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Consider the Galois extension $\mathbb{Q}(\zeta_{7})/\mathbb{Q}$.

I am looking for the decomposition of the prime ideal $5\mathbb{Z}$ in the integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\zeta_{7})$.

Here is what I thought:

First, since the extension is Galois, the ramification index and inertia degree of all the primes $\mathfrak{q_{i}}$ above $5\mathbb{Z}$ are equal and hence we have the formula \begin{equation} efg=6 \end{equation}

Now, suppose I don't know the "description" of the integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\zeta_{7})$. Is it still possible to know these numbers $e$, $f$ and $g$?

How can we use the decomposition and inertia groups to find these numbers? Doesn't it require that we already know the "form" of the prime ideal factors, so that we can tell if it gets fixed or not by the action of the Galois group?

If we can't, then what about passing to a completion? Still can't? If not, I feel like these groups aren't really helping since they require that we already know the decomposition of the ideal, which is what I'm looking for.

Thank you.

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  • $\begingroup$ Have you worked out the norm of $5\mathbb{Z}$? $\endgroup$ – Sam Weatherhog Oct 2 '15 at 0:17
  • $\begingroup$ Factoring the minimal polynomial of $\zeta_7$ mod $5$ should tell you how $\langle 5 \rangle$ factors in the extension. $\endgroup$ – Sam Weatherhog Oct 2 '15 at 0:31
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The (original) argument of Sam for irreducibility of $(x^7-1)/(x-1)$ over $\Bbb F_5$ is not correct. The statement is right, however. Here’s a different approach.

The degree of the irreducible factor(s) of $x^6+x^5+x^4+x^3+x^2+x+1$ is the field extension degree when you adjoin a seventh root of unity to $\Bbb F_5$. The first power of $5$ for which $7|(5^n-1)$ is $n=6$. (You’ll notice that this is the multiplicative order of $5$ in $\Bbb F_7^*$.) Thus $\Bbb F_{5^6}$ has a seventh root of unity, but no smaller field does. So the polynomial in question is irreducible.

As for Shoutre’s original question, it’s by no means necessary to find an integral basis for the ring of integers of $\Bbb Q(\zeta_7)$. The question of the splitting of $5$ is purely local, so you can pass immediately to the same question over $\Bbb Q_5$.

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  • $\begingroup$ Exactly what I was looking for! Thank you! $\endgroup$ – Shoutre Oct 5 '15 at 14:37
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I'm not sure it's possible to know the numbers $e$, $f$, and $g$ without a "description" of the ring of integers of $\mathbb{Q}(\zeta_7)/\mathbb{Q}=K$ (say). To find the factorization of $\langle 5 \rangle$ in $K$, we can use the Dedekind-Kummer theorem which essentially says that the factorization of $\langle 5 \rangle$ in $K$ is the same as the factorization of the minimal polynomial of $\zeta_7$ mod $5$. This calculation gives us:

$$ \begin{split} &&\text{ } x^6+x^5+x^4+x^3+x^2+x+1 \text{ (mod }5\text{)} \\ =&&\text{ } x^2+x+1+x^3+x^2+x+1 \text{ (mod }5\text{)} \\ =&&\text{ } x^3+2x^2+2x+2 \text{ (mod }5\text{)} \end{split} $$

Checking each of $x=0,1,2,3,4$ shows that the above polynomial is irreducible mod $5$ and this implies that $\langle 5 \rangle$ does not factor in $K$.

Although this looks as though it does not rely on $\mathbb{O}_K$ (the ring of integers), the theorem above does not hold for a finite number of primes (those primes which divide the conductor of $\mathbb{O}_K$). This means that we need to check that $5$ is not one of these exceptions and to do this we need to know what $\mathbb{O}_K$ is. In our case we are fine because $\mathbb{O}_K=\mathbb{Z}(\zeta_7)$. In general, I don't think it's that difficult to work out what $\mathbb{O}_K$ is, so there is no problem using it here to work out the factorization (i.e. you should always be able to obtain this information).

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  • $\begingroup$ I always thought that the decomposition and inertia groups were supposed to find these numbers easily when it is hard to see what happens with the ideal. If this Dedekind-Kummer theorem works for almost every prime, why would we use such groups? Only to the primes we can't use the theorem? Besides that, could you explain the passage from line (1) to (2) in the equation? All the other things I think I understand. Thank you. $\endgroup$ – Shoutre Oct 2 '15 at 11:53
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    $\begingroup$ I’m sorry, but I don’t understand the first equality in your display. $\endgroup$ – Lubin Oct 3 '15 at 3:23
  • $\begingroup$ @Shoutre, I think the proof is strongly defective. Unless Sam explains the transition from (1) to (2), I’m going to give it a downvote, an action I rarely take. $\endgroup$ – Lubin Oct 3 '15 at 3:38
  • $\begingroup$ @Lubin and Shoutre, I did think the second line probably deserved more explanation. I have used Fermat's little theorem which says that $x^{p-1}\equiv 1$ mod $p$ to reduce the polynomial. In this case, however, you could directly check that the initial polynomial is irreducible mod $5$. I have always used this theorem for factoring ideals in extensions. $\endgroup$ – Sam Weatherhog Oct 3 '15 at 9:55
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    $\begingroup$ I’m sorry, but Fermat’s Little Theorem does not apply to polynomials, rather only to elements of $\Bbb F_p$. $\endgroup$ – Lubin Oct 3 '15 at 14:32

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