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Thread: Irreducible polynomial

  1. #1
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    Irreducible polynomial

    I need to prove the following:

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

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

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

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

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

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

    Now,

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

    and so,

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

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

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

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

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

    which contradicts the fact that $\displaystyle \mathbb{Q}[X]$ 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; Feb 14th 2009 at 08:59 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 $\displaystyle f(X)$ is irreducible in $\displaystyle \mathbb{Q}[X]$ if and only if $\displaystyle f(\alpha X + \beta)$ is irreducible in $\displaystyle \mathbb{Q}[X]$"

    I want to prove "only if"; that is, $\displaystyle f(\alpha X + \beta)$ irreducible $\displaystyle \Rightarrow f(X)$ 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 $\displaystyle \alpha \not = 0$! Now define $\displaystyle \phi: \mathbb{Q}[X]\to \mathbb{Q}[X]$ by $\displaystyle \phi (f(X)) = f(\alpha X+\beta)$. Since $\displaystyle \alpha\not = 0$ it means $\displaystyle \deg f = \deg \phi (f)$. Say that $\displaystyle f(X) = f_1(X)f_2(X)$ where $\displaystyle \deg f_1,\deg f_2 < \deg f$ and $\displaystyle f_1,f_2\not \in \mathbb{Q}$. This means, $\displaystyle \phi (f) = \phi (f_1)\phi (f_2)$ but $\displaystyle \deg \phi (f) = \deg (f)$, $\displaystyle \deg \phi (f_1) = \deg f_1$, $\displaystyle \deg \phi (f_2) = \deg f_2$. It follows that $\displaystyle \deg \phi(f_1),\deg \phi(f_2) < \deg \phi (f)$ and $\displaystyle \phi(f_1),\phi(f_2) \not \in \mathbb{Q}$, therefore $\displaystyle \phi (f) = f(\alpha X + \beta)$ is not irreducible. We have proved that if $\displaystyle f(X)$ is reducible then $\displaystyle f(\alpha X + \beta)$ is reducible, taking the contrapositive we have proven that if $\displaystyle f(\alpha X + \beta)$ is irreducible then $\displaystyle f(X)$ is irreducible. To prove the converse I am going to give you a hint . Define $\displaystyle \theta : \mathbb{Q}[X] \to \mathbb{Q}[X]$ by $\displaystyle \theta (g(X)) = g\left( \tfrac{1}{\alpha}X - \tfrac{\beta}{\alpha}\right)$ (again $\displaystyle \alpha\not = 0$). Notice that $\displaystyle \phi (\theta(g(X))) = \theta (\phi (g(X))) = g(X)$.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    First of all $\displaystyle \alpha \not = 0$! Now define $\displaystyle \phi: \mathbb{Q}[X]\to \mathbb{Q}[X]$ by $\displaystyle \phi (f(X)) = f(\alpha X+\beta)$. Since $\displaystyle \alpha\not = 0$ it means $\displaystyle \deg f = \deg \phi (f)$. Say that $\displaystyle f(X) = f_1(X)f_2(X)$ where $\displaystyle \deg f_1,\deg f_2 < \deg f$ and $\displaystyle f_1,f_2\not \in \mathbb{Q}$. This means, $\displaystyle \phi (f) = \phi (f_1)\phi (f_2)$ but $\displaystyle \deg \phi (f) = \deg (f)$, $\displaystyle \deg \phi (f_1) = \deg f_1$, $\displaystyle \deg \phi (f_2) = \deg f_2$. It follows that $\displaystyle \deg \phi(f_1),\deg \phi(f_2) < \deg \phi (f)$ and $\displaystyle \phi(f_1),\phi(f_2) \not \in \mathbb{Q}$, therefore $\displaystyle \phi (f) = f(\alpha X + \beta)$ is not irreducible. We have proved that if $\displaystyle f(X)$ is reducible then $\displaystyle f(\alpha X + \beta)$ is reducible, taking the contrapositive we have proven that if $\displaystyle f(\alpha X + \beta)$ is irreducible then $\displaystyle f(X)$ is irreducible. To prove the converse I am going to give you a hint . Define $\displaystyle \theta : \mathbb{Q}[X] \to \mathbb{Q}[X]$ by $\displaystyle \theta (g(X)) = g\left( \tfrac{1}{\alpha}X - \tfrac{\beta}{\alpha}\right)$ (again $\displaystyle \alpha\not = 0$). Notice that $\displaystyle \phi (\theta(g(X))) = \theta (\phi (g(X))) = g(X)$.
    Thanks a lot for your help! When you say $\displaystyle f_1,f_2\not \in \mathbb{Q}$, are you saying that $\displaystyle f_1,f_2\not \in \mathbb{Q}[X]$? 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 $\displaystyle f_1,f_2\not \in \mathbb{Q}$, are you saying that $\displaystyle f_1,f_2\not \in \mathbb{Q}[X]$? I'm not entirely sure what you are trying to say here.
    A polynomial $\displaystyle h(X)$ is in $\displaystyle \mathbb{Q}$ if and only if $\displaystyle h(X)$ 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 $\displaystyle h(X)$ is in $\displaystyle \mathbb{Q}$ if and only if $\displaystyle h(X)$ 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 $\displaystyle p$ be a prime number and consider the polynomial $\displaystyle X^{p-1} + X^{p-2} + ... + X + 1$. This polynomial is always irreducible. See if you can prove this by $\displaystyle X\mapsto X+1$ (i.e. $\displaystyle \alpha = 1 \text{ and }\beta = 1$).
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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 $\displaystyle p$ be a prime number and consider the polynomial $\displaystyle X^{p-1} + X^{p-2} + ... + X + 1$. This polynomial is always irreducible. See if you can prove this by $\displaystyle X\mapsto X+1$ (i.e. $\displaystyle \alpha = 1 \text{ and }\beta = 1$).
    I think in this instance, we have that $\displaystyle f(X) = \frac{X^p - 1}{X-1}$, and so if we set $\displaystyle X\mapsto X+1$, we get

