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.

Let $F=\mathbb{F}_p(X,Y)$ be the field of rational functions in variables $X,Y$ over the finite field of $p$ elements. Let $K=\mathbb{F}_p(X^p,Y^p)$ be a subfield. Note that for any $f\in F$, $f^p\in K$. Deduce from this that $F/K$ is not a simple extension.

share|improve this question
2  
What have you tried? –  JSchlather Jan 12 '13 at 23:49
    
I tried contradiction methods. So assuming $F=K(f)$, $f^p=g\in K$ then $f$ is not separable over $K$. That's all I could think of, and I don't think this is taking me anywhere. –  Montez Jan 12 '13 at 23:51
2  
Maybe if you can prove that the degree of the extension is $p^2$ all is done. Try it! –  user26857 Jan 13 '13 at 0:01
    
I couldn't show the set $\{X,Y,XY,X^2,Y,XY^2,...X^pY^p\}$ is a linearly independent set. I took that this was what you meant by $p^2$. But this is an exercise in a section on separability. –  Montez Jan 13 '13 at 0:21
    
Presumably the reason it's in a section on separability is that this section (or maybe the next section) contains the theorem of the primitive element, and the point of the exercise is to show that this theorem really needs the hypothesis of separability. –  Andreas Blass Jan 13 '13 at 2:11
add comment

1 Answer

up vote 4 down vote accepted

First of all $\mathbb{F}_p(X^p,Y^p)$ is a function field, which means that we may treat $X^p,Y^p$ as variables. The polynomial $(T-X)^p = T^p-X^p$ is irreducible over $\mathbb{F}_p(Y^p)[X^p]$ by Eisenstein's criterion, but then also over the field of fractions $\mathbb{F}_p(X^p,Y^p)$. Hence, $\mathbb{F}_p(X,Y^p)$ has degree $p$ over $\mathbb{F}_p(X^p,Y^p)$. Similarily one proves that $\mathbb{F}_p(X,Y)$ has degree $p$ over $\mathbb{F}_p(X,Y^p)$. Hence, $F=\mathbb{F}_p(X,Y)$ has degree $p^2$ over $K=\mathbb{F}_p(X^p,Y^p)$. But since $F^p \subseteq K$, every element of $F$ has degree $\leq p$ over $K$, so that it cannot generate $F/K$.

share|improve this answer
add comment

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.