The previous article ended with the fact that the kernel of a ring homomorphism is an ideal. Consider the homomorphism \mathbb{R}[x] \to \mathbb{R} that substitutes x = 2. Its kernel is the set of polynomials f with f(2) = 0. If f_1(2) = f_2(2) = 0, then (f_1 + f_2)(2) = 0, and if f(2) = 0, then (gf)(2) = g(2) f(2) = 0 for every polynomial g, whatever value g takes at 2.
The figure shows this second property. The graph of g moves freely and need not pass through (2, 0), but the graph of the product gf passes through (2, 0) at every moment. An ideal of a commutative ring A is a subset I with the same two properties: I contains 0 and is closed under addition and subtraction, and ax \in I for every a \in A and every x \in I [1].
The multiples n\mathbb{Z} of an integer n form an ideal of \mathbb{Z}, and the zero ideal 0 = \{0\} and A itself are ideals of every ring. If an ideal I contains a unit u, then 1 = u^{-1} u \in I and a = a \cdot 1 \in I for every a, so I = A. Unlike a subring, an ideal other than A never contains 1. In a field every nonzero element is a unit, so a field has only two ideals, 0 and the field itself.
Given elements a_1, \ldots, a_n of A, the sums r_1 a_1 + \cdots + r_n a_n with r_1, \ldots, r_n \in A form an ideal, written (a_1, \ldots, a_n) [2]. It contains every a_i, and an ideal that contains every a_i contains all of these sums, so (a_1, \ldots, a_n) is the smallest ideal containing a_1, \ldots, a_n. It is called the ideal generated by a_1, \ldots, a_n.
In \mathbb{Z}, the ideal (4, 6) consists of the integers 4r + 6s. All of them are even, and 2 = 6 - 4 is one of them, so (4, 6) contains (2) and is contained in it, that is, (4, 6) = (2). The figure starts from 0 and adds or subtracts 4 and 6. After two rounds it has reached every even number from -10 to 10, with 2 and -2 reached as 6 - 4 and 4 - 6, and it never reaches an odd number.
In the polynomial ring k[x, y] over a field k, the ideal (x, y) consists of the polynomials xf + yg. These are exactly the polynomials whose constant term is 0, so (x, y) is the kernel of the homomorphism k[x, y] \to k that substitutes x = y = 0. An ideal can have many different sets of generators, and an ideal generated by finitely many elements is called finitely generated.
An ideal generated by a single element is called principal, and (a) = \{ra : r \in A\} consists of the multiples of a. The ideal (b) contains a exactly when a = rb for some r \in A, that is, when b divides a, and then it contains every multiple of a as well. So (a) \subseteq (b) holds exactly when b \mid a, and containment of principal ideals is divisibility read in the opposite direction.
The figure shows the six ideals of \mathbb{Z} that contain 12, with larger ideals placed higher and lines for containment. For each n in the figure, the ideals containing n are the (d) with d \mid n, and they spread upward from (n), the smallest ideal containing n. Two principal ideals are equal when each generator divides the other, which in an integral domain means that the generators differ by a unit factor, as in (6) = (-6) and (2x - 4) = (x - 2).
Every ideal of \mathbb{Z} is principal. If I \neq 0, let d be the smallest positive element of I, and divide any n \in I as n = qd + r with 0 \leq r < d. Then r = n - qd lies in I, so r = 0 by the choice of d, and I = (d). The same argument with degrees shows that every ideal of k[x] is principal, and the ideal of polynomials vanishing at 2 is (x - 2). In k[x, y] this fails: if (x, y) = (f), then f divides x and y, so f is a nonzero constant and (f) contains 1, which (x, y) does not.
For ideals I and J of A, the sum I + J = \{x + y : x \in I, y \in J\} is the smallest ideal containing both, the intersection I \cap J is again an ideal, and the product IJ is the ideal generated by the products xy with x \in I and y \in J [1]. In \mathbb{Z} they are given by the greatest common divisor, the least common multiple and the product: (12) + (18) = (6), (12) \cap (18) = (36) and (12)(18) = (216).
Each product xy lies in both I and J, so IJ \subseteq I \cap J always holds. The two can differ, as (216) \subsetneq (36) shows, but they are equal when I + J = A. In that case 1 = a + b for some a \in I and b \in J, and every c \in I \cap J is the sum c = ca + cb of two elements of IJ. For example, (4) + (9) = (1) and (4)(9) = (36) = (4) \cap (9).
In k[x, y] these operations can be seen through zero sets. Let f = 0 be the circle in the figure and g = 0, h = 0 the two lines, and take I = (fg) and J = (fh). The common zeros of I + J are the points where fg and fh both vanish, namely the circle and the point P where the lines meet. Both I \cap J = (fgh) and IJ = (f^2 gh) vanish exactly on the circle and the two lines, but f appears twice in the product, so for IJ the figure draws the circle as a double line.
For ideals I and J, the ideal quotient (I : J) is the set of a \in A with aJ \subseteq I, that is, the elements whose product with every element of J lies in I [1]. It is an ideal and contains I. In \mathbb{Z}, the quotient ((12) : (8)) consists of the a for which 8a is a multiple of 12. Since 8 already supplies the factor 4 of 12, this means that 3 divides a, so ((12) : (8)) = (3).
