From: hrubin@b.stat.purdue.edu (Herman Rubin) Newsgroups: sci.math Subject: Re: Algebraic proof of the Fundamental Theorem of Algebra? Date: 12 Dec 1998 18:13:35 -0500 In article <74ski0$kk8@mcmail.cis.McMaster.CA>, Zdislav V. Kovarik wrote: >In article <3671b333.0@news.ic.sunysb.edu>, wrote: >>Is an algebraic proof of the Fundamental Theorem of Algebra known? >>The only proofs that I'm aware of use complex analysis. >>Please reply to nmanes@hotmail.com >It depends on where you draw the line between algebra and analysis. A >purely algebraic proof (working only with field operations) would hardly >exist because the condition that (more than) suffices for the existence of >roots of non-constant polynomials is metric completeness of complex >numbers, or Dedekind order completeness of the reals - a property >popularly considered as part of analysis. >There is a proof which sort of reduces the use of analysis to a result >which requires almost nothing from imagination, and very little of Real >Analysis: > "Every real polynomial of odd degree has at least one real root" >(after all, it changes sign for large values (of different signs) of the >variable, and is continuous in between). > From this, the rest can be done purely algebraically, using for example >properties of symmetric polynomials. > The algebra part is actually a theorem: "If a field F containing the >rationals has the property that every polynomial of odd degree over F has >a root in F, but F is still not algebraically closed then F[i] is >algebraically closed." > (It may be possible to drop or weaken the assumption of the presence of >the rationals.) Only a field of characteristic not zero will fail to contain the rationals, and for such fields, the conditions, plus one added one, are not enough. The added condition is that every sum of squares has a square root; this does not follow from odd degree polynomials having roots, as one can consider the algebraic closure of the rationals by successively adjoining one root of an odd degree irreducible polynomial. This proof is due to Gauss, and can be found in van der Waerden, among other places. It proceeds by induction on the highest power of 2 dividing the degree of the polynomial. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399 hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 ============================================================================== From: Robin Chapman Newsgroups: sci.math Subject: Re: WHAT IS: FTA? Date: Thu, 17 Dec 1998 09:49:03 GMT In article , edgar@math.ohio-state.edu (G. A. Edgar) wrote: > > > > It can be proved without the use of Liouville's theorem, Rouch'e's theorem > > the arguement principle and the winding number just using the following > > properties of R: > > (i) R is an ordered field in which each positive element is a square; > > (ii) each polynomial of odd degree over R has a zero in R. > > When you use less analysis you end up using more algebra. There is a trade-off. > > As an analyst, I like the proof in Rudin's text, where he uses > the Bolzano-Weierstrass theorem and a little trigonometry > (but no theory of polynomials or fields, and no complex analysis). > This proof of Rudin's is still essentially a winding number argument: it shows that the function z |--> 1 + b_k z^k + ... + b_n z^n maps small circles about the origin to paths meeting interior of the unit disc. Of course one might still argue that the Fundamental Theorem of Algebra might demand an algebraic proof (even though it's not an algebraic theorem at all :-) ). With Galois theory amd Sylow's theorem there's a slick proof, as given in most texts on Galois theory. On the other hand one can give a much more elementary algebraic proof. It suffices to show that each real poynomial has a complex root. We use induction on the power of 2 dividing the degree d of the polynomial. That is set d = 2^r s with s odd and use induction on r. For r = 0 the polynomial has a real root by property (ii). In general first note that each complex number has a complex square root. This follows from (i) as one can write down a formula for the complex square roots only involving square roots of nonnegative reals. Let f be a real poynomial of even degree d, and let its roots be a_1, a_2, ..., a_d in some finite extension field of C. For a give real number t consider the polynomial of degree d(d-1)/2 whose roots are the a_j + a_k + t a_j a_k. Its coefficients are symmetric functions in the a_j with real coefficients and so also lie in R. But the power of 2 dividing d(d-1)/2 is one less than that dividing d, so by induction for each real t there are j < k with a_j + a_k + t a_j a_k in C. By taking more than d(d-1)/2 values of t we find distinct t_1 and t_2 and some j and k with a_j + a_k + t_1 a_j a_k and a_j + a_k + t_2 a_j a_k in C and so both a_j + a_k and a_j a_k lie in C. Then a_j and a_k are roots of a quadratic with complex coefficients and so are complex themselves. Robin Chapman + "They did not have proper SCHOOL OF MATHEMATICal Sciences - palms at home in Exeter." University of Exeter, EX4 4QE, UK + rjc@maths.exeter.ac.uk - Peter Carey, http://www.maths.ex.ac.uk/~rjc/rjc.html + Oscar and Lucinda, chapter 20 -----------== Posted via Deja News, The Discussion Network ==---------- http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own