A ring lets us add, subtract, and multiply, but quotients are more demanding than subrings. If we try to declare every element of a subset $I \subset R$ equal to zero, addition only asks that $I$ be an additive subgroup. Multiplication asks for a stronger condition: anything in $I$ must stay in $I$ after multiplication by arbitrary elements of the ambient ring. Ideals are the subsets that make this collapse compatible with the whole ring structure.

[example: Why Even Integers Work] In $\mathbb{Z}$, write $a\equiv b \pmod{2\mathbb{Z}}$ to mean $a-b\in 2\mathbb{Z}$. Suppose $a-b\in 2\mathbb{Z}$ and $c-d\in 2\mathbb{Z}$. Then there are integers $m,n\in\mathbb{Z}$ such that \begin{align*} a-b=2m \end{align*} and \begin{align*} c-d=2n. \end{align*} For addition, \begin{align*} (a+c)-(b+d)=(a-b)+(c-d)=2m+2n=2(m+n), \end{align*} so $(a+c)-(b+d)\in 2\mathbb{Z}$. For multiplication, first rewrite the difference as \begin{align*} ac-bd=ac-ad+ad-bd=a(c-d)+d(a-b). \end{align*} Substituting $c-d=2n$ and $a-b=2m$ gives \begin{align*} ac-bd=a(2n)+d(2m)=2(an)+2(dm)=2(an+dm), \end{align*} so $ac-bd\in 2\mathbb{Z}$. The important point is that $a$ and $d$ need not be even; $2\mathbb{Z}$ stays closed under multiplication by arbitrary integers, which is exactly the absorption property isolated by ideals. [/example]

The example is familiar, but it exposes the main point. A modulus is not just a collection of elements; it is a collection stable under every multiplication forced by the ring. Ideals are the ring-theoretic analogue of normal subgroups: they are exactly the pieces that can be collapsed in a quotient.

Throughout this page, a ring means an associative ring with multiplicative identity $1_R$, and a ring homomorphism preserves the multiplicative identity. A commutative ring is always understood in this unital sense. When noncommutative or [Lie algebra](/page/Lie%20Algebra) variants appear later, the relevant hypotheses are stated explicitly.

## Definition

### Absorption by the Ambient Ring

The quotient problem gives the primary definition. We want $a \equiv b \pmod I$ to imply $ra \equiv rb \pmod I$ and $ar \equiv br \pmod I$ for every $r \in R$. The next definition records exactly the additive and multiplicative closure needed for that congruence relation to respect ring operations.

[definition: Ideal] Let $R$ be a ring. A subset $I \subset R$ is an ideal of $R$, written $I \trianglelefteq R$, if the following conditions hold: \begin{align*} I \text{ is an additive subgroup of } (R,+,0_R). \end{align*} \begin{align*} \text{for all } r \in R \text{ and } a \in I, \quad ra \in I \text{ and } ar \in I. \end{align*} [/definition]

The first condition includes $0_R \in I$, additive inverses, and closure under addition, so the empty subset is excluded. The second condition is the absorption condition: $I$ absorbs multiplication from the entire ring, not only from its own elements. In noncommutative ring theory this is also called a two-sided ideal. In a commutative ring the two products $ra$ and $ar$ give the same requirement, so the unqualified word ideal has the same meaning.

Noncommutative rings force us to separate the two sides of multiplication. A subset may be stable under multiplication from the left but not from the right, and such subsets naturally appear as submodules of the regular module. The next definition keeps only the left action because that is the structure needed for left-module arguments.

[definition: Left Ideal] Let $R$ be a ring. A subset $I \subset R$ is a left ideal of $R$ if $I$ is an additive subgroup of $(R,+,0_R)$ and, for all $r \in R$ and $a \in I$, one has $ra \in I$. [/definition]

A left ideal can be used when the ring acts on itself from the left. It need not support a quotient ring because right multiplication may fail to be compatible with the proposed congruence. To track the other regular action, we need the mirror-image definition.

