
Lemma 10.47.1. Let $k$ be a separably algebraically closed field. Let $R$, $S$ be $k$-algebras. If $\mathop{\mathrm{Spec}}(R)$, and $\mathop{\mathrm{Spec}}(S)$ are connected, then so is $\mathop{\mathrm{Spec}}(R \otimes _ k S)$.

Proof. Recall that $\mathop{\mathrm{Spec}}(R)$ is connected if and only if $R$ has no nontrivial idempotents, see Lemma 10.20.4. Hence, by Lemma 10.42.4 we may assume $R$ and $S$ are of finite type over $k$. In this case $R$ and $S$ are Noetherian, and have finitely many minimal primes, see Lemma 10.30.6. Thus we may argue by induction on $n + m$ where $n$, resp. $m$ is the number of irreducible components of $\mathop{\mathrm{Spec}}(R)$, resp. $\mathop{\mathrm{Spec}}(S)$. Of course the case where either $n$ or $m$ is zero is trivial. If $n = m = 1$, i.e., $\mathop{\mathrm{Spec}}(R)$ and $\mathop{\mathrm{Spec}}(S)$ both have one irreducible component, then the result holds by Lemma 10.46.2. Suppose that $n > 1$. Let $\mathfrak p \subset R$ be a minimal prime corresponding to the irreducible closed subset $T \subset \mathop{\mathrm{Spec}}(R)$. Let $I \subset R$ be such that $T' = V(I) \subset \mathop{\mathrm{Spec}}(R)$ is the closure of the complement of $T$. Note that this means that $T' = \mathop{\mathrm{Spec}}(R/I)$ (Lemma 10.16.7) has $n - 1$ irreducible components. Then $T \cup T' = \mathop{\mathrm{Spec}}(R)$, and $T \cap T' = V(\mathfrak p + I) = \mathop{\mathrm{Spec}}(R/(\mathfrak p + I))$ is not empty as $\mathop{\mathrm{Spec}}(R)$ is assumed connected. The inverse image of $T$ in $\mathop{\mathrm{Spec}}(R \otimes _ k S)$ is $\mathop{\mathrm{Spec}}(R/\mathfrak p \otimes _ k S)$, and the inverse of $T'$ in $\mathop{\mathrm{Spec}}(R \otimes _ k S)$ is $\mathop{\mathrm{Spec}}(R/I \otimes _ k S)$. By induction these are both connected. The inverse image of $T \cap T'$ is $\mathop{\mathrm{Spec}}(R/(\mathfrak p + I) \otimes _ k S)$ which is nonempty. Hence $\mathop{\mathrm{Spec}}(R \otimes _ k S)$ is connected. $\square$

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