A [vector space](/page/Vector%20Space) often contains information we want to treat as invisible. In geometry, two vectors may differ by a direction we have decided to collapse. In linear algebra, two solutions of an equation may differ by an element of the kernel and so produce the same output. In algebra, two elements of a module may be indistinguishable modulo a submodule. The quotient space is the construction that turns this deliberate loss of information into a new space rather than a vague instruction to ignore something. The danger is that ignoring a subspace by informal language does not automatically produce algebra. If $v$ and $w$ are to count as the same whenever $v-w$ lies in a chosen subspace $U$, then addition and scalar multiplication must not depend on which representatives are chosen. The quotient construction packages exactly the [equivalence relation](/page/Equivalence%20Relation) for which these operations are well-defined. [example: Lines Parallel to a Plane] Let $V=\mathbb R^3$ and let $U=\operatorname{span}(e_1,e_2)=\{(a,b,0):a,b\in\mathbb R\}$ be the $xy$-plane. We show that two vectors represent the same coset modulo $U$ exactly when they have the same height coordinate. For every $(x,y,z)\in\mathbb R^3$, the displacement from $(0,0,z)$ to $(x,y,z)$ is horizontal: \begin{align*} (x,y,z)-(0,0,z)=(x,y,0)=xe_1+ye_2\in U. \end{align*} Conversely, if $(a,b,c)\in (x,y,z)+U$, then for some $s,t\in\mathbb R$, \begin{align*} (a,b,c)=(x,y,z)+(s,t,0)=(x+s,y+t,z). \end{align*} So every point of $(x,y,z)+U$ has third coordinate $z$, and every point $(r,s,z)$ with third coordinate $z$ lies in that coset because \begin{align*} (r,s,z)=(x,y,z)+(r-x,s-y,0) \end{align*} with $(r-x,s-y,0)\in U$. Hence \begin{align*} (x,y,z)+U=\{(r,s,z):r,s\in\mathbb R\}=(0,0,z)+U. \end{align*} Now define $\pi:\mathbb R^3\to\mathbb R$ by $\pi(x,y,z)=z$. Its kernel is exactly $U$, since \begin{align*} \ker(\pi)=\{(x,y,z):z=0\}=\{(x,y,0):x,y\in\mathbb R\}=U. \end{align*} For each $h\in\mathbb R$, the fibre over $h$ is \begin{align*} \pi^{-1}(\{h\})=\{(x,y,z):z=h\}=\{(x,y,h):x,y\in\mathbb R\}=(0,0,h)+U. \end{align*} Thus the cosets of $U$ are exactly the horizontal planes, and the quotient $\mathbb R^3/U$ records only the common height of all vectors in such a plane. [/example] The example gives the central picture: a quotient is not a subset of the original space. It is a new object whose elements are entire parallel copies of the subspace being collapsed. This distinction prevents a common mistake: a quotient space is not obtained by deleting $U$ from $V$, but by declaring every translate $v+U$ to be a single point. ## Definition The central construction begins with a decision: two vectors should count as the same when their difference lies in the subspace being ignored. The quotient is the space of these equivalence classes, equipped with operations inherited from the original vector space. [definition: Quotient Space] Let $V$ be a vector space over a field $k$, and let $U\subset V$ be a linear subspace. For $v\in V$, write \begin{align*} v+U:=\{v+u:u\in U\}. \end{align*} The quotient space of $V$ by $U$ is the set \begin{align*} V/U:=\{v+U:v\in V\}, \end{align*} equipped with the operations \begin{align*} +:(V/U)\times(V/U)\to V/U,\qquad (v+U,w+U)\mapsto (v+w)+U, \end{align*} and \begin{align*} \cdot:k\times(V/U)\to V/U,\qquad (\lambda,v+U)\mapsto (\lambda v)+U. \end{align*} [/definition] The definition deliberately names the operations before proving that they make sense. That is the main technical point: a quotient space is only legitimate if changing representatives does not change