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?
Primes that ramify in a cyclotomic extension
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0Because 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. – 2011-12-14
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0I don't think the discriminant is 40 because the field is $F(\zeta_{10})$ not $\mathbb{Q}(\zeta_{10})$. – 2011-12-14
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0You can use the relative discriminant over F. – 2011-12-14
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1But doesn't this require me to know the discriminant of $F/\mathbb{Q}$? – 2011-12-14
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0Sorry 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. – 2011-12-15
1 Answers
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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0I 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 $\mathbb{Q}(\zeta_{5})$. – 2011-12-14
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0Sure, give me a minute to write it up. – 2011-12-14
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0Ah, it makes sense now. Thanks! – 2011-12-14
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0@CamMcLeman For several reasons I find this answer amazing. I have just realized that in many situations I have found your "general principle" applied several times but without having been mentioned by the authors. Do you have maybe a reference for a complete proof? The same principal should hold also for the residual degrees right? – 2018-06-29
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0@blacksmith It's been a while since I've thought about this, but I think I recall it being called something like the Ramification Index Shifting Theorem. I'm skeptical about the same being true for residue degrees, but that is a skepticism that comes about after about 30 seconds of thought, so take it with a grain of salt. – 2018-06-29
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0Thank you very much for your time. I think you refer to Koch book. I have found the theorem but I have to say that there is something not completely clear to me about that: the setting for that theorem is $K|F$ where $F$ is a field complete with respect to a discrete valuation. So why is it true also for number fields as you claimed? – 2018-06-29
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0@blacksmith Because the ramification indices of a global field extension equal the ramification indices of the local field extension obtained by completing each field at the respective primes. – 2018-07-01