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Lemma 10.47.14. Let $K/k$ be an extension of fields. Let $\overline{k}/k$ be a separable algebraic closure. Then $\text{Gal}(\overline{k}/k)$ acts transitively on the primes of $\overline{k} \otimes _ k K$.

Proof. Let $K/k'/k$ be the subextension found in Lemma 10.47.13. Note that as $k \subset \overline{k}$ is integral all the prime ideals of $\overline{k} \otimes _ k K$ and $\overline{k} \otimes _ k k'$ are maximal, see Lemma 10.36.20. By Lemma 10.47.7 the map

\[ \mathop{\mathrm{Spec}}(\overline{k} \otimes _ k K) \to \mathop{\mathrm{Spec}}(\overline{k} \otimes _ k k') \]

is bijective because (1) all primes are minimal primes, (2) $\overline{k} \otimes _ k K = (\overline{k} \otimes _ k k') \otimes _{k'} K$, and (3) $K$ is geometrically irreducible over $k'$. Hence it suffices to prove the lemma for the action of $\text{Gal}(\overline{k}/k)$ on the primes of $\overline{k} \otimes _ k k'$.

As every prime of $\overline{k} \otimes _ k k'$ is maximal, the residue fields are isomorphic to $\overline{k}$. Hence the prime ideals of $\overline{k} \otimes _ k k'$ correspond one to one to elements of $\mathop{\mathrm{Hom}}\nolimits _ k(k', \overline{k})$ with $\sigma \in \mathop{\mathrm{Hom}}\nolimits _ k(k', \overline{k})$ corresponding to the kernel $\mathfrak p_\sigma $ of $1 \otimes \sigma : \overline{k} \otimes _ k k' \to \overline{k}$. In particular $\text{Gal}(\overline{k}/k)$ acts transitively on this set as desired. $\square$

Comments (4)

Comment #749 by Keenan Kidwell on

At the end of the proof, would it be simpler to just invoke 037O, which says that the map is a bijection (since all primes are minimal in this situation)?

Comment #8278 by Et on

Why does every prime in being maximal imply that every residue field is isomorphic to ?

Comment #8912 by on

Because then the residue field is a quotient of the ring, whence integral over (because and are integral over ), whence algebraic extension of containing .

There are also:

  • 6 comment(s) on Section 10.47: Geometrically irreducible algebras

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