# A property of polynomials

• June 5th 2010, 05:48 PM
NonCommAlg
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?
• June 5th 2010, 07:14 PM
Bruno J.
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 $\prod_{1 \leq k \leq n}(x-e^{2\pi i k/ n}s) = x^n-s^n$.
• June 5th 2010, 07:30 PM
NonCommAlg
Quote:

Originally Posted by Bruno J.
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 $F$? (Evilgrin)

so this time $p(x)$ is not necessarily factored into linear polynomials because $F$ might not be algebraically closed.
• June 5th 2010, 07:54 PM
Bruno J.
Quote:

Originally Posted by NonCommAlg
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 $F$? (Evilgrin)

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.
• June 5th 2010, 08:25 PM
NonCommAlg
Quote:

Originally Posted by Bruno J.
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 $q(x)=\alpha(x^2)-x \beta(x^2).$
• June 6th 2010, 01:15 PM
Bruno J.
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 $p(x)q(x)=c^3x^6+(b^3-3abc)x^3+a^3$...
• June 6th 2010, 05:15 PM
Bruno J.
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. 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. $p(x)q(x) \in F[x^n]$...
• June 8th 2010, 08:04 AM
NonCommAlg
Quote:

Originally Posted by Bruno J.
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. 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 $q(x)=\prod_{j=1}^{n-1}p(a_jx).$
• June 8th 2010, 08:13 AM
Bruno J.
Quote:

Originally Posted by NonCommAlg
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! (Bow)

Nice problem too.
• June 8th 2010, 08:34 AM
NonCommAlg
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!
• June 8th 2010, 08:59 AM
Bruno J.
Quote:

Originally Posted by NonCommAlg
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!)
• June 8th 2010, 09:53 AM
NonCommAlg
Quote:

Originally Posted by Bruno J.
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 $F(x).$
• June 8th 2010, 05:03 PM
Bruno J.
Can we generalize Luroth's theorem to rational function fields generated by two or more elements? I suppose...