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Math Help - Irreducible polynomial

  1. #1
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    Irreducible polynomial

    I need to prove the following:

    "Prove that f(X) is irreducible in \mathbb{Q}[X] if and only if f(\alpha X + \beta) is irreducible in \mathbb{Q}[X]"

    I want to prove "if"; that is, f(\alpha X + \beta) irreducible \Rightarrow f(X) irreducible.

    So, I think we have that f(X)= a(X)b(X), with one of \deg(a), \deg(b) = \deg(f).

    The hint states that we should argue by contradiction. So, in this instance, we shall assume that f(\alpha X + \beta) is reducible, and hence

    f(\alpha X + \beta) = g(\alpha X + \beta)h(\alpha X + \beta)

    where \deg(g), \deg(h) < \deg(f)

    Now,

    f(\alpha X + \beta) \equiv [f(\alpha X + \beta) - f(X)] + f(X)

    and so,

    [f(\alpha X + \beta) - f(X)] + f(X) = g(\alpha X + \beta)h(\alpha X + \beta)

    we say, f(\alpha X + \beta) - f(X)= \tilde{f}(\alpha X + \beta), and so

    f(X) = g(\alpha X + \beta)h(\alpha X + \beta) - \tilde{f}(\alpha X + \beta)

    with \deg(\tilde{f}) < \deg(g), by the division algorithm. But, this is a contradiction, and so \tilde{f}(\alpha X + \beta) = 0. Thus,

    f(X) = g(\alpha X + \beta)h(\alpha X + \beta) = a(X)b(X)

    which contradicts the fact that \mathbb{Q}[X] is a unique factorisation domain.

    I think this is wrong. I'd be really grateful if anyone could pick out the flaws in this proof, and if possible, provide a better (or, correct) proof.

    Thanks in advance,

    HTale.

    Last edited by HTale; February 14th 2009 at 09:59 AM.
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  2. #2
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    Quote Originally Posted by HTale View Post
    I need to prove the following:

    "Prove that f(X) is irreducible in \mathbb{Q}[X] if and only if f(\alpha X + \beta) is irreducible in \mathbb{Q}[X]"

    I want to prove "only if"; that is, f(\alpha X + \beta) irreducible \Rightarrow f(X) irreducible.

    minor error:

    you are are trying to prove the "if" direction.
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  3. #3
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    Quote Originally Posted by GaloisTheory1 View Post

    minor error:

    you are are trying to prove the "if" direction.
    Silly me! Error corrected; thanks for that. Are there any flaws in that proof, as far as you can see?
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  4. #4
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    First of all \alpha \not = 0! Now define \phi: \mathbb{Q}[X]\to \mathbb{Q}[X] by \phi (f(X)) = f(\alpha X+\beta). Since \alpha\not = 0 it means \deg f = \deg \phi (f). Say that f(X) = f_1(X)f_2(X) where \deg f_1,\deg f_2 < \deg f and f_1,f_2\not \in \mathbb{Q}. This means, \phi (f) = \phi (f_1)\phi (f_2) but \deg \phi (f) = \deg (f), \deg \phi (f_1) = \deg f_1, \deg \phi (f_2) = \deg f_2. It follows that \deg \phi(f_1),\deg \phi(f_2) < \deg \phi (f) and \phi(f_1),\phi(f_2) \not \in \mathbb{Q}, therefore \phi (f) = f(\alpha X + \beta) is not irreducible. We have proved that if f(X) is reducible then f(\alpha X + \beta) is reducible, taking the contrapositive we have proven that if f(\alpha X + \beta) is irreducible then f(X) is irreducible. To prove the converse I am going to give you a hint . Define \theta : \mathbb{Q}[X] \to \mathbb{Q}[X] by \theta (g(X)) = g\left( \tfrac{1}{\alpha}X - \tfrac{\beta}{\alpha}\right) (again \alpha\not = 0). Notice that \phi (\theta(g(X))) = \theta (\phi (g(X))) = g(X).
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    First of all \alpha \not = 0! Now define \phi: \mathbb{Q}[X]\to \mathbb{Q}[X] by \phi (f(X)) = f(\alpha X+\beta). Since \alpha\not = 0 it means \deg f = \deg \phi (f). Say that f(X) = f_1(X)f_2(X) where \deg f_1,\deg f_2 < \deg f and f_1,f_2\not \in \mathbb{Q}. This means, \phi (f) = \phi (f_1)\phi (f_2) but \deg \phi (f) = \deg (f), \deg \phi (f_1) = \deg f_1, \deg \phi (f_2) = \deg f_2. It follows that \deg \phi(f_1),\deg \phi(f_2) < \deg \phi (f) and \phi(f_1),\phi(f_2) \not \in \mathbb{Q}, therefore \phi (f) = f(\alpha X + \beta) is not irreducible. We have proved that if f(X) is reducible then f(\alpha X + \beta) is reducible, taking the contrapositive we have proven that if f(\alpha X + \beta) is irreducible then f(X) is irreducible. To prove the converse I am going to give you a hint . Define \theta : \mathbb{Q}[X] \to \mathbb{Q}[X] by \theta (g(X)) = g\left( \tfrac{1}{\alpha}X - \tfrac{\beta}{\alpha}\right) (again \alpha\not = 0). Notice that \phi (\theta(g(X))) = \theta (\phi (g(X))) = g(X).
    Thanks a lot for your help! When you say f_1,f_2\not \in \mathbb{Q}, are you saying that f_1,f_2\not \in \mathbb{Q}[X]? I'm not entirely sure what you are trying to say here.
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  6. #6
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    Quote Originally Posted by HTale View Post
    Thanks a lot for your help! When you say f_1,f_2\not \in \mathbb{Q}, are you saying that f_1,f_2\not \in \mathbb{Q}[X]? I'm not entirely sure what you are trying to say here.
    A polynomial h(X) is in \mathbb{Q} if and only if h(X) is konstant.
    That is all I meant.
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  7. #7
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    Quote Originally Posted by ThePerfectHacker View Post
    A polynomial h(X) is in \mathbb{Q} if and only if h(X) is konstant.
    That is all I meant.
    Aha, in other words, non-unit polynomials. Thank you once again! I will try to attempt the second part of the proof, and post something on here soon
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  8. #8
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    Quote Originally Posted by HTale View Post
    Aha, in other words, non-unit polynomials. Thank you once again! I will try to attempt the second part of the proof, and post something on here soon
    After you finish posting your proof for the problem you been assigned here is a good application of this theorem. Let p be a prime number and consider the polynomial X^{p-1} + X^{p-2} + ... + X + 1. This polynomial is always irreducible. See if you can prove this by X\mapsto X+1 (i.e. \alpha = 1 \text{ and }\beta = 1).
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  9. #9
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    Quote Originally Posted by ThePerfectHacker View Post
    After you finish posting your proof for the problem you been assigned here is a good application of this theorem. Let p be a prime number and consider the polynomial X^{p-1} + X^{p-2} + ... + X + 1. This polynomial is always irreducible. See if you can prove this by X\mapsto X+1 (i.e. \alpha = 1 \text{ and }\beta = 1).
    I think in this instance, we have that f(X) = \frac{X^p - 1}{X-1}, and so if we set X\mapsto X+1, we get

