# How calculate Galois Group of $\mathbb{Q}_7(\zeta_3,\sqrt{3})/\mathbb{Q}_7$

From local field theory I know that if $L/K$ is a Galois extension of number fields, $\mathfrak{P}$ is a prime ideal of $L$ living above a prime $\mathfrak{p}$ of $K$, then the extension $$L_{\mathfrak{P}}/K_{\mathfrak{p}}$$ is Galois with Galois group $\mathcal{G}=\mathcal{D_{\mathfrak{P}/\mathfrak{p}}}$ whith $\mathcal{D_{\mathfrak{P}/\mathfrak{p}}}$ the decomposition group of $\mathfrak{P}/\mathfrak{p}$.

Now, I would calculate the Galois group of the extension $\mathbb{Q}_7(\zeta_3,\sqrt{3})/\mathbb{Q}_7$. To do this I think I have to find the prime ideals $\mathfrak{P}_i$ of $\mathcal{O}_{\mathbb{Q}(\zeta_3,\sqrt{3})}$ that live above the prime ideal $(7)$ of $\mathcal{O}_{\mathbb{Q}}=\mathbb{Z}$.

Is it correct? How can I find this ideals?

thanks in advance for the support, and if it was possible I would appreciate hints rather than directly solutions.

edit: thanks to peter a g we have $\mathbb{Q}_7(\zeta_3,\sqrt{3})\simeq \mathbb{Q}_7(\sqrt{3})$, since $\hat{\mathbb{Z}}_7 \subset \mathbb{Q}_7$ contains the roots of $x^3-1$ by Hensel's lemma. Moreover since $3$ is not a square $\mathrm{mod}\:7$ then the ideal $7\mathbb{Z}[\sqrt{3}]=7 \mathcal{O}_{\mathbb{Q}(\sqrt{3})}$ is prime with inertia degree $2$. Therefore the decomposition group is $\mathbb{Z}_2= \mathcal{G}$.

I hope the reasoning is correct. However if I take $\mathbb{Q}_7(\zeta_3,\sqrt[3]{3})/\mathbb{Q}_7$ how can i find the primes living over $(7)$ in $\mathbb{Q}(\zeta_3,\sqrt[3]{3})$? Or there exist a faster way to compute the Galois group of finite local fields extension?

For this latest question I made the followin reasoning:

The ideal generated by $7$ in $\mathbb{Z}(\zeta_3)=\mathcal{O}_{\mathbb{Q}(\zeta_3)}$ split in two primes by basic theory.

On the other hand the ideal $7\mathcal{O}_{\mathbb{Q}(\sqrt[3]{3})}$ remains prime since $x^3-3$ is irriducible in $\mathbb{F}_7$. Indeed it can be shown that if $\alpha=\sqrt[3]{m}$ with $m$ cubefree and $p$ is a prime such that $p\neq 3$ and $p^2\nmid m$ then the number of factor and the ramification index that occur in factorization of the ideal $p\mathcal{O}_{\mathbb{Q}(\alpha)}$ depends on the way $x^3-m$ factorize $\mathrm{mod}\:p$. So we have that the inertia degree is $3$.

Therefore we have at least two primes in $\mathbb{Q}(\zeta_3,\sqrt[3]{3})$ living above $(7)$ each with inertia degree at least $3$. But the well known formula that related the degree extension with the splitting of a prime implies that there exist only two primes with inertia degree $3$. In particular we obtain that the decomposition group is $\mathbb{A}_3$.

Does it make sense?

• As an idea for how to start off: look at the roots of $x^3-1$ in $\mathbb F_7$. – peter a g May 9 '16 at 13:05
• The point of the previous - do you know Hensel's lemma? – peter a g May 9 '16 at 13:18
• thanks! Yes I know and since I know that there exist a factorization in $\mathbb{F}_7[x]$ I can lift it in $\mathbb{Z}_7[x]$. In this case the factorization in the residue field is $(x+3)(x+5)(x+6)$. am I right? – user261123 May 9 '16 at 13:25
• Yes - though a bit nicer way of presenting the roots would be as the subgroup of order 3 $\{1,2,4\}$ of the cyclic group of order 6 $\mathbb F^*_7$... In any case, that means your extension is only $\mathbb Q_7(\sqrt 3)$... – peter a g May 9 '16 at 13:34
• You should probably ask another (i.e. submit a separate) question, and state it precisely. Do you want an explicit description of the decomposition of primes (i.e., not just above $7$) in those field extensions? sorry - I have to run now, so will only look again this evening. If just above a fixed prime e.g. $7$, not so hard, obviously (Dedekind, e.,g)... – peter a g May 9 '16 at 15:43