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] The canonical projection sends exactly the elements of $I$ to zero. This is the first sign that quotient rings are not merely a construction: they are the natural codomains of homomorphisms once the kernel has been chosen. ## Well-Defined Multiplication ### Representative Independence The most delicate part of the definition is multiplication. Addition of additive cosets works for any additive subgroup, but multiplication asks for compatibility with all possible representatives. If the class of $a$ equals the class of $a'$ and the class of $b$ equals the class of $b'$, then the class of $ab$ must equal the class of $a'b'$. [quotetheorem:8156] ### Failure Without Ideals This theorem is the point at which the ideal condition pays for itself. If $r_1-r_1' \in I$ and $r_2-r_2' \in I$, then the difference $r_1r_2-r_1'r_2'$ can be expanded into terms absorbed by $I$. The next example shows why a weaker additive condition cannot replace the ideal hypothesis. [example: Representative Dependence in a Non-Ideal Quotient] Let $R=\mathbb{Z}[x]$ and let \begin{align*} H=\{nx : n \in \mathbb{Z}\}. \end{align*} This set is an additive subgroup of $R$. First, $0=0x$, so $0\in H$. If $mx,nx\in H$, then closure under addition follows from \begin{align*} mx+nx=(m+n)x. \end{align*} Since $m+n\in\mathbb{Z}$, we have $(m+n)x\in H$. Closure under additive inverses follows from \begin{align*} -(mx)=(-m)x. \end{align*} Since $-m\in\mathbb{Z}$, this element also lies in $H$. In the additive quotient by $H$, the representatives $0$ and $x$ define the same coset, because \begin{align*} x-0=x=1x\in H. \end{align*} If multiplication of additive cosets by $H$ were well defined, replacing $0$ by the equivalent representative $x$ would not change the product with $x+H$. Using $0$ as the first representative gives \begin{align*} (0+H)(x+H)=0\cdot x+H. \end{align*} Since $0\cdot x=0$, this product is \begin{align*} 0+H=H. \end{align*} Using $x$ instead gives \begin{align*} (x+H)(x+H)=x\cdot x+H. \end{align*} Since $x\cdot x=x^2$ in $\mathbb{Z}[x]$, this product is \begin{align*} x^2+H. \end{align*} These two cosets are not equal. If $x^2+H=H$, then $x^2+H=0+H$, so equality of additive cosets would give \begin{align*} x^2-0\in H. \end{align*} By the definition of $H$, this means that there is some integer $n$ such that \begin{align*} x^2=nx. \end{align*} Writing both sides as polynomials in $\mathbb{Z}[x]$, the coefficient of $x^2$ in $x^2$ is $1$, while the coefficient of $x^2$ in $nx$ is $0$ for every $n\in\mathbb{Z}$. Equality of polynomials requires equality of corresponding coefficients, so no such integer $n$ exists. Hence \begin{align*} x^2\notin H. \end{align*} Therefore \begin{align*} (0+H)(x+H)=H \end{align*} but \begin{align*} (x+H)(x+H)=x^2+H\ne H. \end{align*} The product depends on the chosen representative: although $0+H=x+H$, replacing $0$ by $x$ changes the result of multiplication by $x+H$. The missing property is absorption by multiplication from $R$, since $x=1x\in H$ but \begin{align*} x\cdot x=x^2\notin H. \end{align*} [/example] The quotient construction also needs a practical equality test. Rather than compare cosets as sets, we want a criterion inside the original ring, because most calculations begin with representatives. [quotetheorem:8254] This criterion is the main computational rule for quotient rings. To prove that two quotient classes are equal, return to the original ring and check that their difference lies in the ideal. ## Kernels and Universal Properties Quotient rings become powerful when viewed through maps. A ring homomorphism often forgets information: two elements of the source may have the same image. To quotient by