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 April 14th, 2011, 08:16 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, 09:11 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 counter-example at t=1. Am I missing something? -Ormkärr-
 April 14th, 2011, 10:49 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, 01:36 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, 03:08 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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