History[edit]

Ernst Kummer invented the concept of ideal numbers to serve as the "missing" factors in number rings in which unique factorization fails; here the word "ideal" is in the sense of existing in imagination only, in analogy with "ideal" objects in geometry such as points at infinity.^[1] In 1876, Richard Dedekind replaced Kummer's undefined concept by concrete sets of numbers, sets that he called ideals, in the third edition of Dirichlet's book Vorlesungen über Zahlentheorie, to which Dedekind had added many supplements.^[1]^[2]^[3] Later the notion was extended beyond number rings to the setting of polynomial rings and other commutative rings by David Hilbert and especially Emmy Noether.

In a ring R, the set R itself forms a two-sided ideal of R called the unit ideal. It is often also denoted by $(1)$ since it is precisely the two-sided ideal generated (see below) by the unity $1_{R}$ . Also, the set $\{0_{R}\}$ consisting of only the additive identity 0_R forms a two-sided ideal called the zero ideal and is denoted by $(0)$ . Every (left, right or two-sided) ideal contains the zero ideal and is contained in the unit ideal.^[4]

[note 1]

An (left, right or two-sided) ideal that is not the unit ideal is called a proper ideal (as it is a ).^[5] Note: a left ideal ${\mathfrak {a}}$ is proper if and only if it does not contain a unit element, since if $u\in {\mathfrak {a}}$ is a unit element, then $r=(ru^{-1})u\in {\mathfrak {a}}$ for every $r\in R$ . Typically there are plenty of proper ideals. In fact, if R is a skew-field, then $(0),(1)$ are its only ideals and conversely: that is, a nonzero ring R is a skew-field if $(0),(1)$ are the only left (or right) ideals. (Proof: if $x$ is a nonzero element, then the principal left ideal $Rx$ (see below) is nonzero and thus $Rx=(1)$ ; i.e., $yx=1$ for some nonzero $y$ . Likewise, $zy=1$ for some nonzero $z$ . Then $z=z(yx)=(zy)x=x$ .)

proper subset

The even form an ideal in the ring $\mathbb {Z}$ of all integers, since the sum of any two even integers is even, and the product of any integer with an even integer is also even; this ideal is usually denoted by $2\mathbb {Z}$ . More generally, the set of all integers divisible by a fixed integer $n$ is an ideal denoted $n\mathbb {Z}$ . In fact, every non-zero ideal of the ring $\mathbb {Z}$ is generated by its smallest positive element, as a consequence of Euclidean division, so $\mathbb {Z}$ is a principal ideal domain.^[4]

integers

The set of all with real coefficients that are divisible by the polynomial $x^{2}+1$ is an ideal in the ring of all real-coefficient polynomials $\mathbb {R} [x]$ .

polynomials

Take a ring $R$ and positive integer $n$ . For each $1\leq i\leq n$ , the set of all $n\times n$ with entries in $R$ whose $i$ -th row is zero is a right ideal in the ring $M_{n}(R)$ of all $n\times n$ matrices with entries in $R$ . It is not a left ideal. Similarly, for each $1\leq j\leq n$ , the set of all $n\times n$ matrices whose $j$ -th column is zero is a left ideal but not a right ideal.

matrices

The ring $C(\mathbb {R} )$ of all $f$ from $\mathbb {R}$ to $\mathbb {R}$ under pointwise multiplication contains the ideal of all continuous functions $f$ such that $f(1)=0$ .^[6] Another ideal in $C(\mathbb {R} )$ is given by those functions that vanish for large enough arguments, i.e. those continuous functions $f$ for which there exists a number $L>0$ such that $f(x)=0$ whenever $\vert x\vert >L$ .

continuous functions

A ring is called a if it is nonzero and has no two-sided ideals other than $(0),(1)$ . Thus, a skew-field is simple and a simple commutative ring is a field. The matrix ring over a skew-field is a simple ring.

simple ring

If $f:R\to S$ is a , then the kernel $\ker(f)=f^{-1}(0_{S})$ is a two-sided ideal of $R$ .^[4] By definition, $f(1_{R})=1_{S}$ , and thus if $S$ is not the zero ring (so $1_{S}\neq 0_{S}$ ), then $\ker(f)$ is a proper ideal. More generally, for each left ideal I of S, the pre-image $f^{-1}(I)$ is a left ideal. If I is a left ideal of R, then $f(I)$ is a left ideal of the subring $f(R)$ of S: unless f is surjective, $f(I)$ need not be an ideal of S; see also #Extension and contraction of an ideal below.

ring homomorphism

Ideal correspondence: Given a surjective ring homomorphism $f:R\to S$ , there is a bijective order-preserving correspondence between the left (resp. right, two-sided) ideals of $R$ containing the kernel of $f$ and the left (resp. right, two-sided) ideals of $S$ : the correspondence is given by $I\mapsto f(I)$ and the pre-image $J\mapsto f^{-1}(J)$ . Moreover, for commutative rings, this bijective correspondence restricts to prime ideals, maximal ideals, and radical ideals (see the section for the definitions of these ideals).

