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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?2011-07-12

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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

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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.2011-07-12