Lemma 10.48.1. 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 10.21.4. Hence, by Lemma 10.43.4 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 10.31.6. 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 10.47.2. 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 $T' \subset \mathop{\mathrm{Spec}}(R)$ be the union of the other $n - 1$ irreducible components. Choose an ideal $I \subset R$ such that $T' = V(I) = \mathop{\mathrm{Spec}}(R/I)$ (Lemma 10.17.7). By choosing our minimal prime carefully we may in addition arrange it so that $T'$ is connected, see Topology, Lemma 5.8.17. 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. $\square$

Comment #6546 by WhatJiaranEatsTonight on

If we want to use induction, it still needs to show that R/I has connected spectrum. It cannot be seen easily.

Comment #6594 by on

Thanks very much for catching this rather horrible error. I fixed it by adding a fun topology lemma. See changes here.

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