[definition: Right Ideal] Let $R$ be a ring. A subset $I \subset R$ is a right ideal of $R$ if $I$ is an additive subgroup of $(R,+,0_R)$ and, for all $r \in R$ and $a \in I$, one has $ar \in I$. [/definition]

An ideal in the primary sense is both a left ideal and a right ideal. The distinction disappears in commutative algebra, but it becomes essential in matrix rings and operator algebras. Since quotient rings require a subset that is stable from both sides, the rest of the ring-theoretic discussion focuses on ideals in the primary two-sided sense.

The zero subset and the whole ring always satisfy the axioms. The whole ring gives a degenerate quotient, so many structural questions ask for ideals that do not already contain $1_R$. The next term isolates the ideals that impose a genuine restriction on the ring.

[definition: Proper Ideal] Let $R$ be a ring. A proper ideal of $R$ is an ideal $I \trianglelefteq R$ such that $I \subsetneq R$. [/definition]

Properness is equivalent to $1_R \notin I$ when $R$ has a multiplicative identity. It is the condition that prevents $R/I$ from collapsing to the zero ring. Prime and maximal ideals are both proper by definition.

[example: Ideals in $\mathbb{Z}$] Fix $n\in\mathbb{N}$. The set $n\mathbb{Z}=\{nk:k\in\mathbb{Z}\}$ contains $0=n\cdot 0$. If $a,b\in n\mathbb{Z}$, then $a=nm$ and $b=n\ell$ for some $m,\ell\in\mathbb{Z}$, so \begin{align*} a-b=nm-n\ell=n(m-\ell). \end{align*} Since $m-\ell\in\mathbb{Z}$, this shows $a-b\in n\mathbb{Z}$. If $r\in\mathbb{Z}$ and $a=nm\in n\mathbb{Z}$, then \begin{align*} ra=r(nm)=n(rm). \end{align*} Since $rm\in\mathbb{Z}$, this shows $ra\in n\mathbb{Z}$. Thus $n\mathbb{Z}$ is an ideal of $\mathbb{Z}$. Conversely, let $I\trianglelefteq\mathbb{Z}$. If $I=\{0\}$, then $I=0\mathbb{Z}$. Otherwise $I$ contains a nonzero integer $t$. If $t<0$, then $-t\in I$ because $I$ is an additive subgroup, so $I$ contains a positive integer. Let $n$ be the least positive integer in $I$. Since $n\in I$ and $I$ absorbs multiplication by every integer, every element $qn$ with $q\in\mathbb{Z}$ lies in $I$, so $n\mathbb{Z}\subset I$. For the reverse inclusion, take $x\in I$. By the [division algorithm](/theorems/725), write \begin{align*} x=qn+s \end{align*} with $q\in\mathbb{Z}$ and $0\leq s<n$. Since $x\in I$ and $qn\in I$, the additive subgroup property gives \begin{align*} s=x-qn\in I. \end{align*} If $s>0$, then $s$ is a positive element of $I$ smaller than $n$, contradicting the choice of $n$. Hence $s=0$, so $x=qn\in n\mathbb{Z}$. Therefore $I\subset n\mathbb{Z}$, and $I=n\mathbb{Z}$. The generator is unique in $\mathbb{N}\cup\{0\}$: if $n\mathbb{Z}=m\mathbb{Z}$ with $n,m\geq 0$, then either both ideals are $\{0\}$, giving $n=m=0$, or both contain positive integers and their least positive elements must be equal. Thus every ideal of $\mathbb{Z}$ is exactly one of the ideals $n\mathbb{Z}$ with $n\in\mathbb{N}\cup\{0\}$. [/example]

This example explains why ideals generalise divisibility. Passing from $n$ to $n\mathbb{Z}$ replaces an integer by the set of all its multiples. Ring theory then asks which rings behave like $\mathbb{Z}$ and which rings have more complicated ideals.

### Ideals Generated by Elements