Splitting of primes in an extension with "unknown" ring of integers 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.
 A: 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 $\Bbb Q_5$. 
A: 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).
