# Do $p^n$-th powers determine the field?

This is a question which came to my mind, when seeing A quick question on transcendence

Suppose $F$ is a field of characteristic $p$. Then the set of $p^n$-powers of the elements of $F$ is again a field which I denote by $F^n$ and is a subfield of $F$. Now let $d_n=[F:F^n]$.

My question is, what can be said about $F$ if we know the sequence $(d_n)_n$. Of course for finite fields $d_n$ is monotonically increasing and eventually constant. The constant will be the index degree of $F$ over the prime field $F_p$ and thus $F$ is determined up to isomorphism.

So what happens in general. Of course $d_n$ is monotonically increasing. In general $d_n$ will not be finite. This can be seen by taking $F=F_p(t_1,t_2,\ldots)$ where $t_i$ is transcendental over $F_p(t_1,t_2,\ldots,t_{i-1})$.

So now my questions:

• Can it happen that $d_1$ is finite but $d_n$ is infinite for some $n$?

• Under the assumption that $(d_n)_n$ is finite and the same for a field $F_1$ and $F_2$. Are these fields necessarily isomorphic?

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Wow, sorry about that; I misread parts of your question several times. It might help if you more clearly separated what you already know from your questions. –  Qiaochu Yuan Jul 22 '11 at 15:11
Ok, I will edit –  wood Jul 22 '11 at 15:14
If $F$ is perfect then $d_n=1$ for all $n$. So, the answer of the second question is no. –  user10676 Jul 22 '11 at 15:27
Consider the injection $F^1 \rightarrow F$. The map $Frob : x \mapsto x^p$ yields isomophisms $F \simeq F^1$ and $F^1 \simeq F^2$. The injection becomes $F^2 \rightarrow F^1$, which implies that $[F:F^1] = [F^1:F^2]$. So in fact $d_n = d_1^n$. –  user10676 Jul 22 '11 at 15:44
If you leave that as an answers, I'll accept it. Now I realize that the second question is stupid after all, since my argument above for finite fields is not correct, as your proof shows. So the entire question is a little boring, I hoped for some fancy arguments with curves over finite fields. But $d_n=d_1^n$ settles it. –  wood Jul 22 '11 at 16:02

First, your notation seems potentially misleading: it is standard to use $F^n$ to denote the set of $n$th powers of elements of $F$. So I will use $F^{p^n}$ instead.

Second, your assertions about finite fields are incorrect (perhaps you got confused by your notation, as above?). In fact for any perfect field we have $d_n = 1$ for all $n$.

Here is a key observation: if $E/F$ is a field extension and $\sigma: E \rightarrow E$ is a field homomorphism, then $[E:F] = [\sigma(E):\sigma(F)]$. Indeed, if $\{x_i\}$ is a basis for $E/F$, then $\{\sigma(x_i)\}$ is abasis for $\sigma(E)/\sigma(F)$. Applying this with $F/F^p$ and $\sigma = (x \mapsto x^p)$, we get $[F:F^p] = [F^p:F^{p^2}]$ and thus

$d_2 = [F:F^{p^2}] = [F:F^{p}] [F^{p}:F^{p^2}] = [F:F^p] [F:F^p] = d_1^2$.

Reasoning in a similar way gives $d_n = d_1^n$ for all $n \in \mathbb{Z}^+$. So you are not getting any new information from the entire sequence $\{d_n\}$ that you don't already know from $d_1$.

As one might suspect, there are lots of fields having the same value of $d_1$. For instance, as above, all perfect fields have $d_1 = 1$. Moreover, for any positive integer $d$, any two function fields of transcendence degree $d$ over any algebraically closed field of characteristic $p$ will have $d_1 = p^d$.

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Good to have you back, Pete! –  Alex B. Jul 22 '11 at 16:00