Lemma 10.46.2. Let $k'/k$ be a field extension. The following are equivalent

1. for each $x \in k'$ there exists an $n > 0$ such that $x^ n \in k$, and

2. $k' = k$ or $k$ and $k'$ have characteristic $p > 0$ and either $k'/k$ is a purely inseparable extension or $k$ and $k'$ are algebraic extensions of $\mathbf{F}_ p$.

Proof. Observe that each of the possibilities listed in (2) satisfies (1). Thus we assume $k'/k$ satisfies (1) and we prove that we are in one of the cases of (2). Discarding the case $k = k'$ we may assume $k' \not= k$. It is clear that $k'/k$ is algebraic. Hence we may assume that $k'/k$ is a nontrivial finite extension. Let $k'/k'_{sep}/k$ be the separable subextension found in Fields, Lemma 9.14.6. We have to show that $k = k'_{sep}$ or that $k$ is an algebraic over $\mathbf{F}_ p$. Thus we may assume that $k'/k$ is a nontrivial finite separable extension and we have to show $k$ is algebraic over $\mathbf{F}_ p$.

Pick $x \in k'$, $x \not\in k$. Pick $n, m > 0$ such that $x^ n \in k$ and $(x + 1)^ m \in k$. Let $\overline{k}$ be an algebraic closure of $k$. We can choose embeddings $\sigma , \tau : k' \to \overline{k}$ with $\sigma (x) \not= \tau (x)$. This follows from the discussion in Fields, Section 9.12 (more precisely, after replacing $k'$ by the $k$-extension generated by $x$ it follows from Fields, Lemma 9.12.8). Then we see that $\sigma (x) = \zeta \tau (x)$ for some $n$th root of unity $\zeta$ in $\overline{k}$. Similarly, we see that $\sigma (x + 1) = \zeta ' \tau (x + 1)$ for some $m$th root of unity $\zeta ' \in \overline{k}$. Since $\sigma (x + 1) \not= \tau (x + 1)$ we see $\zeta ' \not= 1$. Then

$\zeta ' (\tau (x) + 1) = \zeta ' \tau (x + 1) = \sigma (x + 1) = \sigma (x) + 1 = \zeta \tau (x) + 1$

implies that

$\tau (x) (\zeta ' - \zeta ) = 1 - \zeta '$

hence $\zeta ' \not= \zeta$ and

$\tau (x) = (1 - \zeta ')/(\zeta ' - \zeta )$

Hence every element of $k'$ which is not in $k$ is algebraic over the prime subfield. Since $k'$ is generated over the prime subfield by the elements of $k'$ which are not in $k$, we conclude that $k'$ (and hence $k$) is algebraic over the prime subfield.

Finally, if the characteristic of $k$ is $0$, the above leads to a contradiction as follows (we encourage the reader to find their own proof). For every rational number $y$ we similarly get a root of unity $\zeta _ y$ such that $\sigma (x + y) = \zeta _ y\tau (x + y)$. Then we find

$\zeta \tau (x) + y = \zeta _ y(\tau (x) + y)$

and by our formula for $\tau (x)$ above we conclude $\zeta _ y \in \mathbf{Q}(\zeta , \zeta ')$. Since the number field $\mathbf{Q}(\zeta , \zeta ')$ contains only a finite number of roots of unity we find two distinct rational numbers $y, y'$ with $\zeta _ y = \zeta _{y'}$. Then we conclude that

$y - y' = \sigma (x + y) - \sigma (x + y') = \zeta _ y(\tau (x + y)) - \zeta _{y'}\tau (x + y') = \zeta _ y(y - y')$

which implies $\zeta _ y = 1$ a contradiction. $\square$

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