
Lemma 10.46.2. Let $k$ be a separably closed field. Let $R$, $S$ be $k$-algebras. If $R$, $S$ have a unique minimal prime, so does $R \otimes _ k S$.

Proof. Let $k \subset \overline{k}$ be a perfect closure, see Definition 10.44.5. 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 10.45.7. 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 10.44.6 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 10.42.4 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 10.33.1. Moreover, the set of maximal ideals is dense in the spectrum of $R$ since $\mathop{\mathrm{Spec}}(R)$ is Jacobson, see Lemma 10.34.2. Hence we see that Lemma 10.46.1 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. $\square$

Comment #3395 by Matthieu Romagny on

Remove "algebraically" in the statement of the Lemma.

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