The Stacks project

Proof. The assumptions (b) and (c) imply that the map on spectra is open, see Proposition 10.41.8. Hence the lemma follows from Topology, Lemma 5.8.14. $\square$

Proof. Let $k \subset \overline{k}$ be a perfect closure, see Definition 10.45.5. By assumption $\overline{k}$ is algebraically closed. The ring maps $R \to R \otimes _ k \overline{k}$ and $S \to S \otimes _ k \overline{k}$ and $R \otimes _ k S \to (R \otimes _ k S) \otimes _ k \overline{k} = (R \otimes _ k \overline{k}) \otimes _{\overline{k}} (S \otimes _ k \overline{k})$ satisfy the assumptions of Lemma 10.46.7. Hence we may assume $k$ is algebraically closed.

We may replace $R$ and $S$ by their reductions. Hence we may assume that $R$ and $S$ are domains. By Lemma 10.45.6 we see that $R \otimes _ k S$ is reduced. Hence its spectrum is reducible if and only if it contains a nonzero zerodivisor. By Lemma 10.43.4 we reduce to the case where $R$ and $S$ are domains of finite type over $k$ algebraically closed.

Note that the ring map $R \to R \otimes _ k S$ is of finite presentation and flat. Moreover, for every maximal ideal $\mathfrak m$ of $R$ we have $(R \otimes _ k S) \otimes _ R R/\mathfrak m \cong S$ because $k \cong R/\mathfrak m$ by the Hilbert Nullstellensatz Theorem 10.34.1. Moreover, the set of maximal ideals is dense in the spectrum of $R$ since $\mathop{\mathrm{Spec}}(R)$ is Jacobson, see Lemma 10.35.2. Hence we see that Lemma 10.47.1 applies to the ring map $R \to R \otimes _ k S$ and we conclude that the spectrum of $R \otimes _ k S$ is irreducible as desired. $\square$

Proof. It is clear that (1) implies (2).

Assume (2) and let $\overline{k}$ is the separable algebraic closure of $k$. Suppose $\mathfrak q_ i \subset R \otimes _ k \overline{k}$, $i = 1, 2$ are two minimal prime ideals. For every finite subextension $\overline{k}/k'/k$ the extension $k'/k$ is separable and the ring map $R \otimes _ k k' \to R \otimes _ k \overline{k}$ is flat. Hence $\mathfrak p_ i = (R \otimes _ k k') \cap \mathfrak q_ i$ are minimal prime ideals (as we have going down for flat ring maps by Lemma 10.39.19). Thus we see that $\mathfrak p_1 = \mathfrak p_2$ by assumption (2). Since $\overline{k} = \bigcup k'$ we conclude $\mathfrak q_1 = \mathfrak q_2$. Hence $\mathop{\mathrm{Spec}}(R \otimes _ k \overline{k})$ is irreducible.

Assume (3) and let $\overline{k}$ be the algebraic closure of $k$. Let $\overline{k}/\overline{k}'/k$ be the corresponding separable algebraic closure of $k$. Then $\overline{k}/\overline{k}'$ is purely inseparable (in positive characteristic) or trivial. Hence $R \otimes _ k \overline{k}' \to R \otimes _ k \overline{k}$ induces a homeomorphism on spectra, for example by Lemma 10.46.7. Thus we have (4).

Assume (4). Let $k'/k$ be an arbitrary field extension and let $\overline{k}$ be the algebraic closure of $k$. We may choose a field $F$ such that both $k'$ and $\overline{k}$ are isomorphic to subfields of $F$. Then

and hence we see from Lemma 10.47.2 that $R \otimes _ k F$ has a unique minimal prime. Finally, the ring map $R \otimes _ k k' \to R \otimes _ k F$ is flat and injective and hence any minimal prime of $R \otimes _ k k'$ is the image of a minimal prime of $R \otimes _ k F$ (by Lemma 10.30.5 and going down). We conclude that there is only one such minimal prime and the proof is complete. $\square$

Proof. Immediate from the remark following Definition 10.47.4. $\square$

Proof. Let $S' \subset S$ be a subalgebra. Then for any extension $k'/k$ the ring map $S' \otimes _ k k' \to S \otimes _ k k'$ is injective also. Hence (1) follows from Lemma 10.30.5 (and the fact that the image of an irreducible space under a continuous map is irreducible). The second and third property follow from the fact that tensor product commutes with colimits. $\square$

