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!

  • 2
    $\begingroup$ A root of $h$ gives a tower of two quadratic extensions, not hard to find $e,f$. $\endgroup$
    – reuns
    Apr 10, 2021 at 23:51
  • 2
    $\begingroup$ 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 $e$. $\endgroup$
    – reuns
    Apr 11, 2021 at 0:00
  • $\begingroup$ 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? $\endgroup$ Apr 11, 2021 at 23:09

1 Answer 1


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 $e_h = f_h = 2$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .