December 5th, 2007, 07:59 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)(Xs)^2 for some g(X)∈R[X]. (hint. divide f(X) by (Xs)^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, 02:21 PM  #2  
Member Joined: Nov 2007 Posts: 50 Thanks: 0  Re: ring Quote:
By long division you get the identity f(x)=g(x)(xs)^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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