email@example.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'. I think we can make this work in non-commutative rings too. But certainly it works over integral domains.