Inverse Galois Problem.

During this lecture we will discuss some historical (and some modern) re- sults concerning the Inverse Galois Problem. To begin with, what is the Inverse Galois Problem? In we see many familiar groups arise as auto- morphism groups of a field L that fix some subfield E ⊂ L. The Inverse Galois problem is concerned with classifying which finite groups can be realized as such automorphism groups. There has been considerable progress made on this question when we take the base field E to be Q. Historical results concerning the Inverse Galois Problem.

• In the 1800’s it was shown that it holds for all finite abelian groups.,

• In 1937 Scholz and Reichardt proved that all odd p-groups can be realized as the of some number field over Q. The proof involved solving special central embedding problems of the form

1 → Z/pZ → G¯ → G → 1

where G is the Galois group of some number field L over Q, Z/pZ is the Galois group of the algebraic closure of L (L¯) over L, and, G¯ is the Galois group of L¯ over Q. • In 1954 Shafarevich showed that every is a Galois group over Q. There was a flaw pointed out in this proof in 1989, but it was resolved by Shafarevich in the same year.

• The question is still open for non-solvable groups. However, many specific examples of non-solvable groups are known to occur as Galois groups over Q (ex. all An and Sn. This follows by Hilbert’s irreducibility theorem which states that if a group G can be recognized as a Galois group of an extension of Q(t) with basefield Q(t), then G can also be recognized as a Galois group of a number field over Q.)

• For simple groups we know that An, Z/pZ, and 25 of the 26 sporadic simple groups occur as Galois groups over Q. The question of whether simple Lie type groups and/or the last M23 occur as Galois groups over Q is still open.

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The Ridgidity Method. It is classically known that the Inverse Galois Problem holds for C(t). Using this own can show that the same is true over Q¯ (t) (that is, every finite group occurs as the Galois group of some field over Q¯ (t)).The ridgity method is realizing that groups satisfying a technical condi- tion on their conjugacy classes can be realized as Galois over Q(t). By Hilbert’s irreducibility theorem this tells us that the groups satisfying this technical con- dition are actually Galois groups over Q. A more refined question. Given a set of primes S ⊂ Z, what finite groups appear as Galois groups of some extension K/Q unramified at precisely the primes outside S? Note that the case S = {All Primes} is just the Inverse Galois Problem in all generality. The case S = ∅ is the only case that has a complete answer (only the trivial group because there are no unramified ex- tensions of Q).

In the local setting where there is only one prime to study, we understand quite well the structure of Galois extensions. Thiss suggests that studying the above question for S = {p} (some individual prime) might be more tractable than the normal Inverse Galois Problem (S = {All Primes}. However, even when we consider S = {p} things are quite different and more difficult than the local setting. We will state a useful theorem for examining this case. Kronecker Weber Theorem. Let S ⊂ Z be a set of primes. The maximum abelian extension of Q that is unramified at finite primes outside S is the com- posiutum of all fields Q(ζpr ) for all p ∈ S, r ∈ N.

So if we have S = {p} with p odd then the abelian groups that occur are just the cyclic groups of order dividing pr−1(p − 1) as r varies over N. Let’s do an example to demonstrate the use of the Kronecker Weber Theorem and see how the local case is easier than picking S = {p}. If we take p = 101, then by the discussion we just had the abelian groups that are Galois over Q corre- sponding to S = {101} are cyclic of order dividing 101r−1(100) = 22 ·52 ·101r−1. In particular, there is no Z3Z extension for S = {101}. However we know that 3 th adjoinging the 101 −1 roots of unity to Q101 gives a Z/3Z extension of Q101.

As another demonstration of the difference between the local setting and the S = {p} setting, we have seen that all Galois extensions of Q101 are solvable. 5 4 3 2 However, the splitting field of x −x = 6x +x +18x−4 gives an S5 extension of Q that is unsolvable where 101 is the only prime that ramifies.

Remark: There exist simple extensions of Q with only 1 finite ramified prime. 5 3 2 For example, the splitting field of x + 3x − 6x + 2x − 1 gives an A5 exten- sion that ramifies only at 653. It’s conjectured that every simple group can be 3 realized as the Galois group of some extension of Q with 1 ramified prime.

In the local setting we saw that we got the following groups for the follow- ing extensions:

tame Qp

Q ` `6=p Z unramified Qp

ˆ Z Qp For Q the situtation seems much more random, and an equally nice description (if it even exists) is far off. Here are some results in that direction:

• Harbater, 1994: If p < 23 is prime and K/Q is a tame extension unrami- fied at finitely many primes outside {p}, then K ⊆ Q(ζp). In particular, there are only finitely many tame extensions.

• If one assumes the Generalized Riemmann Hypothesis, it has been shown that ther are only finitely many extensions of Q unramified at one prime for p < 40 (although Q(ζp) is no longer the maximal extension for p ≥ 23).

• Schoof, 1987: Q(ζ877) has an infinite class field tower. That is, the class group of Q(ζ877 is non trivial, so we can extend it to get another field H, a tamely ramified extension only ramified at 877 that also has non-trivial class group, and this process if continued will never yield an extension with a trivial class group.