exactly the information forgotten by a homomorphism, we first name the elements that the homomorphism sends to zero. [definition: Kernel of a Ring Homomorphism] Let $\varphi: R \to S$ be a ring homomorphism. The kernel of $\varphi$ is \begin{align*} \ker\varphi = \{r \in R : \varphi(r)=0_S\}. \end{align*} [/definition] For kernels to lead to quotient rings, the set of elements killed by a homomorphism must be stable under the two operations that make quotient multiplication well defined. Addition and additive inverses are forced by preservation of sums, but multiplication creates the real obstruction: if $r$ maps to zero, then every product with $r$ must also map to zero. The following result checks exactly this absorption property, so a kernel is always a legitimate ideal to collapse. [quotetheorem:8157] Having recognized kernels as ideals, we need to know what the quotient by a kernel represents. The answer is that it recovers exactly the image of the homomorphism, with no duplicated representatives left over. [quotetheorem:851] The theorem explains why quotient rings appear whenever a ring maps onto another ring. If $\varphi$ is surjective, then $S \cong R/\ker\varphi$. A more flexible version is needed when the homomorphism kills a chosen ideal but perhaps kills more as well. [quotetheorem:8255] This universal property is often the best definition to remember. A quotient ring is the most economical ring receiving a map from $R$ in which every element of $I$ has become zero. [example: Evaluating a Polynomial and Taking a Quotient] Let $k$ be a field and define the evaluation homomorphism \begin{align*} \operatorname{ev}_0: k[x] \to k, \qquad f(x) \mapsto f(0). \end{align*} For a polynomial $f(x)=a_0+a_1x+\cdots+a_dx^d$, evaluation at $0$ gives \begin{align*} f(0)=a_0+a_1\cdot 0+a_2\cdot 0^2+\cdots+a_d\cdot 0^d. \end{align*} Every term $a_i\cdot 0^i$ with $i\ge 1$ is $0$, so \begin{align*} f(0)=a_0. \end{align*} Thus $f\in\ker(\operatorname{ev}_0)$ exactly when $a_0=0$. If $f\in\ker(\operatorname{ev}_0)$, then $a_0=0$, so \begin{align*} f(x)=a_1x+a_2x^2+\cdots+a_dx^d. \end{align*} Factoring one copy of $x$ from each term gives \begin{align*} f(x)=x(a_1+a_2x+\cdots+a_dx^{d-1}). \end{align*} Since $a_1+a_2x+\cdots+a_dx^{d-1}\in k[x]$, this shows $f\in (x)$. Conversely, if $f\in (x)$, then $f=xg$ for some $g\in k[x]$, and evaluating at $0$ gives \begin{align*} f(0)=(xg)(0)=0\cdot g(0)=0. \end{align*} Hence $f\in\ker(\operatorname{ev}_0)$. Therefore \begin{align*} \ker(\operatorname{ev}_0)=(x). \end{align*} By the *[First Isomorphism Theorem for Rings](/theorems/851)*, \begin{align*} k[x]/\ker(\operatorname{ev}_0)\cong \operatorname{im}(\operatorname{ev}_0). \end{align*} Since $\ker(\operatorname{ev}_0)=(x)$, this becomes \begin{align*} k[x]/(x)\cong \operatorname{im}(\operatorname{ev}_0). \end{align*} The image is all of $k$: for every $a\in k$, the constant polynomial $a\in k[x]$ satisfies \begin{align*} \operatorname{ev}_0(a)=a. \end{align*} Thus \begin{align*} \operatorname{im}(\operatorname{ev}_0)=k. \end{align*} Combining the two equalities gives \begin{align*} k[x]/(x)\cong k. \end{align*} In the quotient, $x+(x)=0+(x)$ because \begin{align*} x-0=x\in (x). \end{align*} For $f(x)=a_0+a_1x+\cdots+a_dx^d$, subtracting the constant term gives \begin{align*} f(x)-a_0=a_1x+a_2x^2+\cdots+a_dx^d. \end{align*} Factoring out $x$ gives \begin{align*} f(x)-a_0=x(a_1+a_2x+\cdots+a_dx^{d-1}). \end{align*} Therefore $f(x)-a_0\in (x)$, so \begin{align*} f(x)+(x)=a_0+(x). \end{align*} Quotienting by $(x)$ collapses every positive power of $x$ to zero and keeps exactly the constant term. [/example] ## Imposing Equations A quotient ring is often best understood as a ring where chosen equations