In \mathbb{Z}/4 the element 2 is not 0, but 2^2 = 4 = 0. An element a of a ring A is nilpotent if a^n = 0 for some n \geq 1, and the least such n is called its index. In \mathbb{Z}/27 every multiple of 3 is nilpotent; for example 6^2 = 36 = 9 and 6^3 = 216 = 0. In k[x]/(x^n) the class of x is nilpotent of index exactly n [1].
A nonzero nilpotent is a zero divisor: if a \neq 0 has index n, then n \geq 2 and a \cdot a^{n-1} = 0 with a^{n-1} \neq 0. So an integral domain has no nilpotent element other than 0. The converse fails: \mathbb{Z}/6 has the zero divisors 2 and 3, but no power of 2 or of 3 is a multiple of 6, so its only nilpotent element is 0.
The nilpotent elements form an ideal, the nilradical \mathfrak{N}(A). If a^n = 0, then (ra)^n = r^n a^n = 0. If a^m = 0 and b^n = 0, expand (a + b)^{m+n-1} by the binomial theorem: each term is a multiple of a^i b^j with i + j = m + n - 1, so i \geq m or j \geq n, and the term is 0. The expansion uses ab = ba, and indeed a sum of two nilpotent matrices need not be nilpotent.
If the class of x in A/\mathfrak{N}(A) is nilpotent, some x^n is nilpotent, and so is x. Thus 0 is the only nilpotent element of A/\mathfrak{N}(A); such a ring is called reduced. In the figure the leaves are the elements of \mathbb{Z}/27, those under the vertex 0 of depth 1 form \mathfrak{N}(\mathbb{Z}/27) = (3), and the quotient sends each leaf to its vertex of depth 1, leaving the field \mathbb{Z}/3.
For an ideal I of A, the radical \sqrt{I} is the set of elements a with a^n \in I for some n \geq 1. The same use of the binomial theorem shows that \sqrt{I} is an ideal containing I. An element lies in \sqrt{I} exactly when its class in A/I is nilpotent, so \sqrt{I}/I = \mathfrak{N}(A/I) and \sqrt{(0)} = \mathfrak{N}(A) [1].
In \mathbb{Z} some power of a is a multiple of 72 = 2^3 \cdot 3^2 exactly when 2 and 3 divide a, so \sqrt{(72)} = (6): the prime factors remain and the exponents are lost. In the same way \sqrt{(x^2(x - 1)^3)} = (x(x - 1)) in k[x] keeps the roots 0 and 1 but forgets their multiplicities. An ideal with I = \sqrt{I} is called radical; prime ideals are radical, since a^n \in \mathfrak{p} forces a \in \mathfrak{p} [2].
In the figure f = x^2(x - 1)^3. A polynomial is a multiple of f when it vanishes to order at least 2 at 0 and at least 3 at 1; near each root its graph then stays in a region such as |y| \leq 3x^2 or |y| \leq 3|x - 1|^3. For h = x(x - 1) the graph of h^k fits at 0 from k = 2 and at 1 from k = 3, so h \in \sqrt{(f)}, while for h = x we have h^k(1) = 1 for every k, so x \notin \sqrt{(f)}.
A nilpotent element lies in every prime ideal, because a^n = 0 \in \mathfrak{p} forces a \in \mathfrak{p}. Conversely, let a be an element that is not nilpotent. Then S = \{1, a, a^2, \ldots\} does not contain 0, and Zorn's lemma gives an ideal \mathfrak{p} that is maximal among the ideals disjoint from S, by the same argument that shows that maximal ideals exist.
This \mathfrak{p} is prime. If x \notin \mathfrak{p} and y \notin \mathfrak{p}, the larger ideals \mathfrak{p} + (x) and \mathfrak{p} + (y) meet S, say in a^i and a^j, and then a^{i+j} \in \mathfrak{p} + (xy), so xy \notin \mathfrak{p}. Since a \notin \mathfrak{p}, \mathfrak{N}(A) is the intersection of all prime ideals of A; likewise \sqrt{I} is the intersection of the prime ideals containing I [1].