the resulting class. The next theorem is therefore not a routine afterthought; it is the result that turns the set of cosets into a usable vector space. [quotetheorem:9489] This theorem is the algebraic license to compute with representatives. Once the quotient has been formed, cosets may be added by adding any representatives, because all choices produce the same coset. ## Representatives and Well-Defined Operations The quotient is easiest to use when we understand when two names describe the same new point. The notation $v+U$ is not meant to remember $v$; it remembers the whole translate of $U$ through $v$. [definition: Coset of a Subspace] Let $V$ be a vector space over a field $k$, and let $U\subset V$ be a linear subspace. For $v\in V$, the coset of $U$ represented by $v$ is the subset \begin{align*} v+U:=\{v+u:u\in U\}\subset V. \end{align*} [/definition] A coset is the algebraic version of a parallel affine copy of $U$. If $U$ is a plane through the origin in $\mathbb R^3$, its cosets are planes parallel to it. The representative $v$ is not part of the data of the coset; many different vectors can name the same translate. This creates the first practical problem in any quotient calculation: expressions such as $v+U=w+U$ are statements about two subsets of $V$, but computations usually start with the representatives $v$ and $w$. We need a representative-level criterion for equality, otherwise every calculation in $V/U$ would require comparing two entire translates. The obstruction is that equality of subsets is too large a test for routine algebra. What is needed is an intrinsic condition on the two representatives: exactly how far apart may $v$ and $w$ be while still naming the same collapsed point? In a quotient, the permitted error should be precisely a vector of $U$. [remark: Representative Criterion for Subspace Cosets] For $v,w\in V$, the cosets $v+U$ and $w+U$ are equal exactly when their difference lies in the collapsed subspace: \begin{align*} v+U=w+U \quad\Longleftrightarrow\quad v-w\in U. \end{align*} Thus changing a representative by an element of $U$ does not change the coset, and every equality of cosets comes from such a change of representative. [/remark] This is the moment where the quotient stops being mysterious. Equality in $V/U$ means equality after forgetting all directions in $U$. [example: Quotient by a Line in the Plane] Let $V=\mathbb R^2$ and let $U=\operatorname{span}(1,1)=\{(t,t):t\in\mathbb R\}$. For $(a,b)\in\mathbb R^2$, its coset is \begin{align*} (a,b)+U=\{(a,b)+(t,t):t\in\mathbb R\}=\{(a+t,b+t):t\in\mathbb R\}. \end{align*} Thus each coset is a line parallel to the line spanned by $(1,1)$. We now compute exactly when two representatives give the same coset. If $(a,b)-(c,d)\in U$, then there is $t\in\mathbb R$ such that \begin{align*} (a-c,b-d)=t(1,1)=(t,t). \end{align*} So $a-c=t$ and $b-d=t$, hence \begin{align*} a-b=(c+t)-(d+t)=c-d. \end{align*} Conversely, if $a-b=c-d$, then $a-c=b-d$. Setting $t=a-c=b-d$ gives \begin{align*} (a,b)-(c,d)=(a-c,b-d)=(t,t)=t(1,1)\in U. \end{align*} Therefore $(a,b)$ and $(c,d)$ determine the same coset exactly when $a-b=c-d$. Define $T:\mathbb R^2\to\mathbb R$ by $T(a,b)=a-b$. The calculation above shows that $T$ is constant on each coset, so the rule \begin{align*} \widetilde T((a,b)+U)=a-b \end{align*} is well-defined on $\mathbb R^2/U$. Its kernel is trivial in the quotient: if $\widetilde T((a,b)+U)=0$, then $a-b=0$, so $a=b$ and \begin{align*} (a,b)=a(1,1)\in U. \end{align*} Hence $(a,b)+U=U$, the zero coset. The map is also surjective, since for every $r\in\mathbb R$, \begin{align*} \widetilde