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Let $F$ be a number field and consider the cyclotomic extension $E = F(\zeta_{10})$ where $\zeta_{10}$ is a primitive 10th root of unity. Why is it true that the only primes of $F$ that ramify in in $E$ lies above 2 and 5?

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Because every prime that ramifies must divide the discriminant of the field, which in this case is 40? I am not sure (already forgotten) about how to prove this proposition, but certain it is that from the standard course of algebraic number theory, especially the theory of Kummer it comes. If the errors hide here, inform me, thanks. – awllower Dec 14 '11 at 15:43
I don't think the discriminant is 40 because the field is $F(\zeta_{10})$ not $\mathbb{Q}(\zeta_{10})$. – user21232 Dec 14 '11 at 17:20
You can use the relative discriminant over F. – Noah Snyder Dec 14 '11 at 17:38
But doesn't this require me to know the discriminant of $F/\mathbb{Q}$? – user21232 Dec 14 '11 at 17:55
Sorry for misunderstanding the question, thus giving an incorrect account for this question. I apologize here for the inconvenience as a consequence of my careless reading. – awllower Dec 15 '11 at 13:48
up vote 7 down vote accepted

In fact, the only primes that can ramify in $E/F$ are primes above 5.

This is because $E$ is the compositum of $F$ and $\mathbb{Q}(\zeta_{10})=\mathbb{Q}(\zeta_5)$ (this is true more generally for $n\equiv 2\pmod{4}$), and only 5 ramifies in $\mathbb{Q}(\zeta_5)/\mathbb{Q}$.

Added in response to the comment:

Here's the general principle in play: If $K$ is a number field, and $F$ and $G$ are two extensions of $K$ with compositum $E=FG$, and $\mathfrak{p}$ is a prime of $F$ above $p$ in $K$, then there is the following relationship between the ramification indices: $$ e_{\mathfrak{p}}(E/F)\leq e_p(G/K). $$

In your example (with $K=\mathbb{Q}$ and $G=\mathbb{Q}(\zeta_5)$) gives $$ e_{\mathfrak{p}}(F(\zeta_5)/F)\leq e_p(\mathbb{Q}(\zeta_5)/\mathbb{Q})=1 $$ for all $p\neq 5$.

The "$\leq$" occurs because $F/K$ can eat up/absorb/render redundant some of the ramification that occurs in $L/K$. Most notably, you could take $F=\mathbb{Q}(\zeta_5)$ itself when $p=5$ to get an obvious counter-example to equality.

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I was wondering, can you clarify the fact that only 5 ramifies in $\mathbb{Q}(\zeta_{5})/\mathbb{Q}$ imply that the only primes that can ramify in $E/F$ lies above 5? I can't seem to see how this follows from the fact that $E$ is the compositum of $F$ and $\mathbb{Q}(\zeta_{5})$. – user21232 Dec 14 '11 at 18:13
Sure, give me a minute to write it up. – Cam McLeman Dec 14 '11 at 18:49
Ah, it makes sense now. Thanks! – user21232 Dec 14 '11 at 19:22

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