```Date: Jul 24, 2013 10:39 AM
Author: David C. Ullrich
Subject: Re: can someone point me to the proof that

On 23 Jul 2013 19:15:56 GMT, Bart Goddard <goddardbe@netscape.net>wrote:>dullrich@sprynet.com wrote in news:ppiqu8d1micop548h956stpcfgoopcufjv@>4ax.com:> >> Looking at the difference of our two polynomials,>> say p(t) = 0 for all t in our infinite field. So p>> has zero constant term (hence p(t) = t q(t) for>> some polynomial t and we're done, hence the>> "not that it matters" above). How does>> it follow that p'(t) = 0?>>If the two polynomials are f(X) and g(X), then let>(*) F(X,Y) = (f(X)-f(Y))/(X-Y) and G(X,Y)= (g(X)-g(Y))/(X-Y).>>Since f and g are equal for all values of X, so are F and G>for all values of X and Y.  Since f'(X) = F(X,X) and>g'(X)=G(X,X), it follows that f'=g'. Hmm. Regardless of your answer to my objectionbelow, the "other" argument seems much simpler.Anyway: Of course (*) is literally true onlyfor X <> Y. I'm not trying to be pedantic, I doknow what you mean by F(X,X).But: What's evident is(1) F(X,Y) = G(X,Y) if X <> Y.We need to get from there to(2) F(X,Y) = G(X,Y) for all X, Y.How does  (1) imply (2)? Note that (2) is notliterally trivial from (*), since (*) is literally trueonly for X <> Y.Of course (1) does imply (2), but the only argumentto that effect that I see offhand _uses_ the factthat we're trying to prove! (Fix Y. The polynomialH(X) = F(X,Y) - G(X,Y) has infinitely many zeroes,hence it must be the zero polynomial.)Hmm^2. I just realized that the argumentshowing that (1) implies (2) can't be too trivial,in particular it _must_ use the fact that we'retalking about polynomials over an infinite field.Example:Consider polynomials over Z_2, the field withtwo elements. Let f(x) = x^2, g(x) = x.Then f(X) = g(X) for every X, although f <> g.In this case we have F(X,Y) = X+Y andG(X,Y) = 1. And so (1) holds but (2) is false.What argument did you have in mind to showthat (1) implies (2)?> I think we can make>this work in non-commutative rings too.  But certainly >it works over integral domains.>>B.
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