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Suppose $F$ is an algebraically closed field. Find all monic polynomials $f\left(x\right)\in F\left[x\right]$ with distinct roots such that the set of roots of $f$ is closed under multiplication.

In $\mathbb{C}$ we have that the set of polynomials of the type $x^n-a$ have this property where the roots are of the form $\sqrt[n]{a}$ times a power of the n-th root of unity. But how do we work with the general case?

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Not quite: for $x^2-9$, $(-3)(+3)=-9$, which is certainly not a root. You need a condition on $a$... –  anon Jun 4 '12 at 2:01
    
do we need that $a$ itself not be an nth root? –  Galois Jun 4 '12 at 2:06
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up vote 3 down vote accepted

You can't take $a$ arbitrary ; the only $a$ that works must be $1$.

Assume that the roots are closed under multiplication, and let $\alpha \neq 0$ be a root of $f(x)$ and $n = \deg f$. Then the set $$ \{ \alpha, \alpha^2, \alpha^3, \dots, \alpha^n, \alpha^{n+1}\} $$ must contain two elements that are the same, i.e. there must exists $1 \le i,j \le n+1$ such that $i < j$ and $\alpha^i = \alpha^j$, which means $\alpha^{j-i} = 1$. Therefore the roots of $f$ are roots of unity.

Furthermore, the set of roots is finite, closed under multiplication and inverses (since if $\alpha^d = 1$, $\alpha^{-1} = \alpha^{d-1}$ which is in the set of roots because $\alpha^{d-1}$ is a product of roots of $f$). This means that the set of roots forms a group under multiplication. Since this group is a finite subgroup of $F^{\times}$, it is a cyclic group, hence generated by one element $\alpha$. Let $d$ denote the order of $\alpha$. Then the roots of $f(x)$ are $\alpha, \alpha^2, \dots, \alpha^{d-1}, \alpha^d = 1$.

Since the roots are distinct, we now have two possibilities : either $0$ is not a root of $f$ and we have $f(x) = x^d -1$, $d = n$, or $0$ is a root of $f$ and we have $f(x) = x^{d+1} - x$, $d = n$. (Thanks to anon for the comment!)

Since $F$ is algebraically closed, all the polynomials $x^n -1$ do split (with $n \neq \mathrm{char}(F)$ obviously), hence the polynomials you are looking for must be $x^n-1$ for $n \in \Bbb N$ and $\mathrm{char}(F) \nmid n$.

Hope that helps,

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Technically doesn't $f(x)=x(x^n-1)$ also work? @Galois: No, every field has an algebraic closure, including those of positive/finite characteristic. –  anon Jun 4 '12 at 2:13
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@Galois : There exists algebraically closed fields of characteristic $p$ for every prime $p$, namely the algebraic closures of $\mathbb F_p$. We need to say that $n \neq \mathrm{char}(F)$. –  Patrick Da Silva Jun 4 '12 at 2:15
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@Galois : The characteristic of every field is finite, by the way... $\mathbb Q$ does not have characteristic $\infty$, it has characteristic $0$. –  Patrick Da Silva Jun 4 '12 at 2:17
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The characteristic of $F$ should probably be thought of as a generator for the kernel of the unique homomorphism $\mathbf Z \to F$. –  Dylan Moreland Jun 4 '12 at 2:42
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@Galois You're confusing the characteristic and the number of elements in the field. What is true is that it is not possible for a finite field to be algebraically closed. –  fpqc Jun 4 '12 at 4:04
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Hint: If $\alpha$ is a root of $f(x)$, then $\alpha^2, \alpha^3, \dots$ are all roots of $f$. Conclude that $\alpha$ is a root of unity (why?).

Conversely, if $\alpha$ is a root of unity, show that such an $f$ exists with $f(\alpha) = 0$ (this should be straightforward).

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You also want to cover the possibility of $0$ being a root. –  anon Jun 4 '12 at 2:15
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