have been declared true. In a polynomial ring, quotienting by an ideal generated by polynomials forces those polynomials to vanish. The simplest version imposes a single relation, so it is useful to name this special quotient. [definition: Principal Quotient] Let $R$ be a commutative ring and let $f \in R$. The principal quotient by $f$ is the quotient ring \begin{align*} R/(f), \end{align*} where $(f)=\{rf : r \in R\}$ is the principal ideal generated by $f$. [/definition] The principal quotient $R/(f)$ is the ring obtained from $R$ by imposing $f=0$. If $R=k[x]$, this means that every polynomial may be reduced using the relation $f(x)=0$. [example: Dual Numbers from a Quotient] Let $k$ be a field, and write $\varepsilon$ also for the class $\varepsilon+(\varepsilon^2)$ in \begin{align*} k[\varepsilon]/(\varepsilon^2). \end{align*} We first show that every class has a unique representative of the form $a+b\varepsilon$ with $a,b\in k$. If $f(\varepsilon)=a_0+a_1\varepsilon+\cdots+a_d\varepsilon^d$ and $d\le 1$, then $f$ already has this form. If $d\ge 2$, then \begin{align*} f(\varepsilon)-(a_0+a_1\varepsilon)=a_2\varepsilon^2+a_3\varepsilon^3+\cdots+a_d\varepsilon^d. \end{align*} Factoring out $\varepsilon^2$ gives \begin{align*} a_2\varepsilon^2+a_3\varepsilon^3+\cdots+a_d\varepsilon^d=\varepsilon^2(a_2+a_3\varepsilon+\cdots+a_d\varepsilon^{d-2}). \end{align*} Since $a_2+a_3\varepsilon+\cdots+a_d\varepsilon^{d-2}\in k[\varepsilon]$, this difference lies in $(\varepsilon^2)$, so \begin{align*} f(\varepsilon)+(\varepsilon^2)=(a_0+a_1\varepsilon)+(\varepsilon^2). \end{align*} For uniqueness, suppose \begin{align*} a+b\varepsilon+(\varepsilon^2)=c+d\varepsilon+(\varepsilon^2). \end{align*} Then the difference of the representatives lies in $(\varepsilon^2)$: \begin{align*} (a+b\varepsilon)-(c+d\varepsilon)=(a-c)+(b-d)\varepsilon\in(\varepsilon^2). \end{align*} Thus there is some $q(\varepsilon)\in k[\varepsilon]$ such that \begin{align*} (a-c)+(b-d)\varepsilon=\varepsilon^2q(\varepsilon). \end{align*} If $q(\varepsilon)\ne 0$, then $\varepsilon^2q(\varepsilon)$ has degree at least $2$, while $(a-c)+(b-d)\varepsilon$ has degree at most $1$ unless it is the zero polynomial. This is impossible, so $q(\varepsilon)=0$. Therefore \begin{align*} (a-c)+(b-d)\varepsilon=0. \end{align*} Equality of polynomials over $k$ gives $a-c=0$ and $b-d=0$, so $a=c$ and $b=d$. Now multiply two representatives by expanding in $k[\varepsilon]$: \begin{align*} (a+b\varepsilon)(c+d\varepsilon)=ac+ad\varepsilon+bc\varepsilon+bd\varepsilon^2. \end{align*} Combining the two terms of degree $1$ gives \begin{align*} ac+ad\varepsilon+bc\varepsilon+bd\varepsilon^2=ac+(ad+bc)\varepsilon+bd\varepsilon^2. \end{align*} The difference between this product and $ac+(ad+bc)\varepsilon$ is \begin{align*} (ac+(ad+bc)\varepsilon+bd\varepsilon^2)-(ac+(ad+bc)\varepsilon)=bd\varepsilon^2. \end{align*} Since $bd\varepsilon^2=\varepsilon^2(bd)\in(\varepsilon^2)$, the two representatives define the same quotient class. Hence in $k[\varepsilon]/(\varepsilon^2)$, \begin{align*} (a+b\varepsilon)(c+d\varepsilon)=ac+(ad+bc)\varepsilon. \end{align*} The class of $\varepsilon$ is nonzero. Indeed, if $\varepsilon+(\varepsilon^2)=0+(\varepsilon^2)$, then $\varepsilon\in(\varepsilon^2)$, so $\varepsilon=\varepsilon^2q(\varepsilon)$ for some $q(\varepsilon)\in k[\varepsilon]$. If $q(\varepsilon)=0$, this says $\varepsilon=0$, which is false in $k[\varepsilon]$; if $q(\varepsilon)\ne 0$, the right side has degree at least $2$, while the left side has degree $1$. Thus $\varepsilon\notin(\varepsilon^2)$. But \begin{align*} \varepsilon^2+(\varepsilon^2)=0+(\varepsilon^2) \end{align*} because $\varepsilon^2-0=\varepsilon^2\in(\varepsilon^2)$. Therefore