It is said that there is a proof of fundamental theorem of algebra using Lie Theory. I have seen this claim at various places. But I could never find such a proof. Can anybody help me out?
Fundamental theorem of algebra using Lie Theory
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$\begingroup$
complex-analysis
lie-groups
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0By who? I've never heard anybody make this claim and it's far from clear to me what it could mean. And why is this tagged complex-analysis? – 2010-08-12
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0@Qiaochu: the claim sounds dimly familiar to me too, and I even thought that it had come up recently on MO. I looked there just now without success (but I am not very good at searching for things on MO...). – 2010-08-12
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0@Qiaochu: I wanted the tag complex-variables as Fun. Thm. is a theorem about complex numbers. But I am a new user and I cannot create tags; so I went for complex-analysis. – 2010-08-12
2 Answers
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The references can be found in the comments here:
For those who don't know it, the (well, a) Lie-theoretic proof of Fund. Thm. of Algbera due to Witt is on p. 245 of the book "Numbers" by Ebbinghaus et al. – KConrad Aug 6 at 4:44
Witt's Lie-theoretic proof of Fund. Thm. of Algbera seems to be Witt (Ernst), Über einen Satz von Ostrowski, Arch. Math. (Basel) 3, (1952). 334. – Chandan Singh Dalawat Aug 6 at 6:11
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0There it is. If you found it by searching rather than sheer memory, would you mind divulging your methods? I have a tough time finding things on MO. – 2010-08-12
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1Professor Clark: I remember vaguely that KConrad mentioned it, so it was easier to find. – 2010-08-12
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0I looked in page 245 of Ebbinghaus et. al. and the theorem is not there. A chapter/section name and number might have been more helpful. – 2010-08-13
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7George: See p. 245 on Google Books! Here is the Lie-theoretic proof Dick Gross found as a graduate student: if R^n were a field for n > 2 then the Lie algebra of its (comm.) unit group would be trivial. Since R^n and its unit group would be n-dimensional connected and simply connected real Lie groups with trivial Lie algebra, the bijection between connected and simply connected real Lie groups and real Lie algebras makes the field R^n and its unit group isom., but one has torsion and the other doesn't (-1 is in a field of char. 0). So n is at most 2. – 2010-08-13
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1KCd: +x! To unpack a couple of statements: "if R^n were a degree n extension field R[x]/(p(x)=0 then (...) "; unit group is connected and, for n>2, simply connected because it is (R^n minus one point). – 2010-08-13
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Muad: Are you talking about this www.fc.up.pt/mp/jcsantos/PDF/artigos/FTA.pdf
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0Thanks Chandru .. It could be the one, as it uses the exponential map. – 2010-08-12
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0@George S: Hey no problem! Please vote, for everyone – 2010-08-12
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0@Chandru1, @George: Removed discussion of an interesting, but clearly off-topic paper. – 2010-08-14