Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $g(x)?$

share|improve this question
1  
$p(x)$ is irreducible over $E$, right? Doesn't that do it? –  Dylan Moreland Jul 12 '11 at 2:24

2 Answers 2

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).

share|improve this answer

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 $q \neq p$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.