Prove that $Char(F) \neq 0$ given that for all $x \in E$ there exists $n$ such that $x^n \in F$

Question

The following is a problem in our exam which I couldn't get really far with.

Let $F$ be a field and $E$ a non-trivial extension of $F$ such that for all $x \in E$ there exists $n_x \in \mathbb{N}$ such that $x^{n_x} \in F$, prove that $Char(F) \neq 0$

My progress:

Let $x \in E/F$ and $x^p \in F$, there exists $n \in \mathbb{N}$ such that $(x+1)^n,(x+2)^n,\dots,(x+p)^n \in F$, this means the $p$ fold iteration of the discrete derivative on $f(x)=x^n$ gives $0$.

This was as far as I went, Hints and solutions are appreciated.

Answer 1

I’m pretty sure that this argument isn’t optimal, but it gives a complete solution. I think you can also extend it to show more generally that in general, if $F$ is infinite, then $E/F$ is purely inseparable.

Assume that $F$ has characteristic zero and $E \neq F$. Note that $E/F$ is algebraic.

Let $x \in E \backslash F$. Then, there is a $F$-embedding $\sigma: E \rightarrow \Omega$ (where $\Omega$ is algebraically closed) such that $\sigma(x) \neq x$.

By the hypothesis, we see that $\sigma(x)=\omega x$ for some root of unity $\omega \neq 1$.

For the same reason, we know that $\omega x+1=\sigma(x+1)=\omega’ (x+1)$ for some root of unity $\omega’ \neq 1$. It follows that $\omega \neq \omega’$ and thus $x=\frac{\omega’-1}{\omega-\omega’}$.

In particular, since $E \neq F$, $E$ is contained in the field generated by the roots of unity over $\mathbb{Q}$. Hence $E/F$ is automatically abelian.

Next, pick some $y \in E \backslash F$ (which is algebraic over $\mathbb{Q}$) and let $E’=\mathbb{Q}[y]$, $F’=\mathbb{Q}[y] \cap F$. It’s easy to see that the extension $E’/F’$ still works (and thus is still abelian).

The interest of this construction is that $E’$ has finitely many roots of unity.

Since $y \notin F’$, there exists $\sigma \in \mathrm{Gal}(E’/F’)$ such that $\sigma(y) \neq y$.

Then, for every integer $n$, $u_n=\frac{\sigma(y+n)}{y+n}$ is a root of unity contained in $E$ and distinct from $1$.

Choose two distinct integers $n,m$ such that $u_n=u_m$. Then $(\sigma(y)+n)(y+m)=(\sigma(y)+m)(y+n)$, hence $ny+m\sigma(y)=my+n\sigma(y)$, which implies $y=\sigma(y)$, a contradiction.

Answer 2

HINT.- We know that if the characteristic is distinct of $0$ then it is equal to a prime $p$. We assume that $E|F$ is algebraic (cannot be trascendental).

Let $x\in E\setminus F$ so if $d$ is the degree of $x$ over $F$ then the linearly independant set $S=\{1,x,x^2,\cdots,x^{d-1}\}$ generates $F(x)$ and $F\subset F(x)\subset E$. Since $x^d=-(a_{d-1}x^{d-1}+a_{d-2}x^{d-2}+\cdots +a_1x+a_0)$ (it becomes from the minimal polynomial of $x$ over $F$), the powers $x^n$ are all expresed as $$x^n=g_{d-1}(a_0,a_1,\cdots,a_{d-1})x^{d-1}+g_{d-2}(a_0,a_1,\cdots,a_{d-1})x^{d-2}+\cdots+g_0(a_0,a_1,\cdots,a_{d-1})$$ where the $g_i(a_0,a_1,\cdots,a_{d-1})$ are functions of the above finite set of coefficients $a_m$ of $f(x)$.

It follows that the set of powers $x^n$ cannot be infinite (it is infinite only in characteristic zero when $x$ is not a root of unity) From $x^m=x^n$ with $m\gt n$ we have $x^n(x^{m-n}-1)=0$ and because of we are in a field we must have $x^{m-n}-1=0\iff x^{m-n}=1$-

We remark that the property can be strengthened to $x^{n_x}=1$