From MAILER-DAEMON Thu Mar 16 17:50:10 2000 Date: 16 Mar 2000 17:50:10 -0500 From: Mail System Internal Data Subject: DON'T DELETE THIS MESSAGE -- FOLDER INTERNAL DATA X-IMAP: 0953247010 0000000000 Status: RO This text is part of the internal format of your mail folder, and is not a real message. It is created automatically by the mail system software. If deleted, important folder data will be lost, and it will be re-created with the data reset to initial values. From vakil@math.mit.edu Wed Mar 1 16:19:29 2000 -0500 Status: R X-Status: X-Keywords: Received: from schubert (SCHUBERT.MIT.EDU [18.87.0.16]) by math.mit.edu (8.9.3/8.9.3) with ESMTP id QAA18198; Wed, 1 Mar 2000 16:19:22 -0500 (EST) Date: Wed, 1 Mar 2000 16:19:21 -0500 (EST) From: Ravi Vakil X-Sender: vakil@schubert.mit.edu To: bhargava@math.princeton.edu cc: Ravi Vakil Subject: a big one... Message-ID: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Hi Manjul -- Okay, I understand part of this. I don't believe your argument (or at least I don't understand it), but I have one of my own, which may let us ignore more of the unit issue. Let me run some arguments by you. (1) First of all, I claim that w_1 = w_3 = w_4 = w_6, with no conditions on our S_4-extension. (I don't claim that w_{24}=w_1 etc., but we don't need that of course.) Proof: any root of unity of M that isn't in Z must generate the discriminant field, which isn't in Z, F, K, L. Remark. So we didn't need this assumption (call it assumption A) that this the discriminant extension isn't cyclotomic at this stage. I think we can replace it by a weaker assumption B, given below. (2) Second, assume that w_{24} = w_1 etc, i.e. assumption A. Let U be the roots-of-unity (finite abelian) group. Suppose you have an element of the unit lattice of M fixed by the galois group G of M over a field N, where N= Z, F, K or L. Then I claim that the square of this element is in the image of the unit lattice of N (i.e. there is a unit in N with those absolute values). Proof: Let u be any element of M with those absolute values. We want to check that g(u^2) = u^2 for all g in G. Define r:G-->U given by g --> g(u)/u. This is a group homomorphism: (g_1 u / u) (g_2 u / u) = (g_1 u / u) (g_1 (g_2 u) / g_1 u) = g_1 g_2 (u) /u. (Here we use the fact that U is fixed by G.) The image of G is abelian. The only abelian quotients of G=S_4, S_3, D_4, K_4 are 2-groups. (Way to check: the map G--> U factors through G/[G,G]. In these 4 cases, [G,G] = A_4, A_3, C_2, K_4, so the quotients are Z/2, Z/2, Z/2+Z/2, Z/2+Z/2.) especially!) Moreover, any element of these groups are killed by 2. Hence g(u^2)/u^2 = ( g(u)/u )^2 = 1, as desired. Remark: This is all we need -- note that it gives us yet another 2 issue! But you seem to think that gu=u for all g. Could you tell me why this is true? (See below for my confusion about your explanation.) (3) More generally, we want to remove assumption A, and get a similar result. Here it is one version. It is kind of messy, so perhaps you see a way of streamlining it or improving it. Assume that the discriminant extension of Z isn't Z[w], where w is a cube root of unity. (Call this Assumption B.) Suppose you have an element m of the unit lattice of M fixed by the galois group G of M over a field N, where N= Z, F, K or L. Then there is a unit in N whose absolute values correspond to 2m (NOT necessarily the square!). For convenience, I'll do the G=S_4, i.e. N=Z, case. I haven't tried the other cases, and I won't for a bit because you might tell me that (a) we don't care, or (b) that there's an easy way around this. Again, let U be the roots-of-unity finite abelian group. There is an action of G on U. Note that A_4 acts trivially on U (see (1) above; any root of unity not fixed by G must generate the discriminant field). Hence G acts via sign. For g in G, and u in U, for convenience define sgn(g)(u) = g(u)/u, so g(u) = u sgn(g)(u). So for g even, sgn(g)(u) = 1. Let u be any element of M with absolute values given by the element of the unit lattice of M. Define r:G-->U as before given by g --> g(u)/u. This