On 23 Jul 2013 19:15:56 GMT, Bart Goddard <email@example.com> wrote:
>firstname.lastname@example.org 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 objection below, the "other" argument seems much simpler.
Anyway: Of course (*) is literally true only for X <> Y. I'm not trying to be pedantic, I do know 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 not literally trivial from (*), since (*) is literally true only for X <> Y.
Of course (1) does imply (2), but the only argument to that effect that I see offhand _uses_ the fact that we're trying to prove! (Fix Y. The polynomial H(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 argument showing that (1) implies (2) can't be too trivial, in particular it _must_ use the fact that we're talking about polynomials over an infinite field. Example:
Consider polynomials over Z_2, the field with two 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 and G(X,Y) = 1. And so (1) holds but (2) is false.
What argument did you have in mind to show that (1) implies (2)?
> I think we can make >this work in non-commutative rings too. But certainly >it works over integral domains. > >B.