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

    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; Feb 14th 2009 at 03:21 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 $\displaystyle f(X) = \frac{X^p - 1}{X-1}$, and so if we set $\displaystyle X\mapsto X+1$, we get

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

    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 $\displaystyle f(X) = \frac{X^p - 1}{X-1}$. 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 $\displaystyle f(X)(X-1) = X^p - 1$ - this can be easily seen by multiplying out the LHS. Now if you let $\displaystyle X\mapsto X+1$ then you get $\displaystyle Xf(X+1) = (X+1)^p - 1 = X\sum_{k=0}^{p-1}{{p}\choose {k+1}}X^k$. We can "cancel" the $\displaystyle X$ (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 $\displaystyle f(X+1) = \sum_{k=1}^{p-1} {{p}\choose {k+1}}X^k$. Now apply Eisenstein criterion by using the fact that $\displaystyle p$ divides $\displaystyle {p\choose {\ell}}$ if $\displaystyle 1 < \ell < p$.
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  11. #11
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    One other question. If I set, say, $\displaystyle \alpha = 1$, so that our polynomials are monic, could I conclude that this was the case for $\displaystyle \mathbb{Z}[X]$? 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, $\displaystyle \alpha = 1$, so that our polynomials are monic, could I conclude that this was the case for $\displaystyle \mathbb{Z}[X]$? I guess the maps you could express in a proof would be well defined.
    If the polynomials have integer coefficients then you can define $\displaystyle X\mapsto \pm X + \beta$ 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 $\displaystyle \mathbb{Q}$ is replaced by $\displaystyle \mathbb{Z}$. We need to avoid the case $\displaystyle |\alpha| > 1$ because we cannot define an inverse transform because we are in the integers.
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