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April 14th, 2011, 07:16 PM  #1 
Newbie Joined: Apr 2011 Posts: 18 Thanks: 0  irreducible polynomial
F is a field, and f(x) is an irreducible polynomial in F[x], h(x) is in F[x], and there exists a t in F that f(t) = h(t), Can we conclude that f(x)h(x) ? I think this is right, but I cannot prove it. 
April 14th, 2011, 08:11 PM  #2 
Senior Member Joined: Jun 2010 Posts: 618 Thanks: 0  Re: irreducible polynomial
fleur, Are there additional conditions? It seems that f(x) = x²+1 and g(x) = x+1 as polynomials in ?[x] would provide a counterexample at t=1. Am I missing something? Ormkärr 
April 14th, 2011, 09:49 PM  #3 
Newbie Joined: Apr 2011 Posts: 18 Thanks: 0  Re: irreducible polynomial
Ormkärr, Sorry , I made mistake, It should be: F is a field, and f(x) is an irreducible polynomial in F[x], h(x) is in F[x], and there exists t (in F's extension field) that f(t) = 0 and h(t) = 0, 
April 15th, 2011, 12:36 AM  #4 
Senior Member Joined: Aug 2010 Posts: 195 Thanks: 5  Re: irreducible polynomial
Conceptually, it may be easier to think of the minimal polynomial in F[x] (by degree) for which t has a root. That is, if f(x) is the minimal polynomial in F[x] such that f(t) = 0, and h(x) is a polynomial in F[x} such that h(t) = 0, must f(x)h(x)? Hint: polynomial division It is worth noting that if either your claim, or this claim are true, then so is the other claim and all monic irreducible polynomials with t as a root must be equal, and are exactly the minimal polynomial of t in F[x]. 
April 15th, 2011, 02:08 AM  #5 
Newbie Joined: Apr 2011 Posts: 18 Thanks: 0  Re: irreducible polynomial
Turgul, Thank you very much, your reply made me clear with this question. 

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