Saltman’s Generic Galois Extensions and Problems in Field Theory David Harbater May 16, 2011 Motivation from Inverse Galois Problem Saltman, Advances in Math. (1982) Motivated by (inverse) Galois theory; has motivated further work. IGP: Is every finite group G a Galois group over Q? Strategy: Realize G over Q(x); specialize x to α 2 Q. 2 Example: G = C2. Adjoin y:py = x to Q(x). Can specialize x to 2; get Q[ 2]=Q. Can always specialize: Hilbert’s Irreducibility Theorem: If f (x; y) 2 Q[x; y] is irreducible, then for infinitely many α 2 Q, f (α; y) 2 Q[y] is irreducible. Also works with more variables. So: If G is a Galois group over Q(x1;:::; xn), then G is a Galois group over Q. Similarly over any global field (number field or function field of a curve over a finite field); such fields are Hilbertian. This strategy has been used obtain many groups as Galois groups over Q. 1) Use geometry to get a branched cover of 1 with group G. PC 2) Choosing rational branch points, the cover is defined over Q¯ . 3) Sometimes one can show it’s defined over Q. 4) The extension of function fields is Galois over Q(x); also regular (Q is algebraically closed in the extension of Q(x)). Re 3): Rigidity (Matzat, Thompson, Belyi; Fried, Shih); e.g. monster group is a Galois group over Q. Noether’s problem An early attempt at IGP/Q via IGP/Q(x1;:::; xn), due to Emmy Noether: Given G, let G act transitively on f1;:::; ng (e.g. by regular rep.). G Get an action of G on L := Q(y1;:::; yn); invariants K = L . ∼ Question: Is K = Q(x1;:::; xn)? If so, Hilbert Irreducibility ) G is a Galois group over Q. Example. G = Sn acting on f1;:::; ng. Invariants in Q(y1;:::; yn): Q(σ1; : : : ; σn). So yes. Geometrically: (y ;:::; y ) is the function field of n , with an Q 1 n AQ action of G. Take the quotient n =G. This is unirational. Is it AQ rational? Over an algebraically closed field, it’s hard to find examples of non-rational unirational varieties. There are more examples over non-algebraically closed fields (twists). G Noether’s problem: Determine if Q(y1;:::; yn) is purely transcendental. (If so, G is a Galois group over Q.) Always? No: Swan 1969, Voskresenskii 1970: no if G = C47; C113; C223;::: Lenstra 1974: no if G = C8, or for abelian groups containing C8. Over an algebraically closed field, is the answer to Noether’s problem always yes? Answer: No (Saltman, 1984). If k is algebraically closed, and p 6= char(k), then there is a p-group G of order p9 such that the fixed field is not purely transcendental. This gives new examples of varieties that are unirational but not rational, over an algebraically closed field. Later: related examples by Bogomolov (1987), Plans (2007). The proofs use unramified cohomology (in Saltman’s case, the unramified Brauer group). Generic Galois extensions Notion due to Saltman, related to Noether’s problem. Ask for existence of such an extension for a given group G. The answer is yes more often than for Noether’s problem. Yes ) G is a regular Galois group over Q(x) and so G is a Galois group over Q. Existence of such an extension gives a structural result about all G-Galois extensions of Q, and over all fields containing Q. (As in the earlier discussion, can work over other fields as well.) Generic Galois extensions are related to moduli spaces, but allowing repetition. The definition uses the notion of a Galois extension of rings. Recall: if a finite group G acts on a field L, then a field extension G ∼ L L=K is G-Galois , K = L and L ⊗K L = g2G L. Analogously, if G acts on a ring S, then we say that a ring G ∼ L extension S=R is G-Galois if R = S and S ⊗R S = g2G S. −1 2 Example. G = C2, R = Q[x; x ], S = R[y]=(y − x). 