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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 $a = \alpha ^ e \in K$. 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 $[L : K(\alpha ^ d)] = d$ and that both extensions $K(\alpha ^ d)/K$ and $L/K(\alpha ^ d)$ are extensions as in the lemma.

We will use induction on the pair of integers $([L : L'], [L' : K])$ ordered lexicographically. Let $p$ be a prime number dividing $e$ and set $d = e/p$. If $K(\alpha ^ d)$ is contained in $L'$, then we win by induction, because then it suffices to prove the lemma for $L/L'/K(\alpha ^ d)$. If not, then $[L'(\alpha ^ d) : L'] = p$ and by induction hypothesis we have $L'(\alpha ^ d) = K(\alpha ^ i)$ for some $i | d$. If $i \not= 1$ we are done by induction. Thus we may assume that $[L : L'] = p$.

If $e$ is not a power of $p$, then we can do this trick again with a second prime number and we win. Thus we may assume $e$ is a power of $p$.

If the characteristic of $K$ is $p$ and $e$ is a $p$th power, then $L/K$ is purely inseparable. Hence $L/L'$ is purely inseparable of degree $p$ and hence $\alpha ^ p \in L'$. Thus $L' = K(\alpha ^ p)$ and this case is done.

The final case is where $e$ is a power of $p$, the characteristic of $K$ is not $p$, $L/L'$ is a degree $p$ extension, and $L = L'(\alpha ^{e/p})$. Claim: this can only happen if $e = p$ and $L' = K$. The claim finishes the proof.

First, we prove the claim when $K$ contains a primitive $p$th root of unity $\zeta $. In this case the degree $p$ extension $K(\alpha ^{e/p})/K$ is Galois with Galois group generated by the automorphism $\alpha ^{e/p} \mapsto \zeta \alpha ^{e/p}$. On the other hand, since $L$ is generated by $\alpha ^{e/p}$ and $L'$ we see that the map

\[ K(\alpha ^{e/p}) \otimes _ K L' \longrightarrow L \]

is an isomorphism of $K$-algebras (look at dimensions). Thus $L$ has an automorphism $\sigma $ of order $p$ over $K$ sending $\alpha ^{e/p}$ to $\zeta \alpha ^{e/p}$. Then $\sigma (\alpha ) = \zeta ' \alpha $ for some $e$th root of unity $\zeta '$ (as $\alpha ^ e$ is in $K$). Then on the one hand $(\zeta ')^{e/p} = \zeta $ and on the other hand $\zeta '$ has to be a $p$th root of $1$ as $\sigma $ has order $p$. Thus $e/p = 1$ and the claim has been shown.

Finally, suppose that $K$ does not contain a primitive $p$th root of $1$. Choose a primitive $p$th root $\zeta $ in some algebraic closure $\overline{L}$ of $L$. Consider the diagram

\[ \xymatrix{ K(\zeta ) \ar[r] & L(\zeta ) \\ K \ar[u] \ar[r] & L \ar[u] } \]

By Lemma 9.24.2 the vertical extensions have degree prime to $p$. Hence $[L(\zeta ) : K(\zeta )]$ is divisible by $e$. On the other hand, $L(\zeta )$ is generated by $\alpha $ over $K(\zeta )$ and hence $[L(\zeta ) : K(\zeta )] \leq e$. Thus $[L(\zeta ) : K(\zeta )] = e$. Similarly we have $[K(\alpha ^{e/p}, \zeta ) : K(\zeta )] = p$ and $[L(\zeta ) : L'(\zeta )] = p$. Thus the fields $K(\zeta ), L'(\zeta ), L(\zeta )$ and the element $\alpha $ fall into the case discussed in the previous paragraph we conclude $e = p$ as desired. $\square$

Comments (4)

Comment #5407 by Yuzhou Gu on

The first line of the proof should have and .

Comment #5961 by Yuzhou Gu on

There is still a typo. The first line should be , not .

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  • 2 comment(s) on Section 9.24: Kummer extensions

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