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 December 5th, 2007, 06:59 PM #1 Newbie   Joined: Oct 2007 Posts: 3 Thanks: 0 ring Let R be any ring, let s be an element in R, and suppose we have some f(X)∈R[X] with f(s)=0, and D[f(X)] also giving value 0 at s. Prove that f(X)=g(X)(X-s)^2 for some g(X)∈R[X]. (hint. divide f(X) by (X-s)^2 with remainder, and compute D[f(X)] from that expression.) i understand the hint, but i am not sure what to do from there.
December 6th, 2007, 01:21 PM   #2
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 Originally Posted by Frazier001 Let R be any ring, let s be an element in R, and suppose we have some f(X)∈R[X] with f(s)=0, and D[f(X)] also giving value 0 at s. Prove that f(X)=g(X)(X-s)^2 for some g(X)∈R[X]. (hint. divide f(X) by (X-s)^2 with remainder, and compute D[f(X)] from that expression.) i understand the hint, but i am not sure what to do from there.
You need to know the rules how to manipulate with the formal derivative.

By long division you get the identity
f(x)=g(x)(x-s)^2+r(x).
What is the degree of r(x)?
Can you express from this equality (and the known rules for the formal derivative) Df(x)?
Using the fact that Df(x)=0, what you can say about r(x)?

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