Consider the $5$-adic quadratic extension $\mathbb{Q}_{5}(\sqrt{5})/\mathbb{Q}_{5}$.

I want to check if this extension is unramified, where unramified means that the corresponding extension of residue class fields is separable and has degree equal to the degree of the field extension.

Since the extension is finite, the residue class field extension $\lambda/\mathbb{F}_{5}$ is also finite and hence separable, because $\mathbb{F}_{p}$ is a perfect field. Then I have to show only the equality of degrees.

I could do that by actually computing valuations and checking if the value group extended a lot, leaving no degree for the residue class field extension extend, but I was wondering if there is a faster way to do that.

For example, I would need a result that said something like: the corresponding residue class field extension for the extension is $\mathbb{F}_{5}(\sqrt{5})/\mathbb{F}_{5}$, and hence I could easily check if it extended or not, for example. Is there such a thing?

  • $\begingroup$ In $\Bbb{F}_5$ you have $5=0$, so adjoining $\sqrt{5}$ seems kinda pointless. A quick rule that comes to mind is that if you adjoin a root of an irreducible Eisenstein polynomial, you automatically get ramification. Works here. $\endgroup$ – Jyrki Lahtonen Nov 16 '15 at 19:03
  • $\begingroup$ I know that adjoining $\sqrt{5}$ would be pointless. What I meant was like, if I know that I'm adjoining some element $\alpha$ to $\mathbb{Q}_{p}$ which reduced mod $p$ would not add anything, like the $\sqrt{5}$ case, can I conclude that the extension is not unramified? In other words, if I adjoin an element to $\mathbb{Q}_{p}$, the reduction of this element mod $\mathfrak{p}$ generates the residue field extension, or not necessarily? $\endgroup$ – Shoutre Nov 16 '15 at 20:45

Observe that if $L/K/\mathbb Q_p$, and $\alpha \in \mathcal O_L$ then $\overline \alpha = \alpha \pmod {\mathfrak p}\in k_L$, where $\mathfrak p$ is the maximal ideal of $\mathcal O_L$ and $L$ has residue field $k_L$.

So if $k$ is the residue field of $K$, then $k(\overline \alpha)\subset k_L$, although you won't have equality in general.

This is sometimes useful. For example, we can see that $\mathbb Q_5(\sqrt 2)/\mathbb Q_5$ is unramified because $X^2-2$ is irreducible in $\mathbb F_5$, and hence $[k:\mathbb F_5]\ge 2$; since $[\mathbb Q_5(\sqrt 2) :\mathbb Q_5] = 2$, it follows that $[k:\mathbb F_5] = 2$, and hence the extension is unramified.

In your case, this won't be of much use, as $X^2 -5$ is reducible over $\mathbb F_5$. In fact, your extension is ramified, which can be seen in two ways:

  1. As Jyrki pointed out, $X^2-5$ is Eisenstein, so the extension must be totally ramified.
  2. Alternatively, you could observe that the unique prime ideal $(5) \subset \mathbb Z_5$ factorises as $(\sqrt 5)^2$ in $\mathcal O_{\mathbb Q(\sqrt 5)}$.
  • $\begingroup$ Or the simple argument that $|\sqrt5|$ is the square root of the generator $|5|$ of the value group of $\Bbb Q_5$. So the ramification index is two. $\endgroup$ – Lubin Nov 18 '15 at 22:41

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