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Let $q=p^n$ where $p$ is a prime number. For which $q$ is the extensión $\mathbb{F}_{q^3}/\mathbb{F}_q$ (i.e the extension of degree $3$) an extension of the form $\mathbb{F}_q(\root 3 \of \alpha )$ where $\alpha \in \mathbb{F}_q$.

The only that I noticed is that char ($\mathbb{F}_q$)=3 ( $q=3^n$) , then since the frobenius automorphism is injective in a finite set, it's also surjective, so all the elements have a cubic root , therefore in characteristic 3, the extension of degree 3, are not "produced" by considering cubic roots. In the case of $char F \ne 3 $ I have no idea what to do.

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For what choice(s) of $q$ is $x^3 - \alpha$ an irreducible cubic polynomial in $\mathbb F_q[x]$? –  Dilip Sarwate Sep 29 '12 at 2:08
    
@DilipSarwate At least as I said, $q\ne 2^n$ i'm not sure if this condition it's also sufficient. I have no idea how can I prove this –  Daniel Sep 29 '12 at 2:10
    
I don't see where you said anything about $q \neq 2^n$ in your original question. Please clarify. –  Dilip Sarwate Sep 29 '12 at 2:15
    
@DilipSarwate Sorry, I mean $q\ne 3^n$ .... –  Daniel Sep 29 '12 at 2:17

1 Answer 1

up vote 4 down vote accepted

If $q\equiv2\pmod3$, then every element of the field of $q$ elements is a cube, so the extension is not an extension by a cube root.

On the other hand, if $q\equiv1\pmod3$, then the field of $q$ elements has non-cubes in it, so the extension can be achieved by a cube root.

You have already addressed the case $q\equiv0\pmod3$.

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+1 Very nice. One might add as explication that the injectiveness, and equivalently surjectiveness, of the group endomorphism $x\mapsto x^3$ of $\mathbb F_q^\times$ is a purely group theoretic question, answered by the condition that the order of the group be non-divisble by $3$, so equivalent to $q\not\equiv1\pmod3$ (the cases $q\equiv0$ and $q\equiv2$ are not much different!). –  Marc van Leeuwen Sep 29 '12 at 6:49
    
@Gerry Myerson Thanks for your answer! But I don't understand how can I prove this results (sorry for being so stupid <.<). At least I can't consider $x\to x^3$ as a homomorphism, since it's not true , if the characteristic is not 3. –  Daniel Sep 29 '12 at 16:09
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As Marc points out, it certainly is a homomorphism, not of the field but of the multiplicative group of nonzero elements of the field. When $q\equiv0,2\pmod3$, that group has order prime to 3, so the homomorphism has trivial kernel, so everything's a cube; when $q\equiv1\pmod3$, the group has order a multiple of 3, so the kernel is nontrivial, so there are noncubes. –  Gerry Myerson Sep 29 '12 at 23:41

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