In general ((m) : (n)) = (m / \operatorname{gcd}(m, n)): the generator is what remains of m after removing the prime factors that n supplies. In the figure, the prime factors of n fill the squares 2, 2, 3 of 12, and the product of the squares left empty generates ((12) : (n)). When n = 24 every square is filled and the quotient is (1) = \mathbb{Z}, and when n = 5 no square is filled and the quotient is (12).
The quotient (0 : J) is called the annihilator of J and written \operatorname{Ann}(J), and \operatorname{Ann}(x) = (0 : (x)) for a single element. In \mathbb{Z}/12, the elements a with 8a = 0 are 0, 3, 6, 9, so \operatorname{Ann}(8) is the ideal generated by 3, the image of ((12) : (8)) = (3). A nonzero element is a zero divisor exactly when it lies in \operatorname{Ann}(x) for some x \neq 0, so in an integral domain \operatorname{Ann}(x) = 0 for every x \neq 0.
Fix an ideal I of A, and regard a and b as the same when a - b \in I. The class of a is the set a + I = \{a + x : x \in I\}, and a + I = b + I exactly when a - b \in I. The set of classes is written A/I, and classes are added and multiplied through their representatives: (a + I) + (b + I) = (a + b) + I and (a + I)(b + I) = ab + I [2].
These operations do not depend on the choice of representatives. If a_1 = a + x and b_1 = b + y with x, y \in I, then a_1 b_1 = ab + (ay + xb + xy), and the terms in parentheses lie in I because I absorbs multiplication by every element of A. This is where the definition of an ideal is needed. With these operations A/I is a commutative ring, the quotient ring, and a \mapsto a + I is a surjective homomorphism A \to A/I with kernel I, so every ideal is a kernel [1].
The ring \mathbb{Z}/n of integers modulo n is the quotient of \mathbb{Z} by (n). In \mathbb{Z}[x]/(x - 2) the element x - 2 becomes 0, so x and 2 become equal and 2x^k can be exchanged for x^{k+1}. The figure starts from the constant 11 and repeatedly exchanges two dots on one line for one dot on the next line. The polynomial changes from 11 through 5x + 1 and 2x^2 + x + 1 to x^3 + x + 1, whose coefficients are the binary digits of 11, while its value at x = 2 stays 11.
Let \varphi : A \to B be a ring homomorphism. Sending the class a + \ker\varphi to \varphi(a) gives a well-defined map A/\ker\varphi \to B, because a - a_1 \in \ker\varphi implies \varphi(a) = \varphi(a_1). It is an injective homomorphism, since \varphi(a) = 0 means a \in \ker\varphi, and its image is \operatorname{im}\varphi, so A/\ker\varphi \cong \operatorname{im}\varphi [2]. To recognize a quotient ring, it is enough to find a homomorphism with the right kernel.
Substituting x = i gives a surjective homomorphism \mathbb{R}[x] \to \mathbb{C} whose kernel is (x^2 + 1), so \mathbb{R}[x]/(x^2 + 1) \cong \mathbb{C}. Substituting x = 2 gives a surjective homomorphism \mathbb{Z}[x] \to \mathbb{Z}, and division by the monic polynomial x - 2 gives f = (x - 2)q + f(2) with q \in \mathbb{Z}[x], so the kernel is (x - 2) and \mathbb{Z}[x]/(x - 2) \cong \mathbb{Z}. In the previous figure, two arrangements of dots give the same element exactly when they have the same value at x = 2.
The homomorphism \mathbb{R}[x, y] \to \mathbb{R}[t] with x \mapsto t and y \mapsto t^2 restricts a polynomial to the parabola y = x^2. It is surjective, and writing y = x^2 + (y - x^2) and expanding gives F(x, y) = F(x, x^2) + (y - x^2)G for some polynomial G, so its kernel is (y - x^2) and \mathbb{R}[x, y]/(y - x^2) \cong \mathbb{R}[t]. In the figure, the level curves of F = x + a(y - x^2) change as a changes, but on the parabola F is equal to x, so the points where the level curves cross the parabola do not move.
The ideals of a quotient ring can be found in the original ring. Let \pi : A \to A/I be the quotient map. For an ideal J of A containing I, the image J/I = \pi(J) is an ideal of A/I, and for an ideal \bar{J} of A/I, the preimage \pi^{-1}(\bar{J}) is an ideal of A containing I. These two operations undo each other and preserve containment, so the ideals of A/I correspond one to one to the ideals of A that contain I [1,2].
Applying the homomorphism theorem to A/I \to A/J, a + I \mapsto a + J, whose kernel is J/I, gives (A/I)/(J/I) \cong A/J. For A = \mathbb{Z} and I = (12), the ideals containing (12) are the six ideals in the figure of Principal Ideals and Divisibility, so \mathbb{Z}/12 has exactly six ideals, and (4), for example, corresponds to \{0, 4, 8\}. Since every ideal of \mathbb{R}[x] is principal, the ideals of \mathbb{R}[x]/(f) correspond in the same way to the monic divisors of f.
In the figure, f = x^3 - 3x + c has eight monic divisors while its three roots are real and distinct, six when two of the roots meet at c = 2, and four when those two become a complex pair, which gives one real factor x^2 + px + q. If f is not a product of two polynomials of lower degree, the quotient has only the ideals 0 and itself. Every a \neq 0 then generates the whole ring and is a unit, so the quotient is a field, like \mathbb{R}[x]/(x^2 + 1) \cong \mathbb{C}. Which ideals give fields is the subject of the next article.