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Topic: can someone point me to the proof that
Replies: 9   Last Post: Jul 24, 2013 10:39 AM

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David C. Ullrich

Posts: 3,555
Registered: 12/13/04
Re: can someone point me to the proof that
Posted: Jul 24, 2013 10:39 AM
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On 23 Jul 2013 19:15:56 GMT, Bart Goddard <>

> wrote in news:ppiqu8d1micop548h956stpcfgoopcufjv@

>> 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.

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.

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