the quotient contains a nonzero element whose square is zero, which is the defining feature of the dual numbers over $k$. [/example] The previous example shows that quotienting can introduce nilpotents. Other quotients instead create fields by forcing an irreducible polynomial to have a root. [example: Constructing $\mathbb{C}$ from a Polynomial Quotient] Consider the quotient ring $\mathbb{R}[x]/(x^2+1)$, and let $i$ denote the class $x+(x^2+1)$. In the quotient, \begin{align*} i^2=x^2+(x^2+1). \end{align*} Since \begin{align*} x^2-(-1)=x^2+1\in (x^2+1), \end{align*} we have \begin{align*} x^2+(x^2+1)=-1+(x^2+1). \end{align*} Thus $i^2=-1$ in $\mathbb{R}[x]/(x^2+1)$. Every class has a representative of the form $a+bx$ with $a,b\in\mathbb{R}$. Indeed, if $f(x)\in\mathbb{R}[x]$, polynomial division by the monic polynomial $x^2+1$ gives \begin{align*} f(x)=q(x)(x^2+1)+r(x), \end{align*} where either $r(x)=0$ or $\deg r<2$. Hence $r(x)=a+bx$ for some $a,b\in\mathbb{R}$, and \begin{align*} f(x)-r(x)=q(x)(x^2+1)\in (x^2+1). \end{align*} Therefore \begin{align*} f(x)+(x^2+1)=(a+bx)+(x^2+1). \end{align*} Now multiply two representatives: \begin{align*} (a+bx)(c+dx)=ac+adx+bcx+bdx^2. \end{align*} Combining the degree-$1$ terms gives \begin{align*} ac+adx+bcx+bdx^2=ac+(ad+bc)x+bdx^2. \end{align*} Since $x^2\equiv -1 \pmod{(x^2+1)}$, we compute the difference \begin{align*} ac+(ad+bc)x+bdx^2-\bigl((ac-bd)+(ad+bc)x\bigr)=bd(x^2+1). \end{align*} Because $bd(x^2+1)\in (x^2+1)$, the two polynomials define the same quotient class. Replacing the class of $x$ by $i$, multiplication in the quotient is therefore \begin{align*} (a+bi)(c+di)=(ac-bd)+(ad+bc)i. \end{align*} This is exactly the usual multiplication rule for complex numbers, with the class $i=x+(x^2+1)$ playing the role of a square root of $-1$. Thus $\mathbb{R}[x]/(x^2+1)$ realizes $\mathbb{C}$ as a polynomial quotient. [/example] The field construction in this example raises the natural question of when a quotient of $k[x]$ has division by every nonzero class. Irreducibility is exactly the condition that prevents nonzero factor classes from multiplying to zero. [quotetheorem:8256] This theorem is one of the first places where quotient rings turn equations into new number systems. The class of $x$ becomes a root of $f$, and the quotient ring provides the smallest algebraic environment generated by that root over $k$. ## Ideals Seen Through Quotients Quotienting changes the ambient ring, but it does not erase the ideal structure above the collapsed ideal. If $I \trianglelefteq R$, then an ideal of the quotient ring $R/I$ corresponds to an ideal $J \trianglelefteq R$ with $I \subseteq J$: the ideal $J$ maps to $J/I$ in the quotient, and every ideal of $R/I$ arises this way by taking its inverse image under the quotient map $R \to R/I$. This correspondence is the structural reason quotient rings remember exactly the ideals lying above the one we collapse. It suggests a test for when the quotient has no nonzero proper ideals at all: there should be no intermediate ideals between the collapsed ideal and the whole ring. [definition: Maximal Ideal] Let $R$ be a commutative ring. An ideal $\mathfrak{m} \trianglelefteq R$ is maximal if $\mathfrak{m} \ne R$ and there is no ideal $J \trianglelefteq R$ satisfying \begin{align*} \mathfrak{m} \subsetneq J \subsetneq R. \end{align*} [/definition] Maximal ideals are designed to make the quotient as small as possible without becoming the zero ring. The familiar rings with no nonzero proper ideals are fields, so maximality should be detected by a field quotient. [quotetheorem:852] Field quotients are only one extreme. To study quotient rings that may not have inverses but still have no