Lemma 10.21.2. Let $R_1$ and $R_2$ be rings. Let $R = R_1 \times R_2$. The maps $R \to R_1$, $(x, y) \mapsto x$ and $R \to R_2$, $(x, y) \mapsto y$ induce continuous maps $\mathop{\mathrm{Spec}}(R_1) \to \mathop{\mathrm{Spec}}(R)$ and $\mathop{\mathrm{Spec}}(R_2) \to \mathop{\mathrm{Spec}}(R)$. The induced map

$\mathop{\mathrm{Spec}}(R_1) \amalg \mathop{\mathrm{Spec}}(R_2) \longrightarrow \mathop{\mathrm{Spec}}(R)$

is a homeomorphism. In other words, the spectrum of $R = R_1\times R_2$ is the disjoint union of the spectrum of $R_1$ and the spectrum of $R_2$.

Proof. Write $1 = e_1 + e_2$ with $e_1 = (1, 0)$ and $e_2 = (0, 1)$. Note that $e_1$ and $e_2 = 1 - e_1$ are idempotents. We leave it to the reader to show that $R_1 = R_{e_1}$ is the localization of $R$ at $e_1$. Similarly for $e_2$. Thus the statement of the lemma follows from Lemma 10.21.1 combined with Lemma 10.17.6. $\square$

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).