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!