Proof. Recall that irreducible components correspond to minimal primes (Lemma 10.26.1). As $R \to R \otimes _ k S$ is flat we see by going down (Lemma 10.39.19) that any minimal prime of $R \otimes _ k S$ lies over a minimal prime of $R$. Conversely, if $\mathfrak p \subset R$ is a (minimal) prime then

by flatness of $R \to R \otimes _ k S$. The ring $\kappa (\mathfrak p) \otimes _ k S$ has irreducible spectrum by assumption. It follows that $R \otimes _ k S/\mathfrak p(R \otimes _ k S)$ has a single minimal prime (Lemma 10.30.5). In other words, the inverse image of the irreducible set $V(\mathfrak p)$ is irreducible. Hence the lemma follows. $\square$

Proof. Assume $k$ is algebraically closed in $K$. By Definition 10.47.4 and Lemma 10.47.3 it suffices to show that the spectrum of $K \otimes _ k k'$ is irreducible for every finite separable extension $k'/k$. Say $k'$ is generated by $\alpha \in k'$ over $k$, see Fields, Lemma 9.19.1. Let $P = T^ d + a_1 T^{d - 1} + \ldots + a_ d \in k[T]$ be the minimal polynomial of $\alpha $. Then $K \otimes _ k k' \cong K[T]/(P)$. The only way the spectrum of $K[T]/(P)$ can be reducible is if $P$ is reducible in $K[T]$. Assume $P = P_1 P_2$ is a nontrivial factorization in $K[T]$ to get a contradiction. By Lemma 10.38.5 we see that the coefficients of $P_1$ and $P_2$ are algebraic over $k$. Our assumption implies the coefficients of $P_1$ and $P_2$ are in $k$ which contradicts the fact that $P$ is irreducible over $k$. $\square$

Proof. By Definition 10.47.4 and Lemma 10.47.3 it suffices to show that the spectrum of $S \otimes _ k k'$ is irreducible for every finite separable extension $k'/k$. Since $K$ is geometrically irreducible over $k$ we see that $K' = K \otimes _ k k'$ is a finite, separable field extension of $K$. Hence the spectrum of $S \otimes _ k k' = S \otimes _ K K'$ is irreducible as $S$ is assumed geometrically irreducible over $K$. $\square$

Proof. Assume (1). Denote $\Omega $ an algebraic closure of $k(t)$. By Definition 10.47.4 we find that the spectrum of

is irreducible. Since $K(t)$ is a localization of $K \otimes _ k k(T)$ we conclude that the spectrum of $K(t) \otimes _{k(t)} \Omega $ is irreducible. Thus by Lemma 10.47.3 we find that $K(t)/k(t)$ is geometrically irreducible.

Assume (2). Let $k'/k$ be a field extension. We have to show that $K \otimes _ k k'$ has a unique minimal prime. We know that the spectrum of

is irreducible, i.e., has a unique minimal prime. Since there is an injective map $K \otimes _ k k' \to K(t) \otimes _{k(t)} k'(t)$ (details omitted) we conclude by Lemmas 10.30.5 and 10.30.7. $\square$

Proof. This follows from Lemma 10.47.10 because the fields $L(x)$ and $M(x)$ are purely transcendental extensions of $L$ and $M$. $\square$

Proof. Assume (1) and let $\alpha \in K$ be separably algebraic over $k$. Then $k' = k(\alpha )$ is a finite separable extension of $k$ contained in $K$. By Lemma 10.47.6 the extension $k'/k$ is geometrically irreducible. In particular, we see that the spectrum of $k' \otimes _ k \overline{k}$ is irreducible (and hence if it is a product of fields, then there is exactly one factor). By Fields, Lemma 9.13.4 it follows that $\mathop{\mathrm{Hom}}\nolimits _ k(k', \overline{k})$ has one element which in turn implies that $k' = k$ by Fields, Lemma 9.12.11. Thus (2) holds.

Assume (2). Let $k' \subset K$ be the subfield consisting of elements algebraic over $k$. By Lemma 10.47.8 the extension $K/k'$ is geometrically irreducible. By assumption $k'/k$ is a purely inseparable extension. By Lemma 10.46.7 the extension $k'/k$ is geometrically irreducible. Hence by Lemma 10.47.9 we see that $K/k$ is geometrically irreducible. $\square$

Proof. The first statement is immediate from Lemma 10.47.12 and the fact that elements separably algebraic over $k'$ are in $k'$ by the transitivity of separable algebraic extensions, see Fields, Lemma 9.12.12. If $K/k$ is finitely generated, then $k'$ is finite over $k$ by Fields, Lemma 9.26.11. $\square$

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

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. $\square$