# Ramification in $\mathbb Q(\zeta_5, \sqrt[5]2)/\mathbb Q(\zeta_5)$

Let $F=\mathbb Q(\zeta_5,\sqrt[5]2)$ and $K=\mathbb Q$ where $\zeta_5$ is a primitive $5$th root of unity and let $p=73$ be a prime in $K$. Fix primes $\mathfrak p$ and $\mathfrak q$ above $73$ in $\mathbb Q(\zeta_5)$ and $F$ respectively.

I know that no ramification occurs in $\mathbb Q(\sqrt[5]2)/\mathbb Q$ since $X^5-2$ has distinct roots mod $73$ but would this imply that $\mathfrak q$ is unramified over $\mathfrak p$?

Yes, the only ramification in sight here is at $2$ and $5$. One can conclude that a prime foreign to these two is unramified in $F$ without looking at the factorization of $X^5-2$. Now, you notice that the first power of $73$ that is $\equiv1\pmod5$ is the fourth, so that the residue field of any prime of $\Bbb Q(\zeta_5)$ above $73$ must have $73^4$ elements. In other words, the residue-field extension degree (“$f$”) is $4$, which means that there’s only one prime of $\Bbb Q(\zeta_5)$ above $73$. The polynomial $X^5-2$ factors completely into linears over that big residue field, and you conclude that there are five primes of $F$ above $73$.
• By the big residue field do you mean $\mathcal O_F/\mathfrak q$? Would I also be right to argue like this: if $\mathfrak q/ \mathfrak p$ is ramified then since $[F:\mathbb Q(\zeta_5)]=5$, by the multiplicativity of $e$, $e_{\mathfrak q/73}$ is divisible by $5$. But since $X^5-2$ has distinct roots mod $73$, there is no ramification in $\mathbb Q(\sqrt[5]2)/\mathbb Q$ and so this is not possible as $[F:\mathbb Q(\sqrt[5]2)]=4$. Apologies for writing it out like this- I am new to the subject and I just want to get the details right. Regards – Gwan May 15 '15 at 16:21
• Yes, that’s what I mean by “big residue field”. If $\mathfrak q/\mathfrak p$ were ramified, then one of the primes of $F$ over $\mathfrak p$ would have ramification greater than $1$. Of course, in this case, the extension is normal, so that all the primes would behave similarly, so the only possibilities are one prime with either $e=5$ or $f=5$, or five primes with $e=f=1$. But I insist: if you adjoin the fifth roots of unity, the only ramification is above $5$; and if you adjoin the fifth roots of something else, the only ramification is above $5$ & the divisors of that something else. – Lubin May 15 '15 at 17:28
• (Sorry, ran out of characters.) As a result, ramification above $73$ was out of the question from the outset. – Lubin May 15 '15 at 17:30