The Stacks project

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. $\square$

Proof. For any extension of fields $k'/k$ the connectivity of the spectrum of $R \otimes _ k k'$ is equivalent to $R \otimes _ k k'$ having no nontrivial idempotents, see Lemma 10.21.4. Assume (2). Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. Using Lemma 10.43.4 we see that (2) is equivalent to $R \otimes _ k \overline{k}$ having no nontrivial idempotents. For any field extension $k'/k$, there exists a field extension $\overline{k}'/\overline{k}$ with $k' \subset \overline{k}'$. By Lemma 10.48.1 we see that $R \otimes _ k \overline{k}'$ has no nontrivial idempotents. If $R \otimes _ k k'$ has a nontrivial idempotent, then also $R \otimes _ k \overline{k}'$, contradiction. $\square$

Proof. Immediate from the remark following Definition 10.48.3. $\square$

Proof. This follows from the characterization of connectedness in terms of the nonexistence of nontrivial idempotents. The second and third property follow from the fact that tensor product commutes with colimits. $\square$

Proof. The second assertion follows from the first combined with Lemma 10.22.2. By Lemmas 10.48.5 and 10.43.4 we may assume that $R$ and $S$ are of finite type over $k$. Then we see that also $R \otimes _ k S$ is of finite type over $k$. Note that in this case all the rings are Noetherian and hence their spectra have finitely many connected components (since they have finitely many irreducible components, see Lemma 10.31.6). In particular, all connected components in question are open! Hence via Lemma 10.24.3 we see that the first statement of the lemma in this case is equivalent to the second. Let's prove this. As the algebra $S$ is geometrically connected and nonzero we see that all fibres of $X = \mathop{\mathrm{Spec}}(R \otimes _ k S) \to \mathop{\mathrm{Spec}}(R) = Y$ are connected and nonempty. Also, as $R \to R \otimes _ k S$ is flat of finite presentation the map $X \to Y$ is open (Proposition 10.41.8). Topology, Lemma 5.7.6 shows that $X \to Y$ induces bijection on connected components. $\square$