# Infinitely many number fields

I wonder if it is known that there are $$\textit{infinitely}$$ many number fields $$F$$ (up to isomorphism) with fixed degree $$[F:\mathbb{Q}]=n$$ and fixed a transitive group $$G$$ of $$S_n$$ such that $$G=\textrm{Gal}(F^c/\mathbb{Q})$$ (if we assume inverse Galois holds for $$G$$).

Obviously, there are finitely many number fields $$F$$ if we make a restriction for $$F$$ such that $$|d_{F}| for some $$M>0$$. This is from Hermite's theorem.

• My guess is that this is an open question, but you may get a better response on Mathoverflow. – Mathmo123 Oct 25 '18 at 6:51
• @Mathmo123 Thank you for your suggestion. – Thomas Oct 26 '18 at 1:06

Basically, no.

One comparison is with (a special case of) Dirichlet's Theorem. For example, given an integer $$n$$, suppose you know there exists a prime $$p \equiv 1 \bmod n$$, then you can ask whether there exists infinitely many primes $$p \equiv 1 \bmod n$$? There is not really any mechanism to go from one to the other.

Of course, there is a variant. If you knew that there existed at least one prime $$p \equiv 1 \bmod n$$ for all $$n$$ then you can deduce that there are infinitely many primes. Proof: Given primes $$p_1, \ldots, p_k$$ which are $$1 \bmod n$$, we are guaranteed one prime $$p \equiv 1 \mod p_1 p_2 \ldots p_k n$$. Certainly $$p \ne p_i$$ for any $$i$$, so we have a new prime $$p \equiv 1 \bmod n$$.

The same thing is true in your case. Clearly if you knew there was at least one extension with Galois group $$G$$ for any $$G$$, you can find infinitely many with Galois group $$G$$. (Consider fields with Galois group $$G \times G \times \ldots G$$ for larger and larger products of $$G$$.)

Note the analogy with Dirichet's theorem is stronger than one might expect: Finding infinitely many fields with Galois group $$\mathbf{Z}/n \mathbf{Z}$$ is exactly the same as finding infinitely many primes $$p \equiv 1 \bmod n$$.

Now there are a few cheap tricks in some cases. Suppose we knew the inverse Galois problem for a class of groups (say abelian groups, which is true). And suppose that $$G$$ has the property that there is a surjection

$$\phi: G \times A \rightarrow G$$

for some abelian group $$A$$ for which the image of $$A$$ is non-trivial. Then given one Galois extension $$F$$ with Galois group $$G$$ one can find infinitely many --- by assumption there are infinitely many fields $$E$$ with Galois group $$A$$, and then using $$\phi$$ we can find a subfield of the compositum $$E.F$$ which has Galois group $$G$$ and is not isomorphic to $$F$$ (using basic Galois theory). For an example, let $$G$$ be any group with a non-trivial center. Then there is a map

$$\phi: G \times Z \rightarrow G$$

sending $$(g,z)$$ to $$gz$$. So this works if $$Z(G) \ne 1$$. This is not unrelated to the following example. Let $$E$$ be an elliptic curve over $$\mathbf{Q}$$, and suppose that the Galois extension of $$\mathbf{Q}(E[p])$$ has Galois group $$G \subset \mathrm{GL}_2(\mathbf{F}_p)$$. If $$p > 2$$, the center of $$G$$ has an element of order $$2$$. Hence there is a map

$$\phi: \mathrm{GL}_2(\mathbf{F}_p) \times \mathbf{Z}/2 \mathbf{Z} \rightarrow \mathrm{GL}_2(\mathbf{F}_p)$$

as above, and one gets a different such extension for each quadratic field. In fact, the corresponding extension also comes from another elliptic curve, namely the twist of $$E$$ by the quadratic character of the quadratic extension giving rise to the degree $$2$$ extension.

OTOH, if $$G$$ is simple, then the only way to get non-trivial maps from $$G \times A \rightarrow G$$ is for $$A$$ to surject onto $$G$$. (p.s. this last construction is known as "crossing" of extensions, and first turned up in Cebotarev's proof of his theorem)

• This is a really great answer! – Mathmo123 Oct 25 '18 at 22:01
• @user608470 Thank you for this careful explanation first! I wonder how to know when you get any $G$-extension number field $F$ in $G\times G\times\cdots$-extension, you will $\textit{not}$ get isomorphic number fields $F$ in your construction. So, you can definitely get infinitely many non-isomorphic fields $F$. Sorry maybe this is dumb. – Thomas Oct 26 '18 at 1:24
• @Thomas It follows from the Galois correspondence. Two subfields will be isomorphic if and only if the correspondence subgroups are conjugate. – Mathmo123 Oct 26 '18 at 7:35
• @Mathmo123 Got you! – Thomas Oct 27 '18 at 8:37