T((r,0)+U)=r-0=r. \end{align*} Thus $\widetilde T:\mathbb R^2/U\to\mathbb R$ is a linear isomorphism. The quotient remembers the transverse quantity $a-b$ and forgets motion in the direction $(1,1)$. [/example] Not every formula on representatives descends to the quotient. A formula is allowed only if it gives the same answer on every representative of the same coset; otherwise it is still a formula on $V$, not a function on $V/U$. [example: A Formula That Does Not Descend] Let $V=\mathbb R^2$ and let $U=\operatorname{span}(1,1)=\{t(1,1):t\in\mathbb R\}$. Define $F:\mathbb R^2\to\mathbb R$ by $F(a,b)=a$. The vectors $(0,0)$ and $(1,1)$ determine the same coset modulo $U$, because \begin{align*} (1,1)-(0,0)=(1,1)=1(1,1)\in U. \end{align*} Equivalently, $(1,1)\in U$, so \begin{align*} (1,1)+U=U=(0,0)+U. \end{align*} However, the proposed value of the rule $(a,b)+U\mapsto a$ depends on which representative of this coset is used: \begin{align*} F(0,0)=0 \end{align*} and \begin{align*} F(1,1)=1. \end{align*} Thus the single coset $U$ would have to be assigned both $0$ and $1$, so the rule $(a,b)+U\mapsto a$ does not define a function $\mathbb R^2/U\to\mathbb R$. By contrast, replacing $(a,b)$ by another representative in the same coset means replacing it by \begin{align*} (a,b)+t(1,1)=(a+t,b+t) \end{align*} for some $t\in\mathbb R$. The quantity $a-b$ is unchanged under this replacement, since \begin{align*} (a+t)-(b+t)=a+t-b-t=a-b. \end{align*} So $(a,b)+U\mapsto a-b$ is well-defined, while $(a,b)+U\mapsto a$ still remembers information in the collapsed direction. [/example] ## The Canonical Projection and Universal Property The quotient should not only exist as a set of cosets; it should come with a canonical way to pass from the original space to the collapsed one. This map is the formal version of forgetting the part of a vector that lies in $U$, and it is the map through which all later constructions factor. ### The Quotient Map Once the cosets have been built, we need a systematic way to send each original vector to the coset that represents it after $U$ has been collapsed. The point of the following definition is to name this canonical passage from $V$ to $V/U$, so that later maps out of the quotient can be compared with ordinary linear maps on $V$. [definition: Quotient Map] Let $V$ be a vector space over a field $k$, and let $U\subset V$ be a linear subspace. The quotient map associated to $U$ is the [linear map](/page/Linear%20Map) \begin{align*} q:V\to V/U,\qquad v\mapsto v+U. \end{align*} [/definition] The quotient map should collapse exactly the subspace we intended and nothing more. Its kernel is therefore the first consistency check on the construction: if some vector outside $U$ also maps to zero, then the quotient has discarded more information than promised; if some vector of $U$ survives, then the intended relations have not actually been imposed. To use $q$ as a reliable bridge from $V$ to $V/U$, we need to know precisely which vectors it identifies with the zero coset. The next structural fact records that the quotient map forgets exactly $U$, so its kernel matches the subspace used to form the quotient. [quotetheorem:9490] ### The Universal Property The quotient map also explains the notation $V/U$: the denominator is not being removed; it is being sent to zero. In the quotient, the zero vector is the coset $U$ itself. This raises the structural question that makes quotient spaces more than notation: suppose another linear map $T:V\to W$ also sends every vector of $U$ to zero. Then $T$ has already forgotten at least the information that the quotient