# Finding inertia degree and ramification index over local fields

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!

• A root of $h$ gives a tower of two quadratic extensions, not hard to find $e,f$. Apr 10, 2021 at 23:51
• The general algorithm is to factor $h$ monic irreducible of degree $n$ in the finite monoid $\Bbb{Z}_p[\zeta_{p^n-1}]/(p^r)[x]_{monic \ of \ \deg \le n}$ with $p^r$ large enough so that the factors lift to $\Bbb{Z}_p[\zeta_{p^n-1}][x]$ (something like $r>2s$ with $h\bmod p^s$ separable). The degree of the monic factors is $e$. Apr 11, 2021 at 0:00
• Thank you for the answer! I managed to find a quadratic subfield by adjoining the square $a:=\alpha^2$ of a root $\alpha$ of $h$. This subextension $\mathbb{Q}_2(\alpha)/\mathbb{Q}_2$ turns out to be unramified, which means the inertia degree of the splitting field of $h$ is at least $2$. And now we have $f=e=2$. Is there a specific source, where I can learn more about this algorithm? Apr 11, 2021 at 23:09

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 $$e_h = f_h = 2$$.