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Math Help - Irreducible polynomials over Q

  1. #1
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    Irreducible polynomials over Q

    Let p be any prime number. Show that

    1 + x + x^2/2! + x^3/3! ... + x^p/p! is irreducible over Q.

    Okay for this one, I know that if the polynomial were degree 2 or 3 I could simply check whether it has roots, but I want to generalize to any prime p.

    I would love to use Eisenstein Criterion, but because of the polynomial is not in Z[x] I cannot. I have spent a long time trying to convert the polynomial into a form that is usable. I came across an equation that said x^p -1/x-1 = 1+x+...+x^p-2+x^p-1. I have had no success so far.

    I would really appreciate the help!
    Last edited by Jameson; May 22nd 2009 at 03:01 PM.
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  2. #2
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    Quote Originally Posted by curiousmuch View Post
    Let p be any prime number. Show that

    1 + x + x^2/2! + x^3/3! ... + x^p/p! is irreducible over Q.

    Okay for this one, I know that if the polynomial were degree 2 or 3 I could simply check whether it has roots, but I want to generalize to any prime p.

    I would love to use Eisenstein Criterion, but because of the polynomial is not in Z[x] I cannot. I have spent a long time trying to convert the polynomial into a form that is usable. I came across an equation that said x^p -1/x-1 = 1+x+...+x^p-2+x^p-1. I have had no success so far.

    I would really appreciate the help!
    Actually you can use Eisenstien, but you need to fix it up first.

    Gauss's lemma is the key that you need.

    Let R be a Unique factorization domain with field of fractions F and let p(x) \in R[x] if p(x) is reducible in F[x] then p(x) is reducible in R[x].

    What you actually want is the contraposative.

    if p(x) is irreducible in R[x] then it is irreducable in F[x]

    So in your case R is the integers and F is the rationals.

    You should be able to finish from here

    Happy hunting
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    Wow that is so useful! (we actually have never learned Gauss's lemma in class). Sorry but I'm still a bit confused how to look at the polynomial in Z[x]. Thanks a lot.
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    Quote Originally Posted by curiousmuch View Post
    Wow that is so useful! (we actually have never learned Gauss's lemma in class). Sorry but I'm still a bit confused how to look at the polynomial in Z[x]. Thanks a lot.

    so you have

    1+x+...+\frac{x^p}{p!}=0

    multiply by p! to get

    p! +p!x+\frac{p!}{2}x^2+...+\frac{P!}{(p-1)!}x^{p-1}+\frac{p!}{p!}x^p=0

    p! +p!x+\frac{p!}{2}x^2+...+px^{p-1}+x^p=0

    Note that all of the coeffeints are integers so now this is eisenstien in p so it is irreducable.
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    TheEmptySet's solution needs a little bit fixing:

    let f(x)=1+x + \cdots + \frac{x^{p-1}}{(p-1)!} + \frac{x^p}{p!}. suppose f(x) was reducible over \mathbb{Q}. then p!f(x) would also be reducible over \mathbb{Q} too. but by Eisenstein's criterion p!f(x) is irreducible over \mathbb{Z},

    and therefore irreducible over \mathbb{Q} as well, by Gauss's lemma. contradiction!
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    Quote Originally Posted by NonCommAlg View Post
    TheEmptySet's solution needs a little bit fixing:

    let f(x)=1+x + \cdots + \frac{x^{p-1}}{(p-1)!} + \frac{x^p}{p!}. suppose f(x) was reducible over \mathbb{Q}. then p!f(x) would also be reducible over \mathbb{Q} too. but by Eisenstein's criterion p!f(x) is irreducible over \mathbb{Z},

    and therefore irreducible over \mathbb{Q} as well, by Gauss's lemma. contradiction!
    Thanks
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    It seems that this is also true if p is any positive integer, not just prime.
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  8. #8
    Super Member Gamma's Avatar
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    To use Eisentstein's criterion it needs to be the case that your polynomial is monic (although a stronger condition can be shown to work so long as p does not divide the leading term) and that p divides every other term, but p^2 does not divide the constant term. This is why it is critical that p be prime.

    If p = 4 for instance you would have 4!p(x)=24 + 24x + 12x^2 + 4x^3 + x^4, so look at the x^3 coefficient, the only prime dividing 4 is 2, but 2^2=4|24 so Eisenstein's criterion fails and this method no longer works.
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