0 Note: The definition rules out ramification. Can’t take R = Q[x] and S0 = R0[y]=(y 2 − x). Geometrically: Spec(S) ! Spec(R) is a G-Galois étale cover. A generic Galois extension for G over a field k is a G-Galois ring extension S=R = k[x1;:::; xn; 1=g] such that 8 field k0 ⊇ k, 8 G-Galois extension `0=k0 (not necessarily a field), 0 0 0 ` =k is a specialization of S=R (by setting xi = αi 2 k ). n Geometrically: Spec(R) = U ⊆ Ak , a Zariski open dense subset. We have a G-Galois étale cover of U, corresponding to a G-Galois n branched cover of Ak with ramification away from U. This cover has the property that it parametrizes all G-Galois extensions of fields k0 ⊇ k, by taking the fibers over k0-points of U. A strong condition; does such an extension exist? Saltman: If Noether’s problem has a positive answer for G, then there is a generic Galois extension for G over Q. So in particular there is a generic Galois extension for Sn. But existence of a generic Galois extension is more general than a positive answer to Noether’s problem: Saltman: If n is odd, then there is a generic Galois extension for Cn. (More generally, if n = 2r s, with s odd, then there is a generic Galois extension for Cn over k iff k(ζ2r )=k is cyclic.) In particular, 9 generic Galois extension for C47 over Q, even though the answer to Noether’s problem is no. How to construct such extensions? Simple case: G = Cn, and work over k 3 ζn. By Kummer theory, every Cn-Galois extension of k is of the form k[y]=(y n − α), with α 2 k. So a generic Galois extension is given by R = k[x; x−1], S = R[y]=(y n − x). What if we don’t have ζn? How to get a generic Galois extension for Cn? Example: C3 over Q.A C3-Galois extension of Q is given by −1 Q(ζ7 + ζ7 ). But this is just one extension. How to get a family? We want an extension of Q[x; 1=g] that specializes to various (even all!) C3-Galois extensions of Q. 0 0 Idea: Build a non-constant C3-Galois extension L =K = Q(ζ3; x) by Kummer theory, such that it descends to a C3-Galois extension L=Q(x). To do this, the action of Γ = Gal(Q(ζ3)=Q) should lift to an action 0 0 on L that commutes with the action of C3 on L . 0 So L =Q(x) is Galois with group C3 × Γ. Then take L = (L0)Γ. 0 L L Γ C3 C3 0 (x) K Γ Q 0 Explicitly: Take x = (x1; x2). Over K = Q(ζ3; x1; x2), take 0 0 3 x1 + ζ3x2 0 L = K [y]=(y − s), where s = −1 2 K . x1 + ζ3 x2 Γ = Gal(Q(ζ3)=Q) = Gal(Q(ζ3; x1; x2))=Q(x1; x2)) −1 −1 The generator of Γ takes ζ3 to ζ3 , and s to s . This extends to an action on L0 by y 7! y −1, 0 commuting with the action of C3 on L . −1 u := s + s is invariant under Γ and so lies in Q(x1; x2); in fact 2 2 u = (2x1 − x2 − 2x1x2)=(x1 − x1x2 + x2 ). −1 0 v := y + y is also invariant under Γ, but not under C3; so v 62 K . 0 Γ So v generates L := (L ) over Q(x1; x2). 3 Its minimal polynomial over Q(x1; x2) is V − 3V − u. This defines a C3-Galois regular extension of Q(x1; x2). This C3-Galois extension of Q(x1; x2) gives a generic extension for C3 over Q, by taking the integral closure of Q[x1; x2] in the extension and inverting the discriminant. Saltman shows how to do this more generally for Cq with q an odd prime power (using more xi ’s). Taking products, can get Cn for all odd n. (Can replace Q by other fields) For C2r this can still be done if adjoining ζ2r gives a cyclic extension. So we get a generic Galois extension for an abelian group A over k r if k(ζ2r )=k is cyclic, where 2 is the highest power of 2 dividing the exponent of A. We can also get generic Galois extensions for many semi-direct products. For example, Saltman showed: 1) If there are generic Galois extensions for N and H over k then there is a generic Galois extension for N o H over k. 2) If A; H have relatively prime order, with A abelian, and if there are generic Galois extensions for A and H over k, then there is a generic Galois extension for A o H over k. 3) If q is an odd prime power and ζq 2 k, and if 9 a generic Galois extension for H over k, then 9 a generic Galois extension for Cq o H over k. (Also for a 2-power q if the image of H in Aut(Cq) is cyclic.) In particular, 9 a generic Galois extension for Dq over k if ζq 2 k. Related results by others: Elena Black, 1999: If q is a power of 2 then there is a Dq generic −1 Galois extension over k of char 6= 2 if q = 4 or 8, or ζq + ζq 2 k.