The set of prime ideals of a ring A is called its spectrum and written \operatorname{Spec} A; its elements are called points. The points of \operatorname{Spec}\mathbb{Z} are (2), (3), (5), \ldots and also (0). For an algebraically closed field k, the points of \operatorname{Spec} k[x] are the ideals (x - a) with a \in k and (0). Since every nonzero ring has a maximal ideal, \operatorname{Spec} A is empty only for the zero ring [1].
In \operatorname{Spec} k[x, y] there are three kinds of points: the maximal ideals (x - a, y - b), which correspond to the points of the plane, the ideals (f) of irreducible polynomials, which correspond to irreducible curves, and (0). A whole curve thus appears as a single point of the spectrum.
For an ideal I let V(I) be the set of prime ideals containing I. Then V(I) \cup V(J) = V(IJ), the intersection of any family V(I_\lambda) is the V of the sum of the I_\lambda, V((0)) = \operatorname{Spec} A, and V(A) is empty. So the sets V(I) are the closed sets of a topology, the Zariski topology [1,2]. Since every prime ideal is radical, V(I) = V(\sqrt{I}).
In \operatorname{Spec}\mathbb{Z} the set V((n)) for n \neq 0 consists of the prime factors of n, so the closed sets are the finite sets of points (p) and the whole space. In the figure V((12)) = V((6)) = \{(2), (3)\}, and V((12)) \cap V((45)) = V((3)) because (12) + (45) = (3). A closed set V(I) containing every (p) has I \subseteq (2) \cap (3) \cap \cdots = (0), so it contains (0) as well.
The closure of a point \mathfrak{p} is V(\mathfrak{p}): a closed set V(I) contains \mathfrak{p} exactly when I \subseteq \mathfrak{p}, and then it contains V(\mathfrak{p}). So the closed points are the maximal ideals. In \operatorname{Spec}\mathbb{Z} the closure of (0) is the whole space, and (0) is called its generic point. In \operatorname{Spec} k[x, y] the closure of the point (f) is the curve f = 0 together with (f) [1].
These spaces are far from Hausdorff. In \operatorname{Spec} k[x] every nonempty open set is the complement of finitely many closed points and contains (0), so any two nonempty open sets meet. Still, of two different points, one always lies in an open set that misses the other.
In the figure the real points of y^2 = x^3 - x form an oval and a branch. Three points of the oval form a closed set, since finite sets of closed points are closed. A polynomial vanishing at infinitely many points of the oval is a multiple of the irreducible y^2 - x^3 + x, so the closure of these points is V((y^2 - x^3 + x)), which contains the branch as well. Points spread over a region have the whole space as their closure.
For f \in A let D(f) be the set of prime ideals not containing f, the complement of V((f)). The complement of V(I) is the union of the D(f) for f \in I, so these sets form a basis of the topology, and D(f) \cap D(g) = D(fg) because a prime ideal contains fg exactly when it contains f or g. By the previous article D(f) is empty exactly when f is nilpotent, and D(f) = \operatorname{Spec} A exactly when f is a unit.
\operatorname{Spec} A is quasi-compact. If it is covered by sets D(f_\lambda), no prime ideal contains all the f_\lambda, so they generate the unit ideal and 1 = a_1f_{\lambda_1} + \cdots + a_nf_{\lambda_n} for finitely many indices. A prime ideal outside all of these n sets would contain 1, so the n sets D(f_{\lambda_i}) already cover the space [1].
In the figure X = \operatorname{Spec}\mathbb{R}[x, y], and D(x) and D(y) leave out the lines x = 0 and y = 0. Their union is X \setminus V((x, y)), the plane without the origin, which is open but not of the form D(f). Adding D(x + y - 1) covers the origin too, which matches 1 = x + y - (x + y - 1).
A ring homomorphism \varphi \colon A \to B sends each prime ideal \mathfrak{q} of B to the prime ideal \varphi^{-1}(\mathfrak{q}) of A: if xy \in \varphi^{-1}(\mathfrak{q}), then \varphi(x)\varphi(y) \in \mathfrak{q}, so \varphi(x) or \varphi(y) lies in \mathfrak{q}. This gives a map \operatorname{Spec} B \to \operatorname{Spec} A in the opposite direction [1].
The map is continuous, since the preimage of V(I) is V(J) for the ideal J of B generated by \varphi(I). For the quotient map A \to A/I it is a homeomorphism from \operatorname{Spec}(A/I) onto V(I), because the prime ideals of A/I correspond to the prime ideals of A containing I [1].
The figure shows the inclusion \mathbb{R}[x] \subseteq \mathbb{C}[x]. The point (x - z) of \operatorname{Spec}\mathbb{C}[x] goes to (x - z) \cap \mathbb{R}[x], which is (x - z) when z is real and ((x - z)(x - \bar{z})) otherwise; for z = 1 + i it is (x^2 - 2x + 2). The plane folds along the real axis, so the closed points of \operatorname{Spec}\mathbb{R}[x] form the closed upper half-plane, and those off the real axis have two preimages.