Let $A$ be a Banach ring, that is, a commutative ring that is complete with respect to a submultiplicative norm $\|\cdot\|$. The Berkovich spectrum ${\mathcal{M}}(A)$ is the set of all bounded multiplicative seminorms on $A$. In other words, a point $x\in{\mathcal{M}}(A)$ corresponds to a function $|\cdot|_{x}\colon A\to{\mathbb{R}}_{\geq 0}$ such that $|\cdot|_{x}\leq\|\cdot\|$, $|1|_{x}=1$, $|f+g|_{x}\leq|f|_{x}+|g|_{x}$ and $|fg|_{x}=|f|_{x}|g|_{x}$ for $f,g\in A$. The spectrum is a nonempty, compact Hausdorff space with respect to the topology of pointwise convergence.

For $x\in{\mathcal{M}}(A)$, denote by ${\mathfrak{p}}_{x}$ the kernel of $|\cdot|_{x}$. This is a prime ideal of $A$, and $|\cdot|_{x}$ defines a multiplicative norm on $A/{\mathfrak{p}}_{x}$. The completion of the fraction field of $A/{\mathfrak{p}}_{x}$ with respect to this norm is a valued field ${\mathcal{H}}(x)$. We write $f(x)$ for the image of $f\in A$ in ${\mathcal{H}}(x)$; then $|f(x)|=|f|_{x}$. The assignment $x\mapsto{\mathfrak{p}}_{x}$ yields a map ${\mathcal{M}}(A)\to\operatorname{Spec}(A)$ that is continuous for the Zariski topology $\operatorname{Spec}(A)$.