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In the following suppose L/K is a finite Galois extension of number fields, (maybe it works for other cases also, I don't know) By the Chebotorev density theorem when Gal(L/K) is cyclic, there are infinitely many primes in K that stay inert during this extension (cf Janus p136, Algerbaic Number Fields.) When L/K is non cyclic, an exercise from Neukirch (somewhere in Chap I) says there are at most finitely many primes that stay inert. I want to say that there are none. The reason is by a cycle description from Janus, p101, Prop 2.8,

In short, that proposition says when $\delta:=Frob(\frac{L/K}{\beta})$, $\beta|p$ is a prime in L, consider $\delta$ act on the cosets of H in G, H=Gal(L/E), $K\subset E\subset L$, then every cycle of length i corresponds to a prime factor in E with residue degree i. In particular, for inert guys we want there is only one cycle in the action. When we take H to be trivial, E=L is Galois over K, and the cosets are just the elements of G themselves. So we want that there exist an element (the Frobenius element above p) act transitively on G, thus G is cyclic.

I wonder if this is true, then more people should have been aware of it. If it is not, is there a counter example?

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The answer to this question is contained in my answer to a previous question of Ben Webster: mathoverflow.net/questions/10940. What's the SOP here? Do we close the question when this happens or not? –  Pete L. Clark Jan 20 '10 at 3:09

3 Answers 3

up vote 9 down vote accepted

If $L/K$ is a finite, Galois extension of number fields such that $Gal(L/K)$ is not cyclic, then no prime of K remains inert L. Indeed, one always has an isomorphism $D_p/I_p\cong Gal(L_p/K_p)$ of the Decomposition group modulo the Inertia group with the Galois group of the corresponding residue field extension. The latter group is the Galois group of a finite extension of finite fields, hence is cyclic. If the prime p were to remain inert in L, then by definition the Inertia group would be trivial and the Decomposition group would be all of $Gal(L/K)$. But this would imply that $L/K$ was a cyclic extension - a contradiction.

[Edit] I can't help but mention a cute application of this. Let $n$ be any positive integer for which $(\mathbb{Z}/n\mathbb{Z})^*$ is not cyclic. Then the cycotomic polynomial $\Phi_n(x)$ is reducible modulo $p$ for every rational prime $p$. Indeed, suppose that $p$ is a rational prime for which $\Phi_n(x)$ is irreducible modulo $p$. Then by the Dedekind-Kummer theorem, $p$ is inert in the cyclotomic field $\mathbb{Q}(\zeta_n)$. Then the Galois group of the residue class field extension, which is cyclic, is isomorphic to the Decomposition group, which in this case is $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong(\mathbb{Z}/n\mathbb{Z})^*$. But the latter group is not cyclic - contradiction. Thus $\Phi_n(x)$ is reducible modulo $p$ for all rational primes $p$.

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Does "inert" mean that there is only one prime over $p$, or does it mean that there is only one prime over $p$ and that prime is unramified? As the other answers have explained, unramified primes can not remain inert. However, it is possible that there is only prime over $p$. For example, let $p$ be an odd prime, take $K = \mathbb{Q}$ and $L= \mathbb{Q}(\sqrt{p}, \sqrt{a})$ where $a$ is not a quadratic residue modulo $p$.

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@David: that's a fair point (enough to make me think the question should remain open to address it). As a matter of terminology, inert does typically imply unramified, but it is helpful to remark that there can still be a unique prime of $L$ lying over a given ramified prime of $K$. (If you drop the Galois condition, it is easy to construct examples of such extensions with arbitrarily many totally ramified primes.) –  Pete L. Clark Jan 20 '10 at 3:17
    
In my parenthetical comment, I should have added: of any given degree $d > 1$. –  Pete L. Clark Jan 20 '10 at 3:35

It's worth noting that if $L/K$ is abelian, then you can easily see, using class field theory, that a non-cyclic extension can have no inert primes: the inertia degree of a prime is the order of it's Frob in the class group.

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What does class field theory have to do with anything? –  Ben Webster Jan 20 '10 at 16:00
    
You're right. . . saying 'class field theory' to mean 'facts about frobenii' is overkill. But I don't think it's that wrong to think 'class field theory' whenever you relate ideals to galois group elements. –  Joel Dodge Jan 20 '10 at 20:41
    
Ah yes, a good reason to not always think this is class field theory is that one can relate ideals to galois elements in a non-abelian field extension also. . . you always have frobenius elements! –  Joel Dodge Jan 20 '10 at 20:46
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A social observation: in algebraic number theory, it's never considered in bad taste to make reference to "class field theory", even if the precise connection is tenuous at best. Also, questions like Ben's can get you angry looks. :) –  Pete L. Clark Jan 21 '10 at 4:32

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