A ring is reduced when its only nilpotent element is 0, and R_{\mathrm{red}} = R/\mathfrak{N}(R) is always reduced. Every prime ideal contains every nilpotent element, so the primes of R and of R_{\mathrm{red}} correspond one-to-one, closed sets included, and \operatorname{Spec} R_{\mathrm{red}} \to \operatorname{Spec} R is a homeomorphism [1]. For the same reason V(I) = V(\sqrt{I}), and radical ideals correspond one-to-one to closed sets.
What reduction forgets never shows up at a point: in \mathbb{R}[x, \varepsilon]/(\varepsilon^2) the elements \alpha + \varepsilon\beta and \alpha, with \alpha and \beta polynomials in x, take the same value everywhere. So the shape of the spectrum depends only on R_{\mathrm{red}}. The first question about that shape is irreducibility: a nonempty space is irreducible if it is not the union of two proper closed subsets, or equivalently, if any two nonempty open subsets meet [2].
Since D(f) \cap D(g) = D(fg) and D(f) is empty exactly when f is nilpotent, \operatorname{Spec} R is irreducible if and only if \mathfrak{N}(R) is prime [1], and likewise V(I) is irreducible if and only if \sqrt{I} is prime. An integral domain is therefore exactly a reduced ring with an irreducible spectrum, for then (0) = \mathfrak{N}(R) is prime.
Let R = \mathbb{R}[x, y, z]/(xz, yz) and X = \operatorname{Spec} R, whose real points are the plane z = 0 and the z-axis. In R we have (x^2 + y^2)z = x \cdot xz + y \cdot yz = 0, yet neither factor is nilpotent: z does not vanish on the axis, nor x^2 + y^2 on the plane. So D(z) and D(x^2 + y^2) are disjoint nonempty open sets, and X, the union of the closed sets V(z) and V(x, y), is not irreducible.
A maximal irreducible subset is an irreducible component. Components are closed, every irreducible subset lies in one by Zorn's lemma, and together they cover the space. In \operatorname{Spec} R the irreducible closed sets are the V(\mathfrak{p}) with \mathfrak{p} prime, and a larger V(\mathfrak{p}) means a smaller \mathfrak{p}, so the components are the V(\mathfrak{p}) for minimal primes \mathfrak{p} [1]. In particular every prime contains a minimal prime.
For R = \mathbb{R}[x, y, z]/(xz, yz) the minimal primes are (z) and (x, y), and the components are the plane V(z) and the line V(x, y). They meet at the origin, and they do not even have the same dimension. Nothing forces the components of one spectrum to look alike.
A ring is Noetherian if every increasing chain of ideals stops, as in \mathbb{Z} and, by Hilbert's basis theorem, in polynomial rings in finitely many variables over a field. For such a ring, decreasing chains of closed sets in \operatorname{Spec} R, which are increasing chains of radical ideals, stop as well. A space with this property is called Noetherian; it is a union of finitely many irreducible closed subsets, so a Noetherian ring has finitely many minimal primes [1,2].
Without the Noetherian condition the count can be infinite. In k[x_1, x_2, \ldots]/(x_i x_j \mid i \neq j) a prime ideal contains all the variables but at most one, so the minimal primes are \mathfrak{p}_n = (x_j \mid j \neq n) for n = 1, 2, \ldots, and each V(\mathfrak{p}_n) is a line through the point \mathfrak{m} = (x_1, x_2, \ldots). The closed sets \cdots \subsetneq V(x_1, x_2) \subsetneq V(x_1) keep shrinking without end.
In a reduced ring the zero divisors are the elements that vanish on a whole component, that is, the elements of the minimal prime ideals. If fg = 0 with g \neq 0, then g lies outside some minimal prime \mathfrak{p}, because the minimal primes intersect in \mathfrak{N}(R) = 0, and fg \in \mathfrak{p} forces f \in \mathfrak{p}.
The converse is easiest when there are finitely many minimal primes \mathfrak{p}_1, \ldots, \mathfrak{p}_n, as in a Noetherian ring. Let f \in \mathfrak{p}_1. By minimality \mathfrak{p}_1 contains no other \mathfrak{p}_i, so it does not contain the product \mathfrak{p}_2 \cdots \mathfrak{p}_n, and we can pick g in this product outside \mathfrak{p}_1. Then fg lies in every \mathfrak{p}_i, so fg = 0 while g \neq 0.
Take the ellipses q_1 = 0 and q_2 = 0 with q_1 = \tfrac{x^2}{4} + y^2 - 1 and q_2 = x^2 + \tfrac{y^2}{4} - 1, and let R = \mathbb{R}[x, y]/(q_1 q_2), which is reduced because q_1 and q_2 are distinct irreducible polynomials. The element q_1 vanishes on the whole first ellipse and q_1 q_2 = 0. The element y vanishes at only four points, so it is not a zero divisor.
Neither is q_1 + q_2: on each ellipse it equals the equation of the other one, so it vanishes only at the four crossing points. The sum of two zero divisors need not be a zero divisor, and the zero divisors of a reduced ring form a union of prime ideals rather than an ideal.
An idempotent e, an element with e^2 = e, cuts the spectrum in two. Since e(1 - e) = 0 and e + (1 - e) = 1, every prime ideal contains exactly one of e and 1 - e, so \operatorname{Spec} R is the disjoint union of the closed sets V(e) and V(1 - e), each of which is then also open. Conversely, every splitting of \operatorname{Spec} R into two disjoint closed sets comes from an idempotent [1].
An irreducible space is connected, because it cannot contain two disjoint nonempty open sets. The converse fails. Let R = \mathbb{Z}[x]/(x^2 - 1). Its minimal primes are (x - 1) and (x + 1), both quotients are \mathbb{Z}, and so each component is homeomorphic to \operatorname{Spec}\mathbb{Z}. Since (x - 1) + (x + 1) = (2, x - 1), the two components meet in the single point (2, x - 1) over the prime 2.
Hence \operatorname{Spec} R is connected, and indeed R has no idempotent besides 0 and 1: the candidate \tfrac{1 + x}{2} needs \tfrac{1}{2}. Allowing powers of 2 as denominators removes the meeting point. Over \mathbb{Z}[\tfrac{1}{2}] the ideals (x - 1) and (x + 1) are coprime, and the Chinese remainder theorem splits \mathbb{Z}[\tfrac{1}{2}][x]/(x^2 - 1) into \mathbb{Z}[\tfrac{1}{2}] \times \mathbb{Z}[\tfrac{1}{2}] with idempotent e = \tfrac{1 + x}{2}.
The fiber over 2 is \mathbb{Z}[x]/(2, x^2 - 1) = \mathbb{F}_2[x]/((x - 1)^2), since x^2 - 1 and (x - 1)^2 agree modulo 2. Its spectrum is the single point (2, x - 1) where the two components meet, while the ring itself carries the nilpotent x - 1, which the spectrum does not see, just as in the first section.