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Math Help - A property of polynomials

  1. #1
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    A property of polynomials

    Let \mathbb{C}[x] be the set of polynomials with complex coefficients. For every p(x) \in \mathbb{C}[x] and n \in \mathbb{N}, there exist q(x), r(x) \in \mathbb{C}[x] such that p(x)q(x)=r(x^n). True or false?
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    MHF Contributor Bruno J.'s Avatar
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    True!

    For every factor x-s of p(x), and every 1 \leq k \leq n, we multiply p(x) by an appropriate number of factors of the form x-e^{2\pi i k/ n}s, so as to obtain a polynomial which can be written as a product of powers of polynomials of the form \prod_{1 \leq k \leq n}(x-e^{2\pi i k/ n}s) = x^n-s^n.
    Last edited by Bruno J.; June 5th 2010 at 08:40 PM.
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    Quote Originally Posted by Bruno J. View Post
    True!

    For every factor x-s of p(x), and every 1 \leq k \leq n, we multiply p(x) by an appropriate number of factors of the form x-e^{2\pi i k/ n}s, so as to obtain a polynomial which can be written as a product of powers of polynomials of the form \prod_{1 \leq k \leq n}(x-e^{2\pi i k/ n}s) = x^n-s.
    right, or just multiply by x^{n-1}+sx^{n-2} + \cdots + s^{n-1} to get x^n - s^n.

    now, the interesting part of the question: what if \mathbb{C} is replaced by any field F?

    so this time p(x) is not necessarily factored into linear polynomials because F might not be algebraically closed.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by NonCommAlg View Post
    right, or just multiply by x^{n-1}+sx^{n-2} + \cdots + s^{n-1} to get x^n - s^n.

    now, the interesting part of the question: what if \mathbb{C} is replaced by any field F?

    so this time p(x) is not necessarily factored into linear polynomials because F might not be algebraically closed.
    I doubt it holds in any field! I'll try to think of a counter-example.
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    Quote Originally Posted by Bruno J. View Post
    I doubt it holds in any field! I'll try to think of a counter-example.
    let me give you an idea: it is true for n = 2 because obviously every polynomial p(x) can be written as \alpha(x^2) + x \beta(x^2), for some \alpha(x), \beta(x) \in F[x]. now let q(x)=\alpha(x^2)-x \beta(x^2).
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    MHF Contributor Bruno J.'s Avatar
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    Right! And in the case n=3, we can write p(x)=a+bx+cx^2 with a,b,c \in F[x^3]; then we let q(x)=c^2x^4-bcx^3+(b^2-ac)x^2-abx+a^2, so that p(x)q(x)=c^3x^6+(b^3-3abc)x^3+a^3...
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    MHF Contributor Bruno J.'s Avatar
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    We write p(x)=\sum_{j=0}^n \alpha_jx^j, with \alpha_j \in F[x^n]. We let \epsilon be a primitive root of x^{n-1}+x^{n-2}+\dots + 1 = 0 in \overline{F}. Now, we let q(x)=\prod_{1 \leq j<n} p(\epsilon^j x). A quick check shows that the coefficients of q(x) lie in F. Now forget that the \alpha_j are polynomials in x; the coefficients of p(x)q(x), as a polynomial in x, are symmetric polynomials in \epsilon^j \ (0 \leq j < n) with coefficients in F[x^n]. But since \epsilon^n-1=0, only the first and the last symmetric polynomial invariants do not vanish, and hence only powers of x^n remain, i.e. p(x)q(x) \in F[x^n]...
    Last edited by Bruno J.; June 6th 2010 at 08:46 PM.
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    Quote Originally Posted by Bruno J. View Post
    We write p(x)=\sum_{j=0}^n \alpha_jx^j, with \alpha_j \in F[x^n]. We let \epsilon be a primitive root of x^{n-1}+x^{n-2}+\dots + 1 = 0 in \overline{F}. Now, we let q(x)=\prod_{1 \leq j<n} p(\epsilon^j x). A quick check shows that the coefficients of q(x) lie in F. Now forget that the \alpha_j are polynomials in x; the coefficients of p(x)q(x), as a polynomial in x, are symmetric polynomials in \epsilon^j \ (0 \leq j < n) with coefficients in F[x^n]. But since \epsilon^n-1=0, only the first and the last symmetric polynomial invariants do not vanish, and hence only powers of x^n remain, i.e. p(x)q(x) \in F[x^n]...
    very nice! just note that the n-th primitive root of unity might not exist in \overline{F}, for example the second primitive root of unity does not exsist in \overline{F} if char(F)=2.

