Prime spectrum

Let $R$ be a commutative ring. A prime ideal of $R$ is a proper ideal $\mathfrak p\subsetneq R$ such that whenever $ab\in\mathfrak p$ (with $a,b\in R$), then $a\in\mathfrak p$ or $b\in\mathfrak p$.

The prime spectrum of $R$ is the set

An element $\mathfrak p\in \operatorname{Spec}(R)$ is called a point of $\operatorname{Spec}(R)$.

In commutative algebra one usually studies $\operatorname{Spec}(R)$ together with the Zariski topology; this turns $\operatorname{Spec}(R)$ into a topological space whose basic opens are closely related to localizations. For a point $\mathfrak p\in\operatorname{Spec}(R)$, the associated local data are the localization $R_{\mathfrak p}$ and its residue field $\kappa(\mathfrak p)$.

Examples

1. A field has a one-point spectrum. If $k$ is a field, the only prime ideal is $(0)$, so $\operatorname{Spec}(k)=\{(0)\}$.

1. The spectrum of the integers. In $R=\mathbb{Z}$, the prime ideals are $(0)$ and $(p)$ for primes $p$. Thus
   $$\operatorname{Spec}(\mathbb{Z})=\{(0)\}\ \cup\ \{(p)\mid p\ \text{prime}\}.$$
   Under the Zariski topology, the point $(0)$ is a generic point whose closure is all of $\operatorname{Spec}(\mathbb{Z})$.

1. The spectrum of a polynomial ring in one variable. Let $k$ be a field and $R=k[x]$. Then $(0)$ is prime, and every nonzero prime ideal is generated by an irreducible polynomial. So
   $$\operatorname{Spec}(k[x])=\{(0)\}\ \cup\ \{(f)\mid f\in k[x]\ \text{irreducible}\}.$$
   If $k$ is algebraically closed, the maximal ideals are precisely $(x-a)$, and $\operatorname{MaxSpec}(k[x])$ can be identified with the affine line over $k$.