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Suppose that $L$ is a field of characteristic $p$, $E$ is a field extension of $L$, a is a pth root of an element of $L$ such that a is not in $E$. Consider the polynomial $p(x):=x^p-a^p.$

Question: Let $g(x)$ be a polynomial in $E[x]$, suppose that for some $n$, $p(x)$ divides $g(x)^n$, does it follow that $p(x)$ divides $g(x)?$

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$p(x)$ is irreducible over $E$, right? Doesn't that do it? – Dylan Moreland Jul 12 '11 at 2:24

The answer is yes.

You know that $E$ is a field, which implies that $E[x]$ is a Unique Factorization Domain. Therefore $p(x)$ dividing $g(x)^n$ implies $p(x)$ divides $g(x)$ if and only if $p(x)$ is squarefree in $E[x]$ (otherwise let $g(x)$ be the product of the prime divisors of $p(x)$ and let $n$ be the largest of the powers with which any of those primes appear in $p(x)$).

Now, if $a^p=b$, then what you want is that $b\in L$, but $a\not\in E$, which is equivalent to saying that $p(x)=x^p-b$ does not factor in $E[x]$ (or in $L[x]$). Hence, $p(x)$ is in fact irreducible in $E[x]$, and irreducible elements in a UFD are certainly square-free (and also of course prime).

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Yes, this is true since $p(x)$ is irreducible over $E$. This was already mentioned by others, but I'd like to include the proof of this fact. So let $K$ be a field, $p$ a prime number, $a \in K$ and suppose that $a$ is not a $p^{th}$ power in $K$. Then $f(x) = x^p-a$ is irreducible over K. We argue by contradiction. Suppose not, so that $f(x)$ is reducible over $K$. Since $f(x)$ is reducible, let $g(x)$ be a polynomial of degree $d, d<p$, that divides $f(x)$. In a splitting field we may write $f(x) = \prod_{i = 1}^{i= k} (x - b_i)$ where $b_i$ are roots of $f$. We may reindex the $b_i$'s if necessary so that $g(x) = \prod_{i = 1}^{i=d} (x-b_i).$ (in other words, we may assume that $b_1,..,b_d$ are roots of $g$). Define $z = b_1b_2..b_d$. Note that $z$ is the constant coefficient (or the negative of the constant coefficient of $g$), hence $z \in K$. then $z^p = b_1^p...b_d^p = a^d$ since each $b_i$ is a root of $f$ (hence $b_i^p = a$ for all $i$). Since $d<p$, $p$ and $d$ are coprime, so we may write $1 = up + vd$ for some integers $u,v$. But now $a = a^1 = a^{up+vd} = (a^u)^p (a^d)^v = (a^u)^p (z^p)^v = (a^u z^v)^p$, so that $a^uz^v$ is a root of $f(x)$ that is in $K$. But this means that $a$ is a $p^{th}$ power in $K$, contradicting our hypothesis. hence $f(x)$ is irreducible over $K$. An additional observation is that this proof works in any characteristic, so it does not matter if $K$ has characteristic $0$, or $p$, or $q$ for some prime $q \neq p$.

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Your contradiction is unnecessary: your argument is that $x^p-a$ in $K[x]$ reducible implies $a$ is a $p^\text{th}$ power in $K$, which is the contrapositive. Otherwise this is very nice. – Vladimir Sotirov Jul 12 '11 at 7:07

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