Dear Readers,

In spite of orals closing in a little more every day, I clearly haven’t been updating so much recently. I’d started a post about using Minkowski’s geometry of numbers to think about class numbers and unit groups and such things… but honestly that stuff is quite well-worn and at this point it wouldn’t be a good use of time to think carefully about how best to choose my words and explain this to the world when better expositors like William Stein or James Milne have already done so. Instead, I will talk about the expository part of my exam where I will talk about a particular case of the inverse Galois Problem.

**Theorem(Shih,1974):** If 2,3 or 7 is not a square in then is the Galois group of a polynomial with integral coefficients.

The genesis of this theorem is a theorem of Shimura on Automorphic forms and modular curves. It uses a -torsion Galois representation . Namely if is an elliptic curve over , fix generators for . Then if , let where , .

To approach this problem in a systematic way we use the following universal elliptic curve with -invariant

If we view this as an elliptic curve over the field of rational functions of one variable, , we have the following result of Shimura:

**Theorem:** For all , is Galois over with group . Furthermore, and acts on by .

A proof of this (and results involving other modular curves) takes up the bulk of chapter 7 of Diamond and Shurman’s book on modular forms or alternately much of David Rohrlich’s article in the BU lectures on modular forms and Fermat’s Last Theorem.

With this in mind, let be a primitive element for with minimal polynomial over . If we can find some rational number such that is irreducible and separable over , we will have a number field with the appropriate Galois group. Under what circumstances can we *specialize* to a good rational number?

**Theorem**: (Hilbert’s Irreducibility Theorem) If is a number field and is a *regular* field extension, one where , then there are infinitely many such that is a Galois field extension.

A proof for this theorem can be found in Serre’s Topics in Galois Theory. Using this theorem, we can get a Galois field extension with group … except the ground field isn’t unless , but . We can however use this to build a regular Galois field extension of with the appropriate Galois group.

Following Serre’s topics in Galois Theory, our first step is to take the representation and compose it with the quotient map . The projective representation has a kernel corresponding to a field extension of where , the unique quadratic subfield of . Now we perform some sleight of hand to make this ramification go away, using the theory of twisting.

Certainly if we consider as an elliptic curve over then the -torsion Galois representation of will simply be restricted to . Can we find some which extends the -torsion Galois representation but gives a regular field extension?

To do this, suppose that . Rather than considering the effect of on the projectivization of , we twist by the action of an isogeny. This may seem unnatural at first, but is entirely natural if we ignore that can be defined over .

So let be a quadratic extension with a generator for and an elliptic curve without CM over . If is not necessarily defined over and is a point of then might not be a point of but instead of a conjugate curve . The similarity in notation to the relative Frobenius is intentional, as this can also be described as a fiber product: since is an automorphism, is an automorphism and combining with the structure map we have the diagram:

can also be described in a more down-to-earth way: If then . To send back to , we need an isogeny . That is, we need to be a -curve in the terminology of Gross. A -curve gives a Galois representation as follows:

Consider the isogeny to be cyclic so the kernel is isomorphic to for some . Since does not have CM, the automorphisms of as an elliptic curve are just so the -isogenies are just .

Then we consider the representation which, for a given identification of with , assigns the matrix describing the following action on the projectivization of . If then . If then we take the matrix describing the action of . Note that this is independent of the choice of isogeny by the non-CM condition, hence why we denote it rather than . Through some case-by-case calculation, we can determine that is a homomorphism. Moreover by considering the determinant character we can find out when the image is contained in . If the determinant is the exact same as for , the character defining . However if we get an additional factor of , the character defining . This is exactly what we need since in our case . To get our -isogeny, we consider that can appear in more than one way.

Consider the moduli space of cyclic -isogenies, . This space possesses a natural involution sending an isogeny to the dual isogeny . We can then consider the quotient curve . By the curves/function fields correspondence this describes a degree 2 extension with Galois group generated by . Thus if is an elliptic curve defined over , is -isogenous to . But we don’t want just any quadratic extension, we want and generated by a cyclic -isogeny. That is, we want to be the function field of a twist of by the involutions and some such that . To be explicit, we take the function field of the curve made up of points of such that .

Now the final component of the problem: how can we make into ? Recall briefly that is the function field for . Can we find an isomorphism between and ? Yes, if and only if has genus zero and a rational point. The question of genus for a smooth curve is a geometric one (i.e. dealing with the algebraic closure) so the genus of is the genus of which is zero for (the genus is also zero for 1,4,9,16 and 25, which must be excluded by the condition ).

Then for we can find rational points by finding -rational fixed points of . Shih originally did this by looking at the -function. I prefer the method Serre suggested of using the theory of complex multiplication. That is, if is an elliptic curve with CM by (and for , such an elliptic curve can be defined over ) then it can be shown with relative ease that is a -rational fixed point of .

Now we collect our results. The Galois representation we’ve constructed is, by the isomorphism given above, an onto homomorphism . The kernel of this map gives a finite Galois extension which is regular because the only elements of which is not in the kernel of are those which conjugate . This action comes from the action of the determinant on the -th roots of unity, but now the determinant of is constantly 1. Thus if does not leave fixed, . Then by Hilbert’s irreducibility theorem, there exists some such that is a field and thus is Galois over with Galois Group . There is a corresponding geometric picture: by the curves/function fields correspondence, corresponds to a regular finite etale covering of with automorphism group .