A ring remembers addition, multiplication, and the way the two operations distribute across each other. Sometimes we want to impose an equation inside a ring and keep doing ring arithmetic afterward. For instance, passing from $\mathbb{Z}$ to arithmetic modulo $n$ means deciding that $n$ should behave like $0$; passing from a [polynomial ring](/page/Polynomial%20Ring) $k[x]$ over a field $k$ to a ring where $x^2=0$ means deciding that the polynomial $x^2$ should vanish. The central question is: when can we collapse some elements of a ring to zero without destroying multiplication?

The answer is more restrictive than the additive picture suggests. Additive quotients only need subgroups, but multiplication must be compatible with representatives. If $r$ and $r'$ differ by something we have declared negligible, then multiplying by any element of the ring should not make that difference visible again. This is why ideals, not arbitrary additive subgroups, are the objects by which rings may be quotiented.

[example: Why an Additive Subgroup Is Not Enough] Let $R=\mathbb{Z}$ and let $H=2\mathbb{Z}$. We first check that multiplication of additive cosets is compatible with changing representatives in this case. Suppose \begin{align*} a+H=a'+H \end{align*} and \begin{align*} b+H=b'+H. \end{align*} Then $a-a'\in 2\mathbb{Z}$ and $b-b'\in 2\mathbb{Z}$, so there are integers $m,n\in\mathbb{Z}$ with \begin{align*} a-a'=2m \quad \text{and} \quad b-b'=2n. \end{align*} The difference between the two possible products is \begin{align*} ab-a'b'=ab-a'b+a'b-a'b'. \end{align*} Factoring the first two terms by $b$ and the last two terms by $a'$ gives \begin{align*} ab-a'b'=(a-a')b+a'(b-b'). \end{align*} Substituting $a-a'=2m$ and $b-b'=2n$ gives \begin{align*} ab-a'b'=(2m)b+a'(2n). \end{align*} Using associativity and distributivity in $\mathbb{Z}$, \begin{align*} (2m)b+a'(2n)=2(mb)+2(a'n)=2(mb+a'n). \end{align*} Since $mb+a'n\in\mathbb{Z}$, this shows $ab-a'b'\in 2\mathbb{Z}$. Thus $ab+2\mathbb{Z}=a'b'+2\mathbb{Z}$, so multiplication of cosets is well defined for $\mathbb{Z}/2\mathbb{Z}$. Now take $R=\mathbb{Z}[x]$ and \begin{align*} H=\{nx:n\in\mathbb{Z}\}. \end{align*} This set is an additive subgroup of $\mathbb{Z}[x]$: it contains $0=0x$, and if $mx,nx\in H$, then \begin{align*} mx+nx=(m+n)x\in H \end{align*} because $m+n\in\mathbb{Z}$. Also, \begin{align*} -(mx)=(-m)x\in H \end{align*} because $-m\in\mathbb{Z}$. The representatives $0$ and $x$ determine the same additive coset modulo $H$, since \begin{align*} x-0=1x\in H. \end{align*} If multiplication of additive cosets by $H$ were well defined, replacing the representative $0$ by the equivalent representative $x$ would not change the product with $x+H$. Using $0$ gives \begin{align*} (0+H)(x+H)=0\cdot x+H=0+H=H. \end{align*} Using $x$ instead gives \begin{align*} (x+H)(x+H)=x\cdot x+H=x^2+H. \end{align*} These two cosets are not equal. Equality $x^2+H=H$ would mean $x^2-0\in H$, so there would be some $n\in\mathbb{Z}$ such that \begin{align*} x^2=nx. \end{align*} But the coefficient of $x^2$ in $x^2$ is $1$, while the coefficient of $x^2$ in $nx$ is $0$ for every integer $n$. Hence no such $n$ exists, and \begin{align*} x^2\notin H. \end{align*} Therefore $(0+H)(x+H)=H$ but $(x+H)(x+H)=x^2+H\ne H$, even though $0+H=x+H$. The product depends on the chosen representative, so this additive quotient does not inherit a well-defined ring multiplication. [/example]

This failure is the whole story in miniature. Quotient rings are the algebraic device that lets us impose equations while preserving the operations of a [ring](/page/Ring). They are used to build modular arithmetic, construct fields from polynomial rings, encode nilpotent infinitesimals, describe affine algebraic sets, and formulate the universal principle behind kernels of ring homomorphisms.

## Definition

### The Quotient Construction

The quotient ring is the object the opening question asks for: a ring in which all elements of a chosen collapsible subset have become zero, while addition and multiplication still make sense. The construction uses additive cosets as its underlying elements, but its meaning is multiplicative as well: two representatives are treated as the same precisely when their difference belongs to the ideal being collapsed.

[definition: Quotient Ring] Let $R$ be a ring and let $I$ be a two-sided ideal of $R$, written $I \trianglelefteq R$. The quotient ring $R/I$ is the set of additive cosets \begin{align*} R/I = \{r+I : r \in R\} \end{align*} with operations \begin{align*} (r_1+I)+(r_2+I) = (r_1+r_2)+I \end{align*} and \begin{align*} (r_1+I)(r_2+I) = r_1r_2+I. \end{align*} [/definition]

In this ring, the zero element is $0_R+I=I$, and the multiplicative identity is $1_R+I$. The notation $R/I$ therefore records both the original ring $R$ and the collection $I$ of elements that have been collapsed to zero.