    \frac{1}{X}((X+1)^p -1) = \sum^{p-1}_{r=0} {p \choose r}X^{p-r-1}

    and I think you apply Eisensteins criterion to finish the proof.

    That's just a sketch of what I'd do. If it's correct, then I'll fill in the details and post the other proof and this to this thread.

    HTale
    Last edited by HTale; February 14th 2009 at 04:21 PM.
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  10. #10
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    Quote Originally Posted by HTale View Post
    I think in this instance, we have that f(X) = \frac{X^p - 1}{X-1}, and so if we set X\mapsto X+1, we get

    \frac{1}{X}((X+1)^p -1) = \sum^{p-1}_{r=0} {p \choose r}X^{p-r-1}

    and I think you apply Eisensteins criterion to finish the proof.

    That's just a sketch of what I'd do. If it's correct, then I'll fill in the details and post the other proof and this to this thread.

    HTale
    Yes . The only issue I will take with your proof is that you wrote f(X) = \frac{X^p - 1}{X-1}. This is not a geometric sequence, these are just polynomials (I mean we can add and multiply polynomials we cannot divide them). What I would do to be more formal is say that f(X)(X-1) = X^p - 1 - this can be easily seen by multiplying out the LHS. Now if you let X\mapsto X+1 then you get Xf(X+1) = (X+1)^p - 1 = X\sum_{k=0}^{p-1}{{p}\choose {k+1}}X^k. We can "cancel" the X (I say "cancel" because we are not really dividing since there is no such thing as division of polynomials, however, it is as if we are canceling) to get f(X+1) = \sum_{k=1}^{p-1} {{p}\choose {k+1}}X^k. Now apply Eisenstein criterion by using the fact that p divides {p\choose {\ell}} if 1 < \ell < p.
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  11. #11
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    One other question. If I set, say, \alpha = 1, so that our polynomials are monic, could I conclude that this was the case for \mathbb{Z}[X]? I guess the maps you could express in a proof would be well defined.
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  12. #12
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    Quote Originally Posted by HTale View Post
    One other question. If I set, say, \alpha = 1, so that our polynomials are monic, could I conclude that this was the case for \mathbb{Z}[X]? I guess the maps you could express in a proof would be well defined.
    If the polynomials have integer coefficients then you can define X\mapsto \pm X + \beta to get a transformed polynomial. This transformed polynomial is irreducible if and only if the original polynomial was irreducible by following the exact same proof except \mathbb{Q} is replaced by \mathbb{Z}. We need to avoid the case |\alpha| > 1 because we cannot define an inverse transform because we are in the integers.
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