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April 14th, 2011, 07:16 PM   #1
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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.
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April 14th, 2011, 08:11 PM   #2
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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-
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April 14th, 2011, 09:49 PM   #3
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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,
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April 15th, 2011, 12:36 AM   #4
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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].
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April 15th, 2011, 02:08 AM   #5
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Re: irreducible polynomial

Turgul,
Thank you very much, your reply made me clear with this question.
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