forgets, so it ought to be recoverable from a map whose domain is $V/U$ rather than $V$. The only possible formula would send a coset $v+U$ to $T(v)$, but that formula is vulnerable to the same representative problem as quotient addition: if $v+U=w+U$, the two chosen representatives must give the same value in $W$. The needed condition is exactly that $T$ kills every displacement lying in $U$, and uniqueness should follow because every coset has some representative from $V$. [quotetheorem:9491] This theorem is the precise replacement for the phrase “define the map on equivalence classes.” The hypothesis $U\subset\ker(T)$ is exactly the condition that the proposed formula does not depend on representatives. Its real power is that it lets us recognize quotients without inspecting cosets one by one, which is why the next example can identify a polynomial quotient by a familiar evaluation map. [example: Polynomial Evaluation as a Quotient] Here $k$ denotes a field and $k[x]$ denotes the vector space, and also the ring, of polynomials in one variable with coefficients in $k$. For $a\in k$, the notation $(x-a)k[x]$ means the set of all polynomial multiples of $x-a$: \begin{align*} (x-a)k[x]=\{(x-a)q(x):q(x)\in k[x]\}. \end{align*} As a vector subspace of $k[x]$, this is the collection of polynomials that vanish at $a$. Let $k$ be a field, let $V=k[x]$, fix $a\in k$, and define $E_a:k[x]\to k$ by $E_a(p)=p(a)$. The map $E_a$ is linear because for $p,q\in k[x]$ and $\lambda\in k$, \begin{align*} E_a(p+\lambda q)=(p+\lambda q)(a)=p(a)+\lambda q(a)=E_a(p)+\lambda E_a(q). \end{align*} We first identify its kernel. For any $p\in k[x]$, division by the monic polynomial $x-a$ gives polynomials $q\in k[x]$ and a constant $r\in k$ such that \begin{align*} p=(x-a)q+r. \end{align*} Evaluating at $a$ gives \begin{align*} p(a)=(a-a)q(a)+r=0\cdot q(a)+r=r. \end{align*} Thus $p(a)=0$ exactly when $r=0$, and this is exactly when $p=(x-a)q$ for some $q\in k[x]$. Hence \begin{align*} \ker(E_a)=\{p\in k[x]:p(a)=0\}=(x-a)k[x]. \end{align*} Now define \begin{align*} \Phi:k[x]/(x-a)k[x]\to k,\qquad \Phi(p+(x-a)k[x])=p(a). \end{align*} This rule is well-defined: if $p+(x-a)k[x]=q+(x-a)k[x]$, then $p-q\in (x-a)k[x]$, so $p-q=(x-a)s$ for some $s\in k[x]$, and therefore \begin{align*} p(a)-q(a)=(p-q)(a)=((x-a)s)(a)=(a-a)s(a)=0. \end{align*} So $p(a)=q(a)$. The map $\Phi$ is linear because \begin{align*} \Phi((p+(x-a)k[x])+\lambda(q+(x-a)k[x]))=\Phi((p+\lambda q)+(x-a)k[x])=(p+\lambda q)(a)=p(a)+\lambda q(a). \end{align*} It is injective: if $\Phi(p+(x-a)k[x])=0$, then $p(a)=0$, so $p\in (x-a)k[x]$, and hence \begin{align*} p+(x-a)k[x]=(x-a)k[x]. \end{align*} It is surjective: for any $c\in k$, the constant polynomial $c$ satisfies \begin{align*} \Phi(c+(x-a)k[x])=c(a)=c. \end{align*} Therefore \begin{align*} k[x]/(x-a)k[x]\cong k \end{align*} by the isomorphism $p+(x-a)k[x]\mapsto p(a)$. The quotient records only the value of a polynomial at $a$; all multiples of $x-a$ have been collapsed to zero. [/example] The example suggests a general principle: whenever a linear map has already collapsed some directions, the quotient by its kernel should be the part of the domain that remains visible. The [first isomorphism theorem](/theorems/791) states this principle as an intrinsic identification with the range. [quotetheorem:384] In practice, this theorem often tells us what a quotient is without listing all its cosets. If a familiar linear map has kernel $U$, then $V/U$ is naturally the range of that map. ## Dimension, Bases, and Complements For finite-dimensional vector spaces, quotienting subtracts the dimension of