Finding inertia degree and ramification index over local fields

algebraic-number-theorygalois-theorynumber theoryp-adic-number-theory

I am working with the splitting field $L/\mathbb{Q}_2$ of the polynomial
$$
h=x^4-2x^2+4\in\mathbb{Q}_2[x]
$$
and want to find the inertia degree and ramification index of $L$.

Let $w$ be the unique extension of the $p$-adic valuation to $L$. Using the discriminant and resolvent cubic, I found that the Galois group of $h$ is the Klein four-group $C_2\times C_2$, and by considering the Newton polygon, it is $w(\alpha)=1/2$ for any root $\alpha$ of $h$. This means that the ramification index $e$ is at least $2$ and thus for the inertia degree it is $f\leq 4/e=2$. Additionally, the inertia group $I$ is nontrivial, because $L/K$ is ramified. There is a canonical isomorphism $G/I\to \mathrm{Gal}\left(\lambda/\mathbb{F}_2\right)$ and hence
$$
|G|/|I|=\left|\mathrm{Gal}\left(\lambda/\mathbb{F}_2\right)\right|=\left[\lambda:\mathbb{F}_2\right]=f.
$$

I also know that $\left(|I|/|R|,2\right)=1$ for the ramification group $R\subset I$ and thus $|I|=|R|$. Here I used that we have an injection $I/R\hookrightarrow \lambda^*$. Furthermore, because of the injections $G_k/G_{k+1}\hookrightarrow\lambda,~k\geq 1$ the factor groups $G_k/G_{k+1}$ are abelian and have order a power of $2$.

However, with all this information I still couldn't determine whether $I=C_2\times C_2$ or $I=C_2$.

In Ramification in local fields it is suggested to look at the reduction $h\equiv x^4\mod 2$, which doesn't help either (or I don't see how it would). This is the case for any polynomial whose Newton polygon has nonzero slope, which I encounter multiple times.

Is there a way to determine the inertia group $I$ with the information I provided, or is there a different/better way? I mostly worked with Chapter 2 of Neukirch's Algebraic Number Theory.

In the local fields database https://math.la.asu.edu/~jj/localfields/ we can see that for the polynomial $h$ it is $e=f=2$.

Also, for the polynomial
$$
h_2=x^4+3x^2+1\in\mathbb{Q}_2[x]
$$
we have
$$
h_2\equiv x^4+x^2+1 \equiv \left(x^2+x+1\right)^2 \mod 2.
$$
From how I understood the comments in Ramification in local fields, this already implies $e=f=2$ for the stem field of $h_2$. Is this correct?
If it is correct, could you please suggest a source where I can find the involved results?

Thank you very much for your help!

Best Answer

Your two polynomials are both biquadratic, so you can determine their Galois groups simply by the classification of biquadratic extensions, without any local field theory:

  1. For $h = (x^2 - 1)^2 + 3$, we have $(a,b,c)=(1,-3,4)$. Here $b$ is not a square in $\Bbb Q_2$, and $c$ is a square, and $2(a \pm \sqrt c) = \{6, -2\}$ are not squares in $\Bbb Q_2$, so the Galois group is $C_2 \times C_2$, and the splitting field is $\Bbb Q_2(\sqrt{-2}, \sqrt{-3})$.

  2. For $h_2 = (x^2 + \frac32)^2 - \frac54$, we have $(a,b,c) = (-\frac32, \frac54, 1)$. Again $b$ is not a square, but $c$ is a square, and $2(a\pm\sqrt c) = \{-1,-5\}$ are not squares, so the splitting field is $\Bbb Q_2(\sqrt5, \sqrt{-1})$, and the Galois group is $C_2 \times C_2$.

Since these fields are made from quadratic extensions of $\Bbb Q_2$, we use Kummer theory to study them.

By Kummer theory, quadratic extensions of $\Bbb Q_2$ corresponds to non-trivial elements of $\Bbb Q_2^\times / \Bbb Q_2^{\times 2}$, which are represented by $\{2, 3, 5, 6, 7, 10, 14\}$. Here the number $d$ corresponds to the quadratic extension $\Bbb Q_2(\sqrt d)$.

The unique unramified extension among them is $\Bbb Q_2(\mu_3) = \Bbb Q_2(\sqrt{-3}) = \Bbb Q_2(\sqrt{5})$.

Therefore, $e_h = f_h = e_{h_2} = f_{h_2} = 2$.

Appendix: Classification of biquadratic polynomials

Let $K$ be a field of characteristic $\ne 2$, and let $L$ be the splitting field of $(x^2 - a)^2 - b$ with Galois group $G$. Also, let $c = a^2 - b$. Assume that $b$ is not a square in $K$.

  1. If $bc$ and $c$ are not squares, then $G = D_8$.
  2. If $bc$ is a square (then $c$ is not), then $G = C_4$ and $L = K(\sqrt{a+\sqrt b})$.
  3. If $c$ is a square (then $bc$ is not), and one of $2(a \pm \sqrt c)$ is square, then $G = C_2$ with $L = K(\sqrt b)$.
  4. If $c$ is a square (then $bc$ is not), and none of $2(a \pm \sqrt c)$ is square, then $G = C_2 \times C_2$ with $L = K(\sqrt b, \sqrt{2(a + \sqrt c})$.

Appendix: a trick

For $h$, let $y = -2x^2$, so $y^2 + y + 1 = 0$, so $y$ is a primitive cube root of unity $\omega$. This is the unique unramified quadratic extension of $\Bbb Q_2$. Now $\omega = (\omega^2)^2$ is a square, but $-2$ is not, so it remains to adjoin the square root of $-2$, which gives a ramified quadratic extension.

Therefore $e_h = f_h = 2$.

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