TLDR; Is there an 'elementary' argument to prove the following:

Claim: Given an integer $d>1$, are there infinitely many distinct Galois extensions $K/\mathbb{Q}$ with $[K \colon \mathbb{Q}]= d$?

Elementary in the sense that students new to Field/Galois Theory could easily follow, and that can be explained 'quickly'?

'Brief Background': When discussing field extensions with students, I pointed out that every quadratic extension of $\mathbb{Q}$ is Galois but that this was certainly not generally true. A student then asked the next obvious question,

"What about cubic extensions? Are 'most' of them Galois or not? What about other fields? Are there only finitely many in some degrees? Are fields in some degrees 'more' or 'less' Galois?"

Obviously, pointing out that 'most' extensions are not Galois, I stated that we can at least guarantee one for each degree - for instance, using the construction from the classic proof that every finite abelian group is a Galois group - and commented that there are infinitely many in each degree. Of course, counting them (up to fixed discriminant for example) is a difficult open problem.

I did not prove the latter statement for them, though in the moment I mentally prepped for the question which never came, probably due to them being 'side discussion overwhelmed.' But I think the question is a good one:

Claim: Given an integer $d>1$, are there infinitely many distinct Galois extensions $K/\mathbb{Q}$ with $[K \colon \mathbb{Q}]= d$?

Or more generally for fields $K_0$ with char $K_0=0$, are there infinitely many distinct Galois extensions $K/K_0$ with fixed degree, distinctness here meaning $K \cap K'= K_0$, where $K,K' \subseteq \overline{K_0}$ for some algebraic closure of $K_0$.

The Question: Is there a simpler argument using only 'basic' results from Field/Galois Theory that can be quickly presented and understood by students in their first pass at Abstract Algebra, if only for the case that $K= \mathbb{Q}$?

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    $\begingroup$ I don’t think your idea of using $S_n$ will work. The problem was to find a Galois extension of each degree, but for the most part, $S_n$ has only one normal subgroup, and that one is far too large. $\endgroup$
    – Lubin
    Oct 13, 2020 at 4:31
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    $\begingroup$ @Lubin Taking a look at it a second time, I would tend to agree that this hopeful quick line of thought really cannot be salvaged. You'd probably want to find alternative construction of the abelian case and then use the remaining part of the argument I made, but unfortunately this is really no simpler than the original argument, so I am still left without a good explanation. $\endgroup$ Oct 13, 2020 at 5:29
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    $\begingroup$ If the extension degree is a prime $p$ you know that the Galois group will be cyclic. Hence abelian, hence the field is a subfield of a cyclotomic field by Kronecker-Weber. What I'm gettng at is that in this case you are more or less forced to follow the route in Qiaochu's answer. You need infinitely many primes $\ell$ such that $\ell\equiv1\pmod p$. $\endgroup$ Oct 13, 2020 at 7:02
  • $\begingroup$ @JyrkiLahtonen That is most definitely a great point that I had not given attention to! Between your comment and Qiaochu's excellent answer, I am not hopeful for an elementary solution. Surprising because at a quick glance it seems like something you might see as an exercise and 'feels' like it should be - but most definitely not. $\endgroup$ Oct 15, 2020 at 6:39

1 Answer 1


I think this is not at all an elementary result; it really tells you something nontrivial about the absolute Galois group of $\mathbb{Q}$ (note that it's equivalent to the existence of infinitely many distinct open normal subgroups of index $d$). For example, I think it doesn't hold for $\mathbb{Q}_p$ because the absolute Galois group of $\mathbb{Q}_p$ is topologically finitely generated, which implies there are only finitely many Galois extensions with a fixed Galois group and hence finitely many Galois extensions with a fixed degree.

I think the fastest way to do this "from scratch" is to exhibit subextensions of cyclotomic extensions:

  • For every prime $p$, prove that $\mathbb{Q}(\zeta_p)$ is Galois with Galois group $(\mathbb{Z}/p)^{\times}$.
  • For every positive integer $n$, prove that there are infinitely many primes $p \equiv 1 \bmod n$ (this can be done in an elementary way using the cyclotomic polynomial $\Phi_n(x)$, we don't need Dirichlet's theorem), and hence infinitely many primes $p$ such that the Galois group of $\mathbb{Q}(\zeta_p)$ has a subgroup of index $n$, whose fixed field is an abelian Galois extension of $\mathbb{Q}$ of degree $n$, with Galois group $C_n$.
  • Prove that for distinct primes these extensions are distinct (e.g. by computing their discriminants).

