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 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.