Proposition 2.88 (00JT) · ScalingStacks

00JT

Proposition 2.88. Let $Z=\spec(A_{Z})$ be an affine $k$ -variety, $\vvvert\mathord{\cdot}\vvvert$ an algebra norm on $A_{Z}$ and $\mathcal{A}_{Z}$ be the $k$ -Banach algebra obtained by completing $A_{Z}$ with respect to $\vvvert\mathord{\cdot}\vvvert$ . Then the canonical homomorphism of $k$ -algebras from $A_{Z}$ to $\mathcal{A}_{Z}$ induces a continuous map which embeds the Berkovich spectrum $\mathfrak{M}(\mathcal{A}_{Z})$ into $Z^{an}$ as a compact subspace (and is closed since $Z^{\mathrm{an}}$ is Hausdorff), and the Berkovich topology coincides with the induced topology from $Z^{an}$ .

00JU

Proof. For any $z\in\mathfrak{M}(\mathcal{A}_{Z})$ , the multiplicative algebra seminorm (or the corresponding character) $\chi_{z}$ on $\mathcal{A}_{Z}$ corresponds to a unique multiplicative algebra seminorm on $A_{Z}$ by restriction. Since $A_{Z}$ is dense in $\mathcal{A}_{Z}$ , the family of open sets $\{U(f;p,q),\text{ }f\in A_{Z},\text{ }p,q\in\mathbb{R}\}$ form a basis for topology on $\mathfrak{M}(\mathcal{A}_{Z})$ , hence the inherited topology coincides with the originial topology. So the embedding is continuous, and the image of $\mathfrak{M}(\mathcal{A}_{Z})$ is compact in $Z^{an}$ . Since the topology on $Z^{an}$ is Hausdorff, the image of $\mathfrak{M}(\mathcal{A}_{Z})$ is closed. ∎