zero divisors, we need an ideal condition that watches products rather than all intermediate ideals. [definition: Prime Ideal] Let $R$ be a commutative ring. An ideal $\mathfrak{p} \trianglelefteq R$ is prime if $\mathfrak{p} \ne R$ and for all $a,b \in R$, \begin{align*} ab \in \mathfrak{p} \quad \implies \quad a \in \mathfrak{p} \text{ or } b \in \mathfrak{p}. \end{align*} [/definition] The reason for defining prime ideals this way is that products become zero in $R/\mathfrak{p}$ precisely when their products lie in $\mathfrak{p}$. This sets up the quotient test for being an integral domain. [quotetheorem:853] These two tests are among the most frequently used quotient-ring tools. They convert internal ideal conditions into familiar properties of rings: being a field or being an integral domain. [example: Prime and Maximal Ideals in $\mathbb{Z}$] Let $p$ be a positive prime integer. Since $p\ge 2$, the quotient $\mathbb{Z}/p\mathbb{Z}$ has distinct zero and one classes: if \begin{align*} 1+p\mathbb{Z}=0+p\mathbb{Z}, \end{align*} then $1-0\in p\mathbb{Z}$, so $1=pk$ for some $k\in\mathbb{Z}$, impossible because no multiple of an integer $p\ge 2$ equals $1$. We show that every nonzero class in $\mathbb{Z}/p\mathbb{Z}$ has a multiplicative inverse. Let $a+p\mathbb{Z}$ be nonzero. Nonzero means \begin{align*} a+p\mathbb{Z}\ne 0+p\mathbb{Z}, \end{align*} so $a-0\notin p\mathbb{Z}$, and hence there is no $k\in\mathbb{Z}$ such that $a=pk$. Thus $p\nmid a$. Since $p$ is prime, any positive common divisor of $a$ and $p$ is either $1$ or $p$; the option $p$ is excluded by $p\nmid a$, so \begin{align*} \gcd(a,p)=1. \end{align*} By *Bezout's Identity*, there are integers $u,v\in\mathbb{Z}$ such that \begin{align*} ua+vp=1. \end{align*} Subtracting $vp$ from both sides gives \begin{align*} ua=1-vp. \end{align*} Therefore \begin{align*} ua-1=-vp=p(-v)\in p\mathbb{Z}. \end{align*} Using multiplication in the quotient, \begin{align*} (u+p\mathbb{Z})(a+p\mathbb{Z})=ua+p\mathbb{Z}. \end{align*} Since $ua-1\in p\mathbb{Z}$, the classes $ua+p\mathbb{Z}$ and $1+p\mathbb{Z}$ are equal, so \begin{align*} (u+p\mathbb{Z})(a+p\mathbb{Z})=1+p\mathbb{Z}. \end{align*} Thus $u+p\mathbb{Z}$ is a multiplicative inverse of $a+p\mathbb{Z}$. Every nonzero class has an inverse, so $\mathbb{Z}/p\mathbb{Z}$ is a field. By *Maximal Ideals and Field Quotients*, $p\mathbb{Z}$ is maximal. The ideal $p\mathbb{Z}$ is also prime. Suppose $ab\in p\mathbb{Z}$. Then there is an integer $k$ such that \begin{align*} ab=pk, \end{align*} so $p\mid ab$. Since $p$ is prime, *Euclid's Lemma* gives $p\mid a$ or $p\mid b$. If $p\mid a$, then $a=pm$ for some $m\in\mathbb{Z}$, hence $a\in p\mathbb{Z}$. If $p\mid b$, then $b=pn$ for some $n\in\mathbb{Z}$, hence $b\in p\mathbb{Z}$. Therefore \begin{align*} ab\in p\mathbb{Z}\quad\implies\quad a\in p\mathbb{Z}\text{ or }b\in p\mathbb{Z}, \end{align*} which is exactly the condition that $p\mathbb{Z}$ is prime. By contrast, $6\mathbb{Z}$ is not prime. We have \begin{align*} 2\cdot 3=6=6\cdot 1\in 6\mathbb{Z}. \end{align*} However, $2\notin 6\mathbb{Z}$: if $2=6k$ for some $k\in\mathbb{Z}$, then dividing both sides by $2$ in $\mathbb{Z}$ gives \begin{align*} 1=3k, \end{align*} which is impossible because no multiple of $3$ equals $1$. Similarly, $3\notin 6\mathbb{Z}$: if $3=6k$ for some $k\in\mathbb{Z}$, then \begin{align*} 3=2(3k), \end{align*} so $3$ would be even, which is false. Thus $6\mathbb{Z}$ fails the defining implication for a prime ideal. In quotient language, the same failure appears as a zero divisor calculation. Multiplication of classes gives \begin{align*} (2+6\mathbb{Z})(3+6\mathbb{Z})=(2\cdot 3)+6\mathbb{Z}. \end{align*} Since $2\cdot 3=6$, this becomes \begin{align*} (2+6\mathbb{Z})(3+6\mathbb{Z})=6+6\mathbb{Z}. \end{align*} Also, \begin{align*} 6-0=6=6\cdot 1\in 6\mathbb{Z}, \end{align*} so \begin{align*} 6+6\mathbb{Z}=0+6\mathbb{Z}. \end{align*} The factors are nonzero classes. Indeed, $2+6\mathbb{Z}=0+6\mathbb{Z}$ would imply $2-0\in 6\mathbb{Z}$, contradicting $2\notin 6\mathbb{Z}$, and $3+6\mathbb{Z}=0+6\mathbb{Z}$ would imply $3-0\in 6\mathbb{Z}$, contradicting $3\notin 6\mathbb{Z}$. Hence $\mathbb{Z}/6\mathbb{Z}$ has two nonzero classes whose product is zero, reflecting exactly that $6\mathbb{Z}$ is not prime. [/example] ## Decomposition and the Chinese Remainder Principle Some quotient rings split into simpler pieces. This happens when the imposed congruence conditions are independent. For integers, working modulo $mn$ can be the same as working modulo $m$ and modulo $n$ separately when $m$ and $n$ are coprime. The ideal-theoretic version of coprimality is comaximality. [definition: Comaximal Ideals] Let $R$ be a commutative ring. Ideals $I,J \trianglelefteq R$ are comaximal if \begin{align*} I+J = R. \end{align*} [/definition] Comaximality says that the two quotient conditions do not overlap in a way that loses information. The natural question is whether the single quotient by $I \cap J$ has exactly the same data as the pair of quotients by $I$ and by $J$. [quotetheorem:8160] The theorem turns one quotient by an intersection into a product of quotients. This is not just a classification result; it is a computational method for solving simultaneous congruences. [example: Splitting $\mathbb{Z}/12\mathbb{Z}$] The ideals $3\mathbb{Z}$ and $4\mathbb{Z}$ are comaximal because \begin{align*} 3(-1)+4(1)=-3+4=1. \end{align*} Here $3(-1)\in 3\mathbb{Z}$ and $4(1)\in 4\mathbb{Z}$, so $1\in 3\mathbb{Z}+4\mathbb{Z}$. Since $3\mathbb{Z}+4\mathbb{Z}$ is an ideal of $\mathbb{Z}$ containing $1$, every integer $m$ belongs to it: \begin{align*} m=m\cdot 1\in 3\mathbb{Z}+4\mathbb{Z}. \end{align*} Thus $3\mathbb{Z}+4\mathbb{Z}=\mathbb{Z}$. We next verify that $3\mathbb{Z}\cap 4\mathbb{Z}=12\mathbb{Z}$. If $12k\in 12\mathbb{Z}$, then \begin{align*} 12k=3(4k)\in 3\mathbb{Z}. \end{align*} Also \begin{align*} 12k=4(3k)\in 4\mathbb{Z}. \end{align*} Therefore $12k\in 3\mathbb{Z}\cap 4\mathbb{Z}$, so $12\mathbb{Z}\subseteq 3\mathbb{Z}\cap 4\mathbb{Z}$. Conversely, let $t\in 3\mathbb{Z}\cap 4\mathbb{Z}$. Then $t=3a$ and $t=4b$ for some $a,b\in\mathbb{Z}$. Since $1=4-3$, multiplying by $t$ gives \begin{align*} t=t\cdot 1=t(4-3). \end{align*} Using distributivity in $\mathbb{Z}$, \begin{align*} t(4-3)=4t-3t. \end{align*} Substituting $t=3a$ into the first term gives \begin{align*} 4t=4(3a)=12a. \end{align*} Substituting $t=4b$ into the second term gives \begin{align*} 3t=3(4b)=12b. \end{align*} Hence \begin{align*} t=4t-3t=12a-12b. \end{align*} Factoring out $12$ gives \begin{align*} 12a-12b=12(a-b). \end{align*} Since $a-b\in\mathbb{Z}$, we have $t\in 12\mathbb{Z}$. Thus $3\mathbb{Z}\cap 4\mathbb{Z}\subseteq 12\mathbb{Z}$, and therefore \begin{align*} 3\mathbb{Z}\cap 4\mathbb{Z}=12\mathbb{Z}. \end{align*} By the *[Chinese Remainder Theorem](/theorems/734) for Rings*, the map \begin{align*} \mathbb{Z}/(3\mathbb{Z}\cap 4\mathbb{Z})\to \mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z} \end{align*} given by \begin{align*} m+(3\mathbb{Z}\cap 4\mathbb{Z})\mapsto (m+3\mathbb{Z},m+4\mathbb{Z}) \end{align*} is a ring isomorphism. Since $3\mathbb{Z}\cap 4\mathbb{Z}=12\mathbb{Z}$, this becomes \begin{align*} \mathbb{Z}/12\mathbb{Z}\cong \mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}. \end{align*} For the class $7+12\mathbb{Z}$, the isomorphism sends \begin{align*} 7+12\mathbb{Z}\mapsto (7+3\mathbb{Z},7+4\mathbb{Z}). \end{align*} Modulo $3$, we have \begin{align*} 7-1=6=3\cdot 2\in 3\mathbb{Z}. \end{align*} Therefore \begin{align*} 7+3\mathbb{Z}=1+3\mathbb{Z}. \end{align*} Modulo $4$, we have \begin{align*} 7-3=4=4\cdot 1\in 4\mathbb{Z}. \end{align*} Therefore \begin{align*} 7+4\mathbb{Z}=3+4\mathbb{Z}. \end{align*} So the image of $7+12\mathbb{Z}$ is \begin{align*} (1+3\mathbb{Z},3+4\mathbb{Z}). \end{align*} Thus reducing modulo $12$ is equivalent here to recording the two compatible reductions modulo $3$ and modulo $4$. [/example] Chinese remainders also explain why quotients by products of coprime polynomial factors split into products. This is the algebraic mechanism behind decomposing a polynomial condition into independent factors. ## Quotients in Geometry In commutative algebra, quotient rings are coordinate rings of algebraic sets with equations imposed. If $k[x_1,\ldots,x_n]$ is the ring of polynomial functions on affine $n$-space, quotienting by an ideal records the functions after declaring all polynomials in the ideal to vanish. To make that idea intrinsic to the set of common zeros, we quotient by all polynomials that vanish on the set. [definition: Coordinate Ring of an Affine Algebraic Set] Let $k$ be a field, let $S$ be a subset of $k[x_1,\ldots,x_n]$, and let \begin{align*} V(S) = \{a \in k^n : f(a)=0 \text{ for all } f \in S\}. \end{align*} The coordinate ring associated to $V(S)$ is \begin{align*} k[x_1,\ldots,x_n]/I(V(S)), \end{align*} where \begin{align*} I(V(S)) = \{f \in k[x_1,\ldots,x_n] : f(a)=0 \text{ for all } a \in V(S)\}. \end{align*} [/definition] The coordinate ring keeps polynomial functions only up to agreement on the algebraic set. Two polynomials define the same regular function on $V(S)$ exactly when their difference vanishes on every point of $V(S)$. [example: The Coordinate Ring of the Parabola] Let $k$ be an algebraically closed field and consider the affine parabola \begin{align*} V(y-x^2)=\{(a,b)\in k^2:b=a^2\}. \end{align*} Its coordinate ring is \begin{align*} k[x,y]/(y-x^2). \end{align*} We show that this quotient is isomorphic to $k[x]$ by substituting $y=x^2$. Define \begin{align*} \Phi:k[x,y]\to k[x],\quad \Phi(f(x,y))=f(x,x^2). \end{align*} Substitution preserves sums and products, since \begin{align*} \Phi(f+g)=(f+g)(x,x^2)=f(x,x^2)+g(x,x^2)=\Phi(f)+\Phi(g) \end{align*} and \begin{align*} \Phi(fg)=(fg)(x,x^2)=f(x,x^2)g(x,x^2)=\Phi(f)\Phi(g). \end{align*} For the generator $y-x^2$, \begin{align*} \Phi(y-x^2)=x^2-x^2=0. \end{align*} Thus if $q(x,y)(y-x^2)\in (y-x^2)$, then \begin{align*} \Phi(q(x,y)(y-x^2))=\Phi(q(x,y))\Phi(y-x^2)=\Phi(q(x,y))\cdot 0=0, \end{align*} so $(y-x^2)\subseteq \ker\Phi$. Conversely, let \begin{align*} f(x,y)=\sum_{i,j} c_{ij}x^iy^j, \end{align*} where only finitely many coefficients $c_{ij}$ are nonzero. Then \begin{align*} f(x,x^2)=\sum_{i,j} c_{ij}x^i(x^2)^j=\sum_{i,j} c_{ij}x^{i+2j}. \end{align*} For each $j\ge 1$, \begin{align*} (y-x^2)\sum_{\ell=0}^{j-1}y^{j-1-\ell}x^{2\ell}=\sum_{\ell=0}^{j-1}y^{j-\ell}x^{2\ell}-\sum_{\ell=0}^{j-1}y^{j-1-\ell}x^{2\ell+2}. \end{align*} Reindexing the second sum by $m=\ell+1$ gives \begin{align*} \sum_{\ell=0}^{j-1}y^{j-\ell}x^{2\ell}-\sum_{\ell=0}^{j-1}y^{j-1-\ell}x^{2\ell+2}=\sum_{\ell=0}^{j-1}y^{j-\ell}x^{2\ell}-\sum_{m=1}^{j}y^{j-m}x^{2m}. \end{align*} All middle terms cancel, leaving \begin{align*} \sum_{\ell=0}^{j-1}y^{j-\ell}x^{2\ell}-\sum_{m=1}^{j}y^{j-m}x^{2m}=y^j-x^{2j}. \end{align*} Hence \begin{align*} y^j-x^{2j}\in (y-x^2) \end{align*} for every $j\ge 1$. Therefore \begin{align*} f(x,y)-f(x,x^2)=\sum_{i,j}c_{ij}x^i(y^j-x^{2j})\in (y-x^2), \end{align*} with the $j=0$ terms contributing $x^i-x^i=0$. If $\Phi(f)=0$, then $f(x,x^2)=0$, so \begin{align*} f(x,y)=f(x,y)-f(x,x^2)\in (y-x^2). \end{align*} Thus $\ker\Phi=(y-x^2)$. Now define \begin{align*} \overline{\Phi}:k[x,y]/(y-x^2)\to k[x],\quad f(x,y)+(y-x^2)\mapsto f(x,x^2). \end{align*} This is well defined: if $f+(y-x^2)=g+(y-x^2)$, then $f-g\in (y-x^2)=\ker\Phi$, so \begin{align*} \Phi(f)-\Phi(g)=\Phi(f-g)=0. \end{align*} Hence $\Phi(f)=\Phi(g)$. The map is surjective because every $g(x)\in k[x]$ satisfies \begin{align*} \overline{\Phi}(g(x)+(y-x^2))=g(x). \end{align*} It is injective because if $\overline{\Phi}(f+(y-x^2))=0$, then $\Phi(f)=0$, so $f\in\ker\Phi=(y-x^2)$, and therefore \begin{align*} f+(y-x^2)=0+(y-x^2). \end{align*} Thus \begin{align*} k[x,y]/(y-x^2)\cong k[x]. \end{align*} Under this isomorphism, the class of $y$ maps to $x^2$, so the quotient records the parametrization $a\mapsto(a,a^2)$ at the level of coordinate rings. [/example] This geometric interpretation also gives a reason to care about nilpotents. A quotient such as $k[\varepsilon]/(\varepsilon^2)$ does not describe an ordinary set of points; it records first-order infinitesimal information. Such quotients are indispensable in scheme-theoretic algebraic geometry, where rings carry more structure than their point sets alone. ## Beyond and Connected Topics Quotient rings are the entry point to much of modern algebra. The first continuation is ideal theory: prime ideals, maximal ideals, radicals, and spectra organize a commutative ring by studying all its quotients at once. The space $\operatorname{Spec}(R)$ is built from prime ideals, and the criterion $R/\mathfrak{p}$ being an integral domain is what makes those points algebraically meaningful. This is the viewpoint developed systematically in [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra), where quotient rings sit beside localization and radicals as basic tools for probing a ring locally and globally. A second direction is field construction. Quotients $k[x]/(f)$ are the standard way to adjoin algebraic elements to a field. When $f$ is irreducible, the quotient is a field generated by a root of $f$, connecting quotient rings to field extensions, splitting fields, and Galois theory. This algebraic use of quotients is a natural continuation of the material in [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules), where ideals and homomorphism theorems first become computational objects. A third direction is algebraic geometry. Coordinate rings such as $k[x_1,\ldots,x_n]/I$ translate systems of polynomial equations into ring-theoretic objects. Replacing a geometric set by its quotient ring lets algebraic tools detect components, singularities, and nilpotent thickening. The geometric interpretation belongs naturally with [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry), where quotient rings become coordinate rings and nilpotents begin to carry geometric information. A fourth direction is module theory. If $M$ is an $R$-module and $N$ is a submodule of $M$, the quotient module $M/N$ parallels $R/I$ additively. Many arguments about quotient rings are special cases of module quotients, with multiplication supplying the extra structure. For course-level development, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules) is the natural undergraduate continuation. For the commutative-algebra viewpoint behind spectra, localization, and radicals, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra) is the next step. The geometric interpretation belongs naturally with [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry). ## References Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry). Atiyah and Macdonald, *Introduction to Commutative Algebra* (1969). Dummit and Foote, *Abstract Algebra* (2004). Eisenbud, *Commutative Algebra with a View Toward Algebraic Geometry* (1995).