Let $K/F $ be a separable extension with char $F =p > 0$. Prove that for any given $\alpha \in K , \alpha \in F (K^p) $.

I am not sure how to approach this problem. I think Viete's formula is too comolicated, if not useless; this problem is from field theory anyway. If $K $ is finite then $K=K^p $ and we're done.

EDIT: Here is my solution: let $\alpha \in K $. Then $x^p-\alpha^p = (x-\alpha)^p$ by the binomial formula and the fact that $p $ choose $k $ is divisible by $p$ for $k = 1,2,...,p-1$. Hence the minimal polynomial of $\alpha$, denoted $f \in F [x]$, divides $(x-\alpha)^p$ , and since the extension is separable $f $ must be $x-\alpha $. As $f \in F [x] $ it follows that $\alpha \in F \subset F (K^p) $.

Here is now my updated question: The same proof shows that $K \subset F $, and so $K=F $ if $K/F $ is a separable extension and also $F=K^p $. But I don't think this is true. Is something wrong in my proof?


You were on the right track . . .

Assume $K{/}F $ is a separable extension with $\text{char}(F)=p > 0$, and let $H=F(K^p)$.

Let $\alpha \in K$.

We want to show $\alpha \in H$.

Let $f(x)$ be the monic irreducible polynomial for $\alpha$ over $H$.

Since $K{/}F$ is separable, so is $K{/}H$.

Let $g(x)=x^p-\alpha^p$.

As you noted, we have $x^p-\alpha^p=(x-\alpha)^p$, so $\alpha$ is a root of $g$.

Since $g\in H[x]$, it follows that $f|g$, hence, since $K{/}H$ is separable, we get $f(x)=x-\alpha$.

Hence, since $f\in H[x]$, and $f(x)=x-\alpha$, we get $\alpha\in H$, as was to be shown.

As regards the error in your proof attempt . . .

From the fact that $\alpha$ is a root of $x^p-\alpha^p=(x-\alpha)^p$, you can't claim that the minimal polynomial for $\alpha$ over $F$ divides $x^p-\alpha^p$, since, although $x^p-\alpha^p\in H[x]$, you don't necessarily have $x^p-\alpha^p\in F[x]$.


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