Spectrum of a ring

Remarks

A ring homomorphism $f \colon R \to A$ induces the map $\operatorname{Spec} A \to \operatorname{Spec} R$, $\mathfrak q \mapsto f^{-1}(\mathfrak q)$. It is continuous, since the preimage of $V(E)$ is $V(f(E))$.

Example: $\operatorname{Spec} \Z$ consists of $(0)$ and the ideals $p\Z$ for the prime numbers $p$. The spectrum of a field $k$ is a single point $(0)$. The spectrum of $\Z/n\Z$ for $n \ge 2$ is the finite set of ideals $p\Z/n\Z$ for the primes $p$ dividing $n$.

Non-example: the ring $R$ itself is not a point of $\operatorname{Spec} R$, because a prime ideal is by definition a proper ideal. So the zero ring has empty spectrum.

Every nonzero ring has a maximal ideal, which is prime, so $\operatorname{Spec} R \ne \emptyset$ when $R \ne 0$.