is a _twisted_ group homomorphism: r(g_1 g_2) = (g_1 g_2 u)/u = (g_1 g_2 u)/(g_1 u) (g_1 u/u) = g_1( r(g_2) ) r(g_1) = sgn(g_1)( r(g_2)) r(g_2) r(g_1). (*) This is an element of H^1(G,U), but I don't know enough about group cohomology to know if this tells us a whole lot immediately. If g_1, g_2 are both even, then r(g_1 g_2) = r(g_1) r(g_2). So for the same reason as in (2), the image of A_4 is at most the maximal abelian quotient of A_4, which is A_4/K_4 = Z/3. Let alpha = r( (12) ). Then (substituting g_1 = g_2 = (12) in (*)): 1 = (12)(alpha) times (alpha) so the action of galois on alpha is (12): alpha --> 1/alpha. We have two cases: r( g ) = 1 for all g in A_4, or r( (123)) =w where w is some cube root of 1. Case 1. If r( g ) = 1 for all g in A_4, then for all g in A_4, r( g (12) ) = r(g) r(12) = alpha. Hence r(even) = 1, r(odd) = alpha. Galois acts by g(alpha) = alpha or 1/alpha depending on the sign of g. Then alpha u^2 is fixed by S_4. Proof: it is fixed by A_4; also (12) (alpha u^2) = (1/alpha) (alpha u)^2 = alpha u^2. Case 2. If r( (123) ) = w. Then we can work out the values of r(g) for all g as follows. We know r(g) for any g in A_4. And if g is in A_4, r(g (12)) = r(g) alpha by (*). Here's the table: g r(g) e, (12)(34), (13)(24), (14)(23) 1 (123), (243), (134), (142) w (132), (234), (143), (124) w^2 (12), (34), (2413), (2314) alpha (13), (1432), (1234), (24) alpha w (23), (1342), (1243), (14) alpha w^2 Let's compute the action of Galois on w as follows, using (*). alpha w^2 = r( (23) ) = r( (12) (123) ) = (12)(r(123)) r(12) = (12)(w) alpha Hence (12) sends w to w^2, so w isn't fixed under Galois, contradicting assumption B. Hence we're done. Remark 1. I don't think we can get rid of assumption B! First of all, I haven't checked, but I'm pretty sure the table above showing the possible values of r(g) satisfies (*), so we wouldn't have a contradiction because of that. In this case, I think the best I can do is show that 6 times this unit lattice vector comes from Z. Here's how: Claim: alpha^3 u^6 is fixed by S_4. Proof: (12): u --> alpha u alpha --> 1/alpha w --> w^2 (1234): u --> alpha w u alpha --> 1/alpha w --> w^2 Then (12): alpha^3 u^6 --> alpha^3 u^6, (1234): alpha^3 u^6 --> alpha^3 u^6. Remark 2. I should quickly be able to do this for the other groups S_3, D_4, K_4. For obvious reasons, I'd suspect that we would only need assumption B for S_3. If this seems worthwhile, I'll do the details. I suspect it would be worthwhile, as it seems nice to get rid of big assumption A, and replace it with small assumption B. (Note that assumption B is trivially satisfied if w is in Z!) -- Ravi p.s. A newer tex version is now on the web page. It has a draft of everything up to the proof that R_i h_i is multiplicative in the way we want. On Sat, 26 Feb 2000, Manjul Bhargava wrote: > Hey Ravi-- > > Thanks--concert went well! My fever is gone, but my throat is still > acting up a bit... it should be ok soon, I hope. > > I'll try and take a look at the tex stuff shortly. > Oh, and about the roots of unity stuff--I don't see why what I wrote > doesn't address the problem?? (It's ok to assume the roots of unity are > all in Z.) What I was saying is: if u is any number and a an element of > S_4 such that r=au/u is a root of unity, then conjugates of u include > u, ru, r^2 u, etc. and so if v=u^n, then u is a solution of x^n-v (here > n=order of a). Because the root of unity r lies in Z(v)--since we're > assuming it lies even in Z-- Z(u,v)/Z(v) is a Galois, cyclic extension, > with Galois group generated by u -> ru . (called a "Kummer" extension) > Anyhow, this gives a nontrivial cyclic extension within the S_4-extension, > which doesn't exist really. Here I get concerned. What I think you just said is that there is no nontrivial cyclic subquotient of S_4. But that isn't true -- there are even such things that are order 3, e.g. A_4 / K_4 = Z/3 is a subquotient of S_4. Could you explain more? > I guess I just said the same thing as before-- > why doesn't this address the problem? Am I missing something? > > -Manjul > >