Types of ideals

(For those who know modules) If M is a left R-module and $S\subset M$ a subset, then the $\operatorname {Ann} _{R}(S)=\{r\in R\mid rs=0,s\in S\}$ of S is a left ideal. Given ideals ${\mathfrak {a}},{\mathfrak {b}}$ of a commutative ring R, the R-annihilator of $({\mathfrak {b}}+{\mathfrak {a}})/{\mathfrak {a}}$ is an ideal of R called the ideal quotient of ${\mathfrak {a}}$ by ${\mathfrak {b}}$ and is denoted by $({\mathfrak {a}}:{\mathfrak {b}})$ ; it is an instance of idealizer in commutative algebra.

annihilator

Let ${\mathfrak {a}}_{i},i\in S$ be an of left ideals in a ring R; i.e., $S$ is a totally ordered set and ${\mathfrak {a}}_{i}\subset {\mathfrak {a}}_{j}$ for each $i<j$ . Then the union $\textstyle \bigcup _{i\in S}{\mathfrak {a}}_{i}$ is a left ideal of R. (Note: this fact remains true even if R is without the unity 1.)

ascending chain

The above fact together with proves the following: if $E\subset R$ is a possibly empty subset and ${\mathfrak {a}}_{0}\subset R$ is a left ideal that is disjoint from E, then there is an ideal that is maximal among the ideals containing ${\mathfrak {a}}_{0}$ and disjoint from E. (Again this is still valid if the ring R lacks the unity 1.) When $R\neq 0$ , taking ${\mathfrak {a}}_{0}=(0)$ and $E=\{1\}$ , in particular, there exists a left ideal that is maximal among proper left ideals (often simply called a maximal left ideal); see Krull's theorem for more.

Zorn's lemma

An arbitrary union of ideals need not be an ideal, but the following is still true: given a possibly empty subset X of R, there is the smallest left ideal containing X, called the left ideal generated by X and is denoted by $RX$ . Such an ideal exists since it is the intersection of all left ideals containing X. Equivalently, $RX$ is the set of all the of elements of X over R:
$RX=\{r_{1}x_{1}+\dots +r_{n}x_{n}\mid n\in \mathbb {N} ,r_{i}\in R,x_{i}\in X\}.$

(finite) left R-linear combinations

(For the sake of brevity, some results are stated only for left ideals but are usually also true for right ideals with appropriate notation changes.)

: A proper ideal $I$ is called a maximal ideal if there exists no other proper ideal J with $I$ a proper subset of J. The factor ring of a maximal ideal is a simple ring in general and is a field for commutative rings.^[7]

Maximal ideal

: A nonzero ideal is called minimal if it contains no other nonzero ideal.

Minimal ideal

: A proper ideal $I$ is called a prime ideal if for any $a$ and $b$ in $R$ , if $ab$ is in $I$ , then at least one of $a$ and $b$ is in $I$ . The factor ring of a prime ideal is a prime ring in general and is an integral domain for commutative rings.^[8]

Prime ideal

or semiprime ideal: A proper ideal $I$ is called radical or semiprime if for any a in R, if aⁿ is in $I$ for some n, then a is in $I$ . The factor ring of a radical ideal is a semiprime ring for general rings, and is a reduced ring for commutative rings.

Radical ideal

: An ideal $I$ is called a primary ideal if for all a and b in R, if ab is in $I$ , then at least one of a and bⁿ is in $I$ for some natural number n. Every prime ideal is primary, but not conversely. A semiprime primary ideal is prime.

Primary ideal

: An ideal generated by one element.^[9]

Principal ideal

Finitely generated ideal: This type of ideal is as a module.

finitely generated

: A left primitive ideal is the annihilator of a simple left module.

Primitive ideal

: An ideal is said to be irreducible if it cannot be written as an intersection of ideals that properly contain it.

Irreducible ideal

Comaximal ideals: Two ideals $I$ , $J$ are said to be comaximal if $x+y=1$ for some $x\in I$ and $y\in J$ .