The prime ideals of \mathbb{Z}/72 are (2) and (3), so \mathfrak{N}(\mathbb{Z}/72) = (2) \cap (3) = (6). The figure follows a, a^2, a^3, \ldots in \mathbb{Z}/72 until a power repeats, which traces a tail and a loop; filled points lie in (2) and ringed points in (3). No power of 2 lies in (3), no power of 3 lies in (2), and the powers of 5 avoid both, while the powers of 6 reach 0, which lies in every prime ideal.
If a^n = 0, then (1 + a)(1 - a + a^2 - \cdots + (-1)^{n-1}a^{n-1}) = 1 + (-1)^{n-1}a^n = 1, so 1 + a is a unit: the geometric series for its inverse stops after n terms. For a unit u, u + a = u(1 + u^{-1}a) is again a unit, since u^{-1}a is nilpotent. In \mathbb{Z}/8 the units 1, 3, 5, 7 are 1 plus the nilpotents 0, 2, 4, 6 [1].
Hence a nilpotent a lies in every maximal ideal \mathfrak{m}: otherwise \mathfrak{m} + (a) = A, so m + ra = 1 for some m \in \mathfrak{m} and r \in A, and m = 1 - ra would be a unit in \mathfrak{m}. The intersection of all maximal ideals is the Jacobson radical \operatorname{Jac}(A). An element x lies in it exactly when 1 - xy is a unit for every y \in A, and \mathfrak{N}(A) \subseteq \operatorname{Jac}(A).
The inclusion can be strict. In the ring of fractions with odd denominator, 2 is not nilpotent, but for every y = c/d with d odd, 1 - 2y = (d - 2c)/d has an odd numerator and an odd denominator, so it is a unit, and 2 lies in the Jacobson radical.
In the figure the elements of \mathbb{Z}/10000 are shown with four digits. Since 20^4 = 160000 = 0, the inverse of 21 is 1 - 20 + 400 - 8000 = 2381. The term 20^k is a multiple of 10^k and does not change the last k digits, so each new term fixes one more digit from the right. At the end 21 \times 2381 = 50001, whose fifth digit falls outside, and the display reads 0001.
Over a field k the degree of a product is the sum of the degrees, so the units of k[x] are the nonzero constants. Over \mathbb{Z}/4 this fails, since (1 + 2x)^2 = 1 + 4x + 4x^2 = 1. In general f = a_0 + a_1x + \cdots + a_nx^n is a unit of A[x] exactly when a_0 is a unit and a_1, \ldots, a_n are nilpotent [1].
In that case a_1x + \cdots + a_nx^n is nilpotent, so f is a unit plus a nilpotent, hence a unit. Conversely, let f be a unit and \mathfrak{p} a prime ideal. The image of f in (A/\mathfrak{p})[x] is a unit of a polynomial ring over an integral domain, hence a constant, so a_1, \ldots, a_n \in \mathfrak{p}. This holds for every \mathfrak{p}, so they are nilpotent. In the same way f is nilpotent exactly when all of its coefficients are.
In the ring of formal power series A[[x]] the condition is weaker: f is a unit as soon as a_0 is, because the coefficients of the inverse can be found one after another. For example 1 - x has the inverse 1 + x + x^2 + \cdots in k[[x]] [1].
The figure works in \mathbb{R}[\varepsilon], \varepsilon^2 = 0, and draws f = \alpha(x) + \varepsilon\beta(x) as the band of graphs of f_s = \alpha + s\beta for small s. For \alpha = 2 the inverse is g = 1/2 - \varepsilon\beta/4, and with g_s formed likewise, f_s g_s = 1 - s^2\beta^2/4 is within a multiple of s^2 of 1. If \alpha is not constant, f is not a unit: setting \varepsilon = 0 would give a unit \alpha of \mathbb{R}[x].