The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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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