < 6$.
We see here that these arguments are contradicting each other. Could you please explain me which one is correct/wrong and if one is wrong why that is the case?
Thank you!
Argument 1 is wrong. That the minimal polynomial of some generating element of the field extension reduces to something which has all its roots in the residue field proves nothing about the extension. Actually, by replacing a primitive element $\alpha$ with $p^n \alpha$ for high enough $n$, we can achieve such a minimal polynomial which just reduces to some power of $x$. As an easy example, look at the unramified extension $\mathbb Q_5(\sqrt2)=\mathbb Q_5(5\sqrt2)$ and consider the minimal polynomial $x^2-50$. Modulo $5$ it's just $x^2$.
The same flawed argument came up in the question Line of arguments for showing why some extension is quartic and unramified, compare my more detailed answer there.