Lemma 10.48.6. Let $k$ be a field. Let $S$ be a geometrically connected $k$-algebra. Let $R$ be any $k$-algebra. The map

$R \longrightarrow R \otimes _ k S$

induces a bijection on idempotents, and the map

$\mathop{\mathrm{Spec}}(R \otimes _ k S) \longrightarrow \mathop{\mathrm{Spec}}(R)$

induces a bijection on connected components.

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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).