The previous article ended with the question of which ideals give fields. We start with a weaker question: which ideals give integral domains? An ideal \mathfrak{p} of A is prime if \mathfrak{p} \neq A and ab \in \mathfrak{p} implies a \in \mathfrak{p} or b \in \mathfrak{p} [1].
In \mathbb{Z}, the ideal (5) is prime, because a product divisible by 5 has a factor divisible by 5. The ideal (6) is not, since 2 \cdot 3 \in (6) while 2, 3 \notin (6). The figure marks the integer points on the hyperbolas ab = n, 2n, \ldots, 6n. For n = 5 and n = 7 every point lies on a line where a or b is a multiple of n, while for n = 6 and n = 9 points such as (2, 3) and (3, 3) lie on none of these lines.
For n > 1, the ideal (n) is prime exactly when n is a prime number, and (0) is prime as well. In k[x], the ideal generated by a nonconstant polynomial f is prime exactly when f is not a product of two polynomials of lower degree. The zero ideal of A is prime exactly when A is an integral domain.
In A/\mathfrak{p}, the product (a + \mathfrak{p})(b + \mathfrak{p}) is zero exactly when ab \in \mathfrak{p}, and a + \mathfrak{p} is zero exactly when a \in \mathfrak{p}. So \mathfrak{p} is prime exactly when A/\mathfrak{p} is an integral domain, the condition \mathfrak{p} \neq A matching 1 \neq 0 in the quotient [2]. In the same way, the kernel of a homomorphism into an integral domain is a prime ideal.
Take the field \mathbb{F}_2 = \mathbb{Z}/2, where 1 + 1 = 0. Division by a polynomial f of degree 3 leaves a remainder a + bx + cx^2 with a, b, c \in \mathbb{F}_2, so \mathbb{F}_2[x]/(f) has eight elements, which the figure places at the corners of a cube. Since x^3 + 1 = (x + 1)(x^2 + x + 1), the ring \mathbb{F}_2[x]/(x^3 + 1) has three pairs of nonzero elements with product 0, joined by lines in the figure.
The polynomial x^3 + x + 1 has no root in \mathbb{F}_2, so it is not a product of polynomials of lower degree, and \mathbb{F}_2[x]/(x^3 + x + 1) is an integral domain. Multiplying by x again and again from 1 passes through all seven nonzero elements before returning to 1, so every nonzero element is a power x^k with inverse x^{7 - k}. In general, multiplication by a nonzero element of a finite integral domain is injective, hence surjective, so every finite integral domain is a field.
An ideal \mathfrak{m} of A is maximal if \mathfrak{m} \neq A and no ideal lies strictly between \mathfrak{m} and A, that is, every ideal I with \mathfrak{m} \subseteq I \subseteq A is \mathfrak{m} or A [1].
In \mathbb{Z}, the ideals containing (n) are the (d) with d \mid n, as we saw for n = 12 in the previous article. For n > 1, the ideals strictly between (n) and \mathbb{Z} therefore correspond to the divisors d with 1 < d < n, and (n) is maximal exactly when n is a prime number.
The figure plots the points (n, d) with d \mid n, which line up along the lines d = n/2, n/3, \ldots and along horizontal lines. A vertical line moves through n = 2, 3, \ldots, 30 and colors the points of its column strictly between 1 and n. The columns of prime numbers have only their two ends, and they correspond to the maximal ideals (p) of \mathbb{Z}.
An ideal \mathfrak{m} is maximal exactly when A/\mathfrak{m} is a field. If \mathfrak{m} is maximal and x \notin \mathfrak{m}, then \mathfrak{m} + (x) is strictly larger than \mathfrak{m}, so it is A and 1 = y + ax with y \in \mathfrak{m}. Then (a + \mathfrak{m})(x + \mathfrak{m}) = 1 + \mathfrak{m}, so every nonzero class has an inverse [2]. Conversely, if A/\mathfrak{m} is a field, it has only the ideals 0 and itself, so by the correspondence of ideals nothing lies strictly between \mathfrak{m} and A.
Every field is an integral domain, so every maximal ideal is prime. The converse fails: (0) is prime in \mathbb{Z} but lies inside (2), and (x) is prime in \mathbb{Z}[x], since \mathbb{Z}[x]/(x) \cong \mathbb{Z}, but lies inside (2, x). In \mathbb{Z} and in k[x], where every ideal is principal, every nonzero prime ideal is maximal.
In the figure, each point (b, c) stands for the polynomial f = x^2 + bx + c. The polynomials with a root r satisfy c = -rb - r^2, so they fill a line, and each of these lines touches the parabola c = b^2/4. When the moving point P is below the parabola, two of the lines pass through it, and (x - r_1) and (x - r_2) lie strictly between (f) and \mathbb{R}[x]. When P is above it, no line passes through P, so (f) is maximal and \mathbb{R}[x]/(f) is a field.
Let \varphi : A \to B be a homomorphism and \mathfrak{q} a prime ideal of B. The preimage \varphi^{-1}(\mathfrak{q}) is the kernel of A \to B/\mathfrak{q}, a homomorphism into an integral domain, so it is a prime ideal of A [1]. The preimage of a maximal ideal need not be maximal: for the inclusion \mathbb{Z} \to \mathbb{Q}, the preimage of the maximal ideal 0 of \mathbb{Q} is (0), which is not maximal in \mathbb{Z}.
For the inclusion \mathbb{Z} \to \mathbb{Z}[i], the preimage of an ideal is its intersection with \mathbb{Z}. The element 2 + i is prime in \mathbb{Z}[i] [2], so (2 + i) \cap \mathbb{Z} is a prime ideal of \mathbb{Z} containing 5 = (2 + i)(2 - i), which makes it (5). In the other direction, the ideal of \mathbb{Z}[i] generated by 5 is not prime, since it contains (2 + i)(2 - i) but neither factor.
The figure draws the circle a^2 + b^2 = p and the Gaussian integers a + bi on it. For p = 5, 13, 17 there are eight points in two groups of four, such as the group of 2 + i and the group of 2 - i, and within a group the points differ by the units 1, -1, i, -i. The two groups generate two different prime ideals over (p). For p = 2 the four points form one group, and for p = 3 the circle has no points, while 3 itself, on the circle a^2 + b^2 = 9, stays prime in \mathbb{Z}[i].
Every ring A \neq 0 has a maximal ideal. The proper ideals of A are ordered by inclusion, and the union of a chain of proper ideals is again a proper ideal, since it does not contain 1. Zorn's lemma then gives a maximal element, which is a maximal ideal [1]. The same argument applied to the proper ideals containing a given proper ideal I gives a maximal ideal containing I.
If a is not a unit, then (a) is a proper ideal and lies in some maximal ideal, while a unit lies in no proper ideal. So the units are exactly the elements outside every maximal ideal. In \mathbb{Z}/30 the maximal ideals are (2), (3) and (5), and in the figure each non-unit lies in at least one of the three circles, while the eight units 1, 7, 11, 13, 17, 19, 23, 29 lie outside them.
A ring with exactly one maximal ideal is called a local ring, and in a local ring the non-units are exactly the elements of the maximal ideal. The ring \mathbb{Z}/8 is local, with maximal ideal (2). The ring \mathbb{Z}/12 has the two maximal ideals (2) and (3), and 3 + 4 = 7 shows that its non-units are not closed under addition.