The existence of the cyclotomic extensions $\mathbb{Q}(\zeta_n)$ implies that the absolute Galois group surjects onto $\widehat{\mathbb{Z}}^{\times}$, the group of units of the profinite integers, which is an enormous group and in particular not topologically finitely generated. (When we consider subextensions of $\mathbb{Q}(\zeta_p)$ we are considering the projection of this group to $(\mathbb{Z}_p / p \mathbb{Z}_p)^{\times}$.)

We can ask for elementary methods in each small degree at least. For degree $3$ we want to exhibit infinitely many cubics with Galois group $C_3$ with nonisomorphic splitting fields. This can be done without cyclotomic extensions by proving that an irreducible cubic has Galois group $C_3$ iff its discriminant is a square and exhibiting a family of irreducible cubics with square discriminants. This is already not entirely elementary. The discriminant of $x^3 - px + q$ is $\Delta = 4p^3 - 27 q^2$ so requiring that this is a square involves solving a Diophantine equation

$$4p^3 - 27q^2 = r^2.$$

Equivalently we want to exhibit infinitely many $p$ such that $4p^3$ can be represented by the quadratic form $r^2 + 27q^2$. Recognizing this as the norm of $r + 3q \sqrt{-3} \in \mathbb{Q}(\sqrt{-3})$, we can do this by taking $p = x^2 + 3y^2 = N(x + y \sqrt{-3})$ and writing $4 = N(1 + \sqrt{-3})$, which gives

$$4p^3 = N \left( (1 + \sqrt{-3})(x + y \sqrt{-3})^3 \right) = (x^3 - 9x^2 y - 9 xy^2 + 9y^3)^2 + 3 (x^3 + 3 x^2 y - 9 xy^2 - 3y^3)^2$$

which gives us our infinite family of solutions. Now we have to prove that infinitely many of these cubics are irreducible and that infinitely many of their splitting fields are distinct which seems a bit painful; at this point this seems to me like more work than using cyclotomic fields.

For degree $4$ things are easier because we can piggyback off of the degree $2$ case: we can consider the biquadratic extensions $\mathbb{Q}(\sqrt{a}, \sqrt{b})$ where $\gcd(a, b) = 1$.

For degree $5$ we want to exhibit infinitely many quintics with Galois group $C_5$ with nonisomorphic splitting fields, which without cyclotomic extensions seems hard again. In general things will be hard in prime degree $p$ since the Galois group must be $C_p$ and we have no intermediate fields to piggyback off of. For large $p$, and by "large" here I mean $p \ge 5$, I don't think it's clear a priori that there exists even one such extension.

  • $\begingroup$ I guess over a base field like $\mathbb{C}(t)$ containing all roots of unity things are easier because we can use Kummer extensions. But over $\mathbb{Q}$ this only handles a few cases like the quadratic (and biquadratic, etc.) cases, and doesn’t handle any odd prime case. $\endgroup$ Oct 13, 2020 at 18:14
  • $\begingroup$ Thank you for the excellent answer - though all of your answers are always of high quality. I had thought of the cyclotomic extension route, but it leaned on Dirichlet's Theorem (I had not thought of the cyclotomic route, so a definite +1 for that thought!) but was hoping for a simpler route. I foolishly had not spent much time thinking about what such a route would mean in terms of the absolute Galois group. Between your answer and Jyrki's comment, I am doubtful a 'nice, elementary' solution could be found. I just wish I could upvote twice! $\endgroup$ Oct 15, 2020 at 6:35
  • $\begingroup$ @mathematics2x2life: you're welcome! Once you remove the appeal to Dirichlet's theorem it's really not that bad of an argument, although admittedly I've never tried to present it in a classroom. You would still have to prove that the cyclotomic polynomials are irreducible and that their prime divisors are exactly the primes congruent to $1 \bmod n$ (then there is a Euclid-style proof that there are infinitely many such primes). $\endgroup$ Oct 15, 2020 at 9:18
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    $\begingroup$ The first paragraph, relying on the absolute Galois group of $\mathbf Q_p$ being topologically finitely generated, is overkill. The field $\mathbf Q_p$ (and more generally each local field) has finitely many extensions of a fixed degree inside its algebraic closure, both Galois and not Galois. This is a consequence of the uniqueness of unramified extensions in each degree and a use of Krasner's lemma to show "nearby" irreducibles over a local field have roots generating the same extension fields plus compactness of the ring of integers of a local field. $\endgroup$
    – KCd
    Dec 3, 2020 at 3:43

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