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Thread: A property of polynomials

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    A property of polynomials

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

    For every factor $\displaystyle x-s$ of $\displaystyle p(x)$, and every $\displaystyle 1 \leq k \leq n$, we multiply $\displaystyle p(x)$ by an appropriate number of factors of the form $\displaystyle x-e^{2\pi i k/ n}s$, so as to obtain a polynomial which can be written as a product of powers of polynomials of the form $\displaystyle \prod_{1 \leq k \leq n}(x-e^{2\pi i k/ n}s) = x^n-s^n$.
    Last edited by Bruno J.; Jun 5th 2010 at 07:40 PM.
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    Quote Originally Posted by Bruno J. View Post
    True!

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

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

    so this time $\displaystyle p(x)$ is not necessarily factored into linear polynomials because $\displaystyle F$ 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 $\displaystyle x^{n-1}+sx^{n-2} + \cdots + s^{n-1}$ to get $\displaystyle x^n - s^n.$

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

    so this time $\displaystyle p(x)$ is not necessarily factored into linear polynomials because $\displaystyle F$ 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 $\displaystyle p(x)$ can be written as $\displaystyle \alpha(x^2) + x \beta(x^2),$ for some $\displaystyle \alpha(x), \beta(x) \in F[x].$ now let $\displaystyle q(x)=\alpha(x^2)-x \beta(x^2).$
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    MHF Contributor Bruno J.'s Avatar
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    Right! And in the case $\displaystyle n=3$, we can write $\displaystyle p(x)=a+bx+cx^2$ with $\displaystyle a,b,c \in F[x^3]$; then we let $\displaystyle q(x)=c^2x^4-bcx^3+(b^2-ac)x^2-abx+a^2$, so that $\displaystyle p(x)q(x)=c^3x^6+(b^3-3abc)x^3+a^3$...
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    MHF Contributor Bruno J.'s Avatar
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    We write $\displaystyle p(x)=\sum_{j=0}^n \alpha_jx^j$, with $\displaystyle \alpha_j \in F[x^n]$. We let $\displaystyle \epsilon$ be a primitive root of $\displaystyle x^{n-1}+x^{n-2}+\dots + 1 = 0$ in $\displaystyle \overline{F}$. Now, we let $\displaystyle q(x)=\prod_{1 \leq j<n} p(\epsilon^j x)$. A quick check shows that the coefficients of $\displaystyle q(x)$ lie in $\displaystyle F$. Now forget that the $\displaystyle \alpha_j$ are polynomials in $\displaystyle x$; the coefficients of $\displaystyle p(x)q(x)$, as a polynomial in $\displaystyle x$, are symmetric polynomials in $\displaystyle \epsilon^j \ (0 \leq j < n)$ with coefficients in $\displaystyle F[x^n]$. But since $\displaystyle \epsilon^n-1=0$, only the first and the last symmetric polynomial invariants do not vanish, and hence only powers of $\displaystyle x^n$ remain, i.e. $\displaystyle p(x)q(x) \in F[x^n]$...
    Last edited by Bruno J.; Jun 6th 2010 at 07:46 PM.
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    Quote Originally Posted by Bruno J. View Post
    We write $\displaystyle p(x)=\sum_{j=0}^n \alpha_jx^j$, with $\displaystyle \alpha_j \in F[x^n]$. We let $\displaystyle \epsilon$ be a primitive root of $\displaystyle x^{n-1}+x^{n-2}+\dots + 1 = 0$ in $\displaystyle \overline{F}$. Now, we let $\displaystyle q(x)=\prod_{1 \leq j<n} p(\epsilon^j x)$. A quick check shows that the coefficients of $\displaystyle q(x)$ lie in $\displaystyle F$. Now forget that the $\displaystyle \alpha_j$ are polynomials in $\displaystyle x$; the coefficients of $\displaystyle p(x)q(x)$, as a polynomial in $\displaystyle x$, are symmetric polynomials in $\displaystyle \epsilon^j \ (0 \leq j < n)$ with coefficients in $\displaystyle F[x^n]$. But since $\displaystyle \epsilon^n-1=0$, only the first and the last symmetric polynomial invariants do not vanish, and hence only powers of $\displaystyle x^n$ remain, i.e. $\displaystyle p(x)q(x) \in F[x^n]$...
    very nice! just note that the n-th primitive root of unity might not exist in $\displaystyle \overline{F}$, for example the second primitive root of unity does not exsist in $\displaystyle \overline{F}$ if $\displaystyle char(F)=2.$

    anyway, we really don't need that! just let $\displaystyle a_j, \ 0 \leq j < n, \ a_0=1,$ be the roots of $\displaystyle x^n=1$ in $\displaystyle \overline{F}$ and let $\displaystyle q(x)=\prod_{j=1}^{n-1}p(a_jx).$
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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 $\displaystyle \overline{F}$, for example the second primitive root of unity does not exsist in $\displaystyle \overline{F}$ if $\displaystyle char(F)=2.$

    anyway, we really don't need that! just let $\displaystyle a_j, \ 0 \leq j < n, \ a_0=1,$ be the roots of $\displaystyle x^n=1$ in $\displaystyle \overline{F}$ and let $\displaystyle q(x)=\prod_{j=1}^{n-1}p(a_jx).$
    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 $\displaystyle F$ let $\displaystyle F(x)$ be the field of rational functions in the indeterminate $\displaystyle x,$ i.e. $\displaystyle F(x)=\left \{\frac{p(x)}{q(x)}: \ \ p(x),q(x) \in F[x], \ q(x) \neq 0 \right \}.$

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

    can be written uniquely as $\displaystyle \sum_{j=0}^{n-1} \beta_j x^j, \ \beta_j \in F(x^n).$ thus we get the result $\displaystyle \dim_{F(x^n)} F(x)=n,$ 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 $\displaystyle F$ let $\displaystyle F(x)$ be the field of rational functions in the indeterminate $\displaystyle x,$ i.e. $\displaystyle F(x)=\left \{\frac{p(x)}{q(x)}: \ \ p(x),q(x) \in F[x], \ q(x) \neq 0 \right \}.$

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

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

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

    (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 $\displaystyle \mathbb{C}(z)$ which contains more than $\displaystyle \mathbb{C}$ itself is isomorphic to $\displaystyle \mathbb{C}(z)$.

    (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 $\displaystyle F.$ what it says is that every subfield $\displaystyle K$ of $\displaystyle F(x)$ which contains $\displaystyle F$ is in the form $\displaystyle F(r(x)),$ for some $\displaystyle r(x) \in F(x).$ now if $\displaystyle K \neq F,$ then $\displaystyle r(x) \notin F$ and then it's

    easy to see that $\displaystyle t(r(x)) \mapsto t(x),$ for every $\displaystyle t(x) \in F(x)$, defnes an isomorphism from $\displaystyle K$ to $\displaystyle F(x).$
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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.; Jun 8th 2010 at 07:25 PM.
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