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Say $\mathbb{Q}\subset\mathbb{Q}(\theta)$ is a Galois extension, and $\theta$ is integral over $\mathbb{Z}$. What I'm having a hard time understanding is, if $f(X)=\min_{\theta,\mathbb{Q}}(X)\in\mathbb{Z}[X]$ is the minimal polynomial of $\theta$ over $\mathbb{Q}$ and is irreducible $\mod p$ for some prime, how does this imply that the Galois group is in fact cyclic? Is it isomorphic to $C_p$ somehow or is my hunch off?

I tried writing out $f(X)=g(X)h(X)\pmod{p}$ implies $g(X)$ or $h(X)$ is constant $\pmod{p}$, but I don't know what to say about the Galois group from this. Is there a clever way to see the cyclicity of the Galois group without actually knowing what the minimal polynomial looks like here? Thanks.

Later: I appreciate the answers I received so far, but is it possible to derive this result without much knowledge of algebraic number theory? If not, I guess I know what I have to do next.

yunone
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2 Answers2

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Edit: As countinghaus indicates below, this argument addresses the splitting field of an irreducible polynomial $f$ rather than $\mathbb{Q}[x]/f(x)$ so isn't directly relevant.

No. By Dedekind's theorem, if $f$ factors into (relatively prime) irreducible factors of degrees $d_1, ... d_k$, then there exists an element of the Galois group of cycle type $(d_1, ... d_k)$. In particular, if $f$ is irreducible modulo some prime, there is an element of the Galois group of cycle type $(n)$ (where $n = \deg f$). This does not imply that the Galois group is cyclic; for example $S_n$ always has such an element and is not cyclic for $n \ge 3$. (The converse of Dedekind's theorem is also true and follows from the Frobenius density theorem, a special case of Chebotarev's density theorem.)

As an explicit counterexample, $S_3$ is not cyclic, and an irreducible cubic $x^3 + bx + c$ has Galois group $S_3$ if and only if its discriminant $-4b^3 - 27c^2$ is not a square. So, for example, $x^3 + 2x + 2$ is irreducible by Eisenstein's criterion and its discriminant is negative. Furthermore it is irreducible $\bmod 3$.

The contrapositive of this statement is more interesting: if no element of the Galois group has cycle type $(n)$, then $f$ is reducible modulo every prime (not dividing the discriminant)! This cannot occur when $n = 2, 3$ but can occur for all $n \ge 4$ (for example when $n = 4$ the group $K_4$ is such a group).

Qiaochu Yuan
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  • @Sam: sure, but I would think it's pretty well-known that there exist polynomials with Galois group, say, $S_3$. Anyway, I added an example. – Qiaochu Yuan Apr 27 '12 at 22:08
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    The OP wants the extension generated by $\theta$ to be Galois already, which in this case it isn't - you're taking the splitting field. –  Apr 27 '12 at 22:36
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The way the question is worded, the answer is yes. Since the minimal poly is irreducible mod $p$, the prime $p$ is inert in this Galois extension. The decomposition group of the unique prime above $p$ surjects to the Galois group of the extension of finite fields with kernel the inertia group (this is always true). In this case, inertia is trivial and decomposition is the whole group, so we have an isomorphism between the Galois group and a Galois group of finite fields, but the latter are always cyclic.

If you don't require $\mathbb{Q}(\theta)/\mathbb{Q}$ Galois, this argument goes away, and of course the Galois closure won't be cyclic since it has non-normal subgroups.

  • Thanks for pointing out my mistake! I admit I didn't read the question too carefully. – Qiaochu Yuan Apr 27 '12 at 22:46
  • I actually wish you would put your old answer back up with an edit as I don't know the cycle decomposition business that you were using and I was going to come back and read the converse part. –  Apr 27 '12 at 22:53
  • Dear countinghaus, can you please explain how you see that the inertia group is trivial and the decomposition group is the whole group? I've done more reading about this, and after seeing that I would be glad to accept and award the bounty even. Thank you. – yunone May 01 '12 at 18:23
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    Because the minimal polynomial for $\theta$ mod $p$ has one irreducible factor, we know that its factorization in the field extension if of the form $\mathfrak{p}$, i.e., that there's exactly one prime above it and that prime is unramified. In general Galois permutes the set of primes lying above a fixed prime, and the decomposition group of a particular one is the stabilizer; in this case, that set is a one point set, so the whole group is the decomposition group. Inertia is trivial because the cardinality of inertia is equal to the ramification index. –  May 03 '12 at 07:08