    anyway, we really don't need that! just let a_j, \ 0 \leq j < n, \ a_0=1, be the roots of x^n=1 in \overline{F} and let q(x)=\prod_{j=1}^{n-1}p(a_jx).
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by NonCommAlg View Post
    very nice! just note that the n-th primitive root of unity might not exist in \overline{F}, for example the second primitive root of unity does not exsist in \overline{F} if char(F)=2.

    anyway, we really don't need that! just let a_j, \ 0 \leq j < n, \ a_0=1, be the roots of x^n=1 in \overline{F} and let q(x)=\prod_{j=1}^{n-1}p(a_jx).
    Haha! Thanks. Wouldn't have done it without the hint.

    Thanks for the correction!

    Nice problem too.
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    this problem has a nice result in linear algebra: for a field F let F(x) be the field of rational functions in the indeterminate x, i.e. F(x)=\left \{\frac{p(x)}{q(x)}: \ \ p(x),q(x) \in F[x], \ q(x) \neq 0 \right \}.

    let n \in \mathbb{N}. then F(x^n) is obviously a subfield of F(x) and therefore we can consider F(x) as a vector space over F(x^n). using what we just proved, it's easy to see that every element of F(x)

    can be written uniquely as \sum_{j=0}^{n-1} \beta_j x^j, \ \beta_j \in F(x^n). thus we get the result \dim_{F(x^n)} F(x)=n, which is basically the reason that i created this thread!
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by NonCommAlg View Post
    this problem has a nice result in linear algebra: for a field F let F(x) be the field of rational functions in the indeterminate x, i.e. F(x)=\left \{\frac{p(x)}{q(x)}: \ \ p(x),q(x) \in F[x], \ q(x) \neq 0 \right \}.

    let n \in \mathbb{N}. then F(x^n) is obviously a subfield of F(x) and therefore we can consider F(x) as a vector space over F(x^n). using what we just proved, it's easy to see that every element of F(x)

    can be written uniquely as \sum_{j=0}^{n-1} \beta_j x^j, \ \beta_j \in F(x^n). thus we get the result \dim_{F(x^n)} F(x)=n, which is basically the reason that i created this thread!
    Nice!

    Remind's me of Luroth's theorem : every subfield of \mathbb{C}(z) which contains more than \mathbb{C} itself is isomorphic to \mathbb{C}(z).

    (I don't know if Luroth's theorem generalizes somewhat to rational function fields over other fields - I'm sure you can tell me!)
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    Quote Originally Posted by Bruno J. View Post
    Nice!

    Remind's me of Luroth's theorem : every subfield of \mathbb{C}(z) which contains more than \mathbb{C} itself is isomorphic to \mathbb{C}(z).

    (I don't know if Luroth's theorem generalizes somewhat to rational function fields over other fields - I'm sure you can tell me!)
    Luroth's theorem is true for any field F. what it says is that every subfield K of F(x) which contains F is in the form F(r(x)), for some r(x) \in F(x). now if K \neq F, then r(x) \notin F and then it's

    easy to see that t(r(x)) \mapsto t(x), for every t(x) \in F(x), defnes an isomorphism from K to F(x).
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    MHF Contributor Bruno J.'s Avatar
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    Can we generalize Luroth's theorem to rational function fields generated by two or more elements? I suppose...
    Last edited by Bruno J.; June 8th 2010 at 08:25 PM.
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