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I need to prove that any irreducible polynomial $f$ of degree $d\,\big|\,n$ over $\mathbb F_p$ devides $x^{p^n} - x$. I know that the splitting field of the latter is the finite field with $p^n$ elements, and that if $\alpha$ is the root of $f$, then $[\mathbb F_p(\alpha):\mathbb F_p] = d$. I don't see, why all the roots of $f$ must lie in the splitting field of $x^{p^n} - x$. In fact, the splitting field of $f$ may have degree $d\,!$ over $\mathbb F_p$, which may well be greater than $n$.

Thank you.

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Thank you! Just realized this myself. I'm very embarrassed. I'll delete this question from the post. –  Artem Oct 23 '12 at 12:52

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Q: Why does $\mathbb{F}_{p^n}$ contain all the roots of an irreducible polynomial $f(x)\in\mathbb{F}_{p}[x]$ of degree $d$, $d\mid n$?


One way of seeing this is to observe that the extension $\mathbb{F}_{p^n}/\mathbb{F}_p$ is a Galois extension because it has $n$ distinct automorphisms (powers of the Frobenius automorphism). As a Galois extension it is normal, meaning exactly that any irreducible polynomial with coefficients in the prime field and a root in the bigger field splits there into a product of linear factors.


A possibly more concrete way of seeing the same thing is to observe that if $\alpha$ is a root of $f(x)$, then so are $\alpha^p$, $\alpha^{p^2}$ et cetera. The conjugates $\alpha^{p^i}$ must start repeating at some point, so we get, first $\alpha^{p^i}=\alpha^{p^j}$ for some integers $i,j$ such that $0<i<j$, and then (using injectivity of the $p$th power mapping) that $\alpha=\alpha^{p^{\ell}}$ with $\ell=j-i>0$. Without loss of generality we can assume that $\ell$ is the smallest positive integer with the property $\alpha^{p^\ell}=\alpha$.

At this point we know $\ell$ roots of $f(x)$, namely $\alpha^{p^i},0 \le i<\ell.$ Consider the polynomial $$ g(x)=\prod_{i=0}^{\ell-1}(x-\alpha^{p^i})=x^\ell+\sum_{i=0}^{\ell-1}a_ix^i $$ for some coefficients $a_i\in\mathbb{F}_p[\alpha]$. Because $F:z\mapsto z^p$ is an automorphism of $\mathbb{F}_p[\alpha]$, we see that $$ g^F(x):=x^\ell+\sum_{i=0}^{\ell-1}F(a_i)x^i=\prod_{i=0}^{\ell-1}(x-F(\alpha^{p^i})) =\prod_{i=0}^{\ell-1}(x-(\alpha^{p^i})^p)=\prod_{i=1}^{\ell}(x-\alpha^{p^i})=g(x). $$ Here in the last step we used the fact $\alpha^{p^\ell}=\alpha$. Therefore $F(a_i)=a_i$ for all $i$, in other words $g(x)\in\mathbb{F}_p[x]$. Because $g(x)$ is a non-trivial factor of $f(x)$ with coefficients in the prime field, we must have $g(x)=f(x)$, and hence also that $\ell=d$. This means that all the roots $\alpha_i$ of $f(x)$ are of the form $\alpha^{p^i}$ for some $i$. In particular they all satisfy the equation $\alpha_i^{p^d}=\alpha_i$. Consequently we also have $\alpha_i^{p^n}=\alpha_i$ for all $i$, which is what we wanted to show.

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Thank you so much! Since this is leading up to Galois theory for us, we have to employ some other means (and I don't know any Galois theory at the moment). The hands-on answer is a great help! –  Artem Oct 23 '12 at 12:56

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