: This term has multiple uses. See the article for a list.

Regular ideal

: An ideal is a nil ideal if each of its elements is nilpotent.

Nil ideal

: Some power of it is zero.

Nilpotent ideal

: an ideal generated by a system of parameters.

Parameter ideal

: A proper ideal $I$ in a Noetherian ring $R$ is called a perfect ideal if its grade equals the projective dimension of the associated quotient ring,^[10] ${\textrm {grade}}(I)={\textrm {proj}}\dim(R/I)$ . A perfect ideal is unmixed.

Perfect ideal

: A proper ideal $I$ in a Noetherian ring $R$ is called an unmixed ideal (in height) if the height of I is equal to the height of every associated prime P of R/I. (This is stronger than saying that R/I is equidimensional. See also equidimensional ring.

Unmixed ideal

To simplify the description all rings are assumed to be commutative. The non-commutative case is discussed in detail in the respective articles.

Ideals are important because they appear as kernels of ring homomorphisms and allow one to define factor rings. Different types of ideals are studied because they can be used to construct different types of factor rings.

Two other important terms using "ideal" are not always ideals of their ring. See their respective articles for details:

${\mathfrak {a}}({\mathfrak {b}}+{\mathfrak {c}})={\mathfrak {a}}{\mathfrak {b}}+{\mathfrak {a}}{\mathfrak {c}}$ ,

$({\mathfrak {a}}+{\mathfrak {b}}){\mathfrak {c}}={\mathfrak {a}}{\mathfrak {c}}+{\mathfrak {b}}{\mathfrak {c}}$ .

The sum and product of ideals are defined as follows. For ${\mathfrak {a}}$ and ${\mathfrak {b}}$ , left (resp. right) ideals of a ring R, their sum is

which is a left (resp. right) ideal, and, if ${\mathfrak {a}},{\mathfrak {b}}$ are two-sided,

i.e. the product is the ideal generated by all products of the form ab with a in ${\mathfrak {a}}$ and b in ${\mathfrak {b}}$ .

Note ${\mathfrak {a}}+{\mathfrak {b}}$ is the smallest left (resp. right) ideal containing both ${\mathfrak {a}}$ and ${\mathfrak {b}}$ (or the union ${\mathfrak {a}}\cup {\mathfrak {b}}$ ), while the product ${\mathfrak {a}}{\mathfrak {b}}$ is contained in the intersection of ${\mathfrak {a}}$ and ${\mathfrak {b}}$ .

The distributive law holds for two-sided ideals ${\mathfrak {a}},{\mathfrak {b}},{\mathfrak {c}}$ ,

If a product is replaced by an intersection, a partial distributive law holds:

where the equality holds if ${\mathfrak {a}}$ contains ${\mathfrak {b}}$ or ${\mathfrak {c}}$ .

Remark: The sum and the intersection of ideals is again an ideal; with these two operations as join and meet, the set of all ideals of a given ring forms a complete modular lattice. The lattice is not, in general, a distributive lattice. The three operations of intersection, sum (or join), and product make the set of ideals of a commutative ring into a quantale.

If ${\mathfrak {a}},{\mathfrak {b}}$ are ideals of a commutative ring R, then ${\mathfrak {a}}\cap {\mathfrak {b}}={\mathfrak {a}}{\mathfrak {b}}$ in the following two cases (at least)

(More generally, the difference between a product and an intersection of ideals is measured by the Tor functor: $\operatorname {Tor} _{1}^{R}(R/{\mathfrak {a}},R/{\mathfrak {b}})=({\mathfrak {a}}\cap {\mathfrak {b}})/{\mathfrak {a}}{\mathfrak {b}}$ .^[11])

An integral domain is called a Dedekind domain if for each pair of ideals ${\mathfrak {a}}\subset {\mathfrak {b}}$ , there is an ideal ${\mathfrak {c}}$ such that ${\mathfrak {\mathfrak {a}}}={\mathfrak {b}}{\mathfrak {c}}$ .^[12] It can then be shown that every nonzero ideal of a Dedekind domain can be uniquely written as a product of maximal ideals, a generalization of the fundamental theorem of arithmetic.

