Lemma 9.24.3 (0EXP)—The Stacks project

[Theorem 5.2, Radical]

Lemma 9.24.3. Let $K$ be a field. Let $L/K$ be a finite extension of degree $e$ which is generated by an element $\alpha $ with $a = \alpha ^ e \in K$. If every $e$th root of unity in $L$ is contained in $K$, then any sub extension $L/L'/K$ is generated by $\alpha ^ d$ for some $d | e$.

Proof. Observe that for $d | e$ the subfield $K(\alpha ^ d)$ has $[K(\alpha ^ d) : K] = e/d$ and $[L : K(\alpha ^ d)] = d$. Let $L/L'/K$ be a subextension. Say $d = [L : L']$. If $\alpha ^ d \in L'$, then we have $L' = K(\alpha ^ d)$ for degree reasons. Let $P \in L'[x]$ be the minimal polynomial of $\alpha $ over $L'$. Then $P$ divides $x^ e - a$ and $P$ has degree $d$. Let us write

\[ x^ e - a = \prod \nolimits _{i = 1, \ldots , e} (x - \zeta _ i \alpha ) \]

in a splitting field of $x^ e - a$ over $L$. The $\zeta _ i$ are $e$th roots of unity and after renumbering we have

\[ P = \prod \nolimits _{i = 1, \ldots , d} (x - \zeta _ i \alpha ) \]

The constant term of $P$ is equal to

\[ c = (\prod \nolimits _{i = 1, \ldots , d} \zeta _ i) \alpha ^ d \]

and is in $L' \subset L$. Since $\alpha \in L$ this implies that $\zeta = \prod _{i = 1, \ldots , d} \zeta _ i$ is in $L$ and hence in $K$ by our assumption. Thus $\alpha ^ d = \zeta ^{-1}c \in L'$ and we conclude. $\square$

Comments (7)

Comment #5407 by Yuzhou Gu on July 22, 2020 at 13:16

The first line of the proof should have $[K(\alpha^d):K]=e/d$ and $[L:K(\alpha^d)]=d$ .

Comment #5637 by Johan on November 14, 2020 at 13:54

Thanks and fixed here.

Comment #5961 by Yuzhou Gu on March 08, 2021 at 13:51

There is still a typo. The first line should be $e/d$ , not $d/e$ .

Comment #6142 by Johan on April 30, 2021 at 22:22

THanks and fixed here.

Comment #11049 by thesnakefromthelemma on December 19, 2025 at 12:20

Re Comment #11043, the fundamental error in the given proof seems to be the line "and on the other hand $\zeta'$ has to be a $p$ th root of $1$ as $σ$ has order $p$ "; perhaps we implicitly wrongly assumed that $\sigma$ necessarily fixes $\zeta'$ ?

Note that this lemma, i.e., Lemma 0EXP (fields-lemma-subfields-kummer) is afaict explicitly invoked once, namely in the proof of Lemma 0EYG (pione-lemma-abhyankar-one-divisor). I am not sure what the maximally strong correct statement here should be or to what extent the proof there should be rewritten, but presumably both require attention.

Comments (7)

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