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Sep 28, 2011 at 15:06 comment added paul Monsky R^2+R+G+H lies in S', is stabilized by the automorphisms [j]-->[rj] of S', and has poles of order no more than 12 at each of the l(l-1)(l+1)/24 valuation rings in L/K that do not contain S'. If it's divisible by x^(d+1) it has at least (d+1)(l-1)/2 zeros counted with multiplicity. So it has more zeros than poles, and must vanish. It follows that R=C.
Sep 28, 2011 at 14:58 comment added paul Monsky In the theorem I state above I can now show that d can be replaced by l(l+1). As a result, to verify the analogues of (3) and (4) for any l that is 15 mod 16, it's enough to use 2 terms in the power series expansion of each [j]. And so Ira's calculations which led to h/4 identities for each l<1500 and congruent to 7 mod 16, extended a minute bit further, will do the same for l that are 15 mod 16. Here's a proof of my assertion. If x^(d+1) divides R+C, it divides (R+C)(R+C+1)=R^2+R+G+H. Now my (self-accepted) answer to "Existence of certain identities.." shows that (to be continued)
Sep 24, 2011 at 4:36 history edited paul Monsky CC BY-SA 3.0
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Sep 23, 2011 at 3:02 history answered paul Monsky CC BY-SA 3.0