${\mathfrak {a}}+{\mathfrak {b}}=(z,w,x+z,y+w)=(x,y,z,w)$ and ${\mathfrak {a}}+{\mathfrak {c}}=(z,w,x)$

${\mathfrak {a}}{\mathfrak {b}}=(z(x+z),z(y+w),w(x+z),w(y+w))=(z^{2}+xz,zy+wz,wx+wz,wy+w^{2})$

${\mathfrak {a}}{\mathfrak {c}}=(xz+z^{2},zw,xw+zw,w^{2})$

${\mathfrak {a}}\cap {\mathfrak {b}}={\mathfrak {a}}{\mathfrak {b}}$ while ${\mathfrak {a}}\cap {\mathfrak {c}}=(w,xz+z^{2})\neq {\mathfrak {a}}{\mathfrak {c}}$

In $\mathbb {Z}$ we have

since $(n)\cap (m)$ is the set of integers that are divisible by both $n$ and $m$ .

Let $R=\mathbb {C} [x,y,z,w]$ and let ${\mathfrak {a}}=(z,w),{\mathfrak {b}}=(x+z,y+w),{\mathfrak {c}}=(x+z,w)$ . Then,

In the first computation, we see the general pattern for taking the sum of two finitely generated ideals, it is the ideal generated by the union of their generators. In the last three we observe that products and intersections agree whenever the two ideals intersect in the zero ideal. These computations can be checked using Macaulay2.^[13]^[14]^[15]

${\mathfrak {b}}$ is prime in B $\Rightarrow$ ${\mathfrak {b}}^{c}$ is prime in A.

${\mathfrak {a}}^{ec}\supseteq {\mathfrak {a}}$

${\mathfrak {b}}^{ce}\subseteq {\mathfrak {b}}$

Let A and B be two commutative rings, and let f : A → B be a ring homomorphism. If ${\mathfrak {a}}$ is an ideal in A, then $f({\mathfrak {a}})$ need not be an ideal in B (e.g. take f to be the inclusion of the ring of integers Z into the field of rationals Q). The extension ${\mathfrak {a}}^{e}$ of ${\mathfrak {a}}$ in B is defined to be the ideal in B generated by $f({\mathfrak {a}})$ . Explicitly,

If ${\mathfrak {b}}$ is an ideal of B, then $f^{-1}({\mathfrak {b}})$ is always an ideal of A, called the contraction ${\mathfrak {b}}^{c}$ of ${\mathfrak {b}}$ to A.

Assuming f : A → B is a ring homomorphism, ${\mathfrak {a}}$ is an ideal in A, ${\mathfrak {b}}$ is an ideal in B, then:

It is false, in general, that ${\mathfrak {a}}$ being prime (or maximal) in A implies that ${\mathfrak {a}}^{e}$ is prime (or maximal) in B. Many classic examples of this stem from algebraic number theory. For example, embedding $\mathbb {Z} \to \mathbb {Z} \left\lbrack i\right\rbrack$ . In $B=\mathbb {Z} \left\lbrack i\right\rbrack$ , the element 2 factors as $2=(1+i)(1-i)$ where (one can show) neither of $1+i,1-i$ are units in B. So $(2)^{e}$ is not prime in B (and therefore not maximal, as well). Indeed, $(1\pm i)^{2}=\pm 2i$ shows that $(1+i)=((1-i)-(1-i)^{2})$ , $(1-i)=((1+i)-(1+i)^{2})$ , and therefore $(2)^{e}=(1+i)^{2}$ .

On the other hand, if f is surjective and ${\mathfrak {a}}\supseteq \ker f$ then:

Remark: Let K be a field extension of L, and let B and A be the rings of integers of K and L, respectively. Then B is an integral extension of A, and we let f be the inclusion map from A to B. The behaviour of a prime ideal ${\mathfrak {a}}={\mathfrak {p}}$ of A under extension is one of the central problems of algebraic number theory.

The following is sometimes useful:^[16] a prime ideal ${\mathfrak {p}}$ is a contraction of a prime ideal if and only if ${\mathfrak {p}}={\mathfrak {p}}^{ec}$ . (Proof: Assuming the latter, note ${\mathfrak {p}}^{e}B_{\mathfrak {p}}=B_{\mathfrak {p}}\Rightarrow {\mathfrak {p}}^{e}$ intersects $A-{\mathfrak {p}}$ , a contradiction. Now, the prime ideals of $B_{\mathfrak {p}}$ correspond to those in B that are disjoint from $A-{\mathfrak {p}}$ . Hence, there is a prime ideal ${\mathfrak {q}}$ of B, disjoint from $A-{\mathfrak {p}}$ , such that ${\mathfrak {q}}B_{\mathfrak {p}}$ is a maximal ideal containing ${\mathfrak {p}}^{e}B_{\mathfrak {p}}$ . One then checks that ${\mathfrak {q}}$ lies over ${\mathfrak {p}}$ . The converse is obvious.)