### Ideals as Collapsible Subsets

The definition depends on the hypothesis $I \trianglelefteq R$, so we now isolate the absorption condition that makes the construction legitimate. If an element is to become zero, then multiplying it by any element of the ambient ring should still produce something that becomes zero.

[definition: Two-Sided Ideal] Let $R$ be a ring. A two-sided ideal of $R$ is an additive subgroup $I \subset R$ such that for every $r \in R$ and every $a \in I$, \begin{align*} ra \in I \quad \text{and} \quad ar \in I. \end{align*} We write $I \trianglelefteq R$. [/definition]

### Congruence Notation

The notation $r+I$ emphasizes the additive origin of the construction. For computations, however, constantly mentioning cosets can obscure the basic question: when do two representatives give the same element of the quotient? This motivates a congruence notation that records equality after quotienting.

[definition: Congruence Modulo an Ideal] Let $R$ be a ring and let $I \trianglelefteq R$. For $a,b \in R$, we say that $a$ is congruent to $b$ modulo $I$, written $a \equiv b \pmod I$, if \begin{align*} a-b \in I. \end{align*} [/definition]

Thus $a$ and $b$ determine the same element of $R/I$ exactly when $a \equiv b \pmod I$. The quotient ring is the ring obtained by replacing equality in $R$ with equality modulo $I$.

[example: Modular Arithmetic as a Quotient Ring] Let $n\in\mathbb{N}$ with $n\ge 1$ and let $R=\mathbb{Z}$. The subset $n\mathbb{Z}=\{nk:k\in\mathbb{Z}\}$ is an ideal of $\mathbb{Z}$. It is an additive subgroup because $0=n\cdot 0\in n\mathbb{Z}$, and if $na,nb\in n\mathbb{Z}$, then \begin{align*} na-nb=n(a-b)\in n\mathbb{Z}. \end{align*} If $r\in\mathbb{Z}$ and $na\in n\mathbb{Z}$, then \begin{align*} r(na)=(rn)a=n(ra)\in n\mathbb{Z}. \end{align*} Also, \begin{align*} (na)r=n(ar)\in n\mathbb{Z}, \end{align*} so $n\mathbb{Z}$ absorbs multiplication from both sides. Thus $\mathbb{Z}/n\mathbb{Z}$ is a quotient ring. Every integer $m$ can be written as \begin{align*} m=qn+r \end{align*} with $q,r\in\mathbb{Z}$ and $0\le r<n$. Subtracting $r$ from both sides gives \begin{align*} m-r=qn=nq\in n\mathbb{Z}. \end{align*} Therefore $m$ and $r$ determine the same coset: \begin{align*} m+n\mathbb{Z}=r+n\mathbb{Z}. \end{align*} So every class in $\mathbb{Z}/n\mathbb{Z}$ is represented by one of \begin{align*} 0+n\mathbb{Z},1+n\mathbb{Z},\ldots,(n-1)+n\mathbb{Z}. \end{align*} These representatives are distinct: if $0\le r,s<n$ and $r+n\mathbb{Z}=s+n\mathbb{Z}$, then $r-s\in n\mathbb{Z}$, so $r-s=nk$ for some $k\in\mathbb{Z}$. Since $0\le r,s<n$, we have \begin{align*} -(n-1)\le r-s\le n-1. \end{align*} The only multiple of $n$ in this interval is $0$, so $r-s=0$ and hence $r=s$. For $n=6$, the classes $2+6\mathbb{Z}$ and $3+6\mathbb{Z}$ are both nonzero. If \begin{align*} 2+6\mathbb{Z}=0+6\mathbb{Z}, \end{align*} then $2-0\in 6\mathbb{Z}$, so $2=6k$ for some $k\in\mathbb{Z}$; no such integer exists. Similarly, $3=6k$ has no integer solution, so \begin{align*} 3+6\mathbb{Z}\ne 0+6\mathbb{Z}. \end{align*} Their product is \begin{align*} (2+6\mathbb{Z})(3+6\mathbb{Z})=(2\cdot 3)+6\mathbb{Z}. \end{align*} Since $2\cdot 3=6$, this is \begin{align*} 6+6\mathbb{Z}. \end{align*} Because $6-0=6=6\cdot 1\in 6\mathbb{Z}$, we have \begin{align*} 6+6\mathbb{Z}=0+6\mathbb{Z}. \end{align*} Thus two nonzero classes in $\mathbb{Z}/6\mathbb{Z}$ multiply to zero. The quotient $\mathbb{Z}/6\mathbb{Z}$ has zero divisors even though $\mathbb{Z}$ itself does not, so quotienting can introduce new algebraic behaviour. [/example]

### The Projection Map

After forming a quotient, we need a canonical way to pass from the original ring to the collapsed one. This is the map that performs reduction modulo the ideal and lets quotient rings interact with homomorphisms.

[definition: Canonical Projection] Let $R$ be a ring and let $I \trianglelefteq R$. The canonical projection is the ring homomorphism \begin{align*} \pi: R \to R/I, \qquad r \mapsto r+I. \end{align*} [/definition]