the collapsed subspace. This is not deletion of points; it is removal of independent directions. [quotetheorem:377] The dimension formula is useful, but it only counts the surviving directions; it does not tell us how to name them. For computations, we need actual representatives whose classes form coordinates in the quotient. To build such a basis, start with a basis of the subspace being killed and extend it to a basis of the whole space. The added vectors are precisely the transverse directions that survive. [quotetheorem:9492] This result also warns against a tempting but misleading picture. A quotient is not canonically the span of the surviving basis vectors; that identification depends on having chosen a transverse copy of the quotient inside $V$. In finite-dimensional calculations this choice is often hidden inside an extended basis, but in structural arguments it needs a name: we want a subspace whose vectors provide exactly one representative for each coset and whose overlap with the collapsed subspace is only the zero vector. The definition below isolates that extra choice. It records when $V$ has been split into the discarded directions and a separate transverse subspace. [definition: Complementary Subspace] Let $V$ be a vector space over a field $k$, and let $U\subset V$ be a linear subspace. A subspace $W\subset V$ is a complementary subspace to $U$ in $V$ if \begin{align*} V=U\oplus W. \end{align*} [/definition] A complement provides representatives for every coset, but the representatives are a choice. Different complements give different concrete copies of the same quotient. The next theorem records the payoff of making such a choice: the abstract quotient can be replaced by the chosen transverse subspace. [quotetheorem:9493] [example: A Quotient Basis in $\mathbb R^3$] Let $V=\mathbb R^3$ and let $U=\operatorname{span}(e_1+e_2)$. The list $(e_1+e_2,e_2,e_3)$ is a basis of $\mathbb R^3$: if \begin{align*} \alpha(e_1+e_2)+\beta e_2+\gamma e_3=0, \end{align*} then \begin{align*} \alpha e_1+(\alpha+\beta)e_2+\gamma e_3=0e_1+0e_2+0e_3, \end{align*} so $\alpha=0$, $\alpha+\beta=0$, and $\gamma=0$, hence $\beta=0$ as well. Also, for any $(x,y,z)\in\mathbb R^3$, \begin{align*} (x,y,z)=x(e_1+e_2)+(y-x)e_2+ze_3, \end{align*} because the right-hand side is $(x,x,0)+(0,y-x,0)+(0,0,z)=(x,y,z)$. Therefore the quotient is spanned by the two classes $e_2+U$ and $e_3+U$. Indeed, for any $(x,y,z)\in\mathbb R^3$, \begin{align*} (x,y,z)+U=x(e_1+e_2)+U+(y-x)(e_2+U)+z(e_3+U). \end{align*} Since $x(e_1+e_2)\in U$, its class is the zero coset: \begin{align*} x(e_1+e_2)+U=U. \end{align*} Thus \begin{align*} (x,y,z)+U=(y-x)(e_2+U)+z(e_3+U). \end{align*} The two classes are linearly independent in the quotient. If \begin{align*} a(e_2+U)+b(e_3+U)=U, \end{align*} then \begin{align*} ae_2+be_3+U=U, \end{align*} so $ae_2+be_3\in U$. Hence there is $t\in\mathbb R$ such that \begin{align*} ae_2+be_3=t(e_1+e_2). \end{align*} In coordinates this says \begin{align*} (0,a,b)=(t,t,0), \end{align*} so $t=0$, $a=t=0$, and $b=0$. Therefore $(e_2+U,e_3+U)$ is a basis of $\mathbb R^3/U$. The class of $e_1$ is not a new independent direction, because \begin{align*} e_1-(-e_2)=e_1+e_2\in U. \end{align*} Equivalently, \begin{align*} e_1+U=-e_2+U. \end{align*} The quotient has kept two independent transverse directions and collapsed the direction $e_1+e_2$. [/example] ## Exact Sequences and Imposed Relations Quotients are the language of imposed relations. Saying that elements of $U$ become zero is the same as saying that every relation encoded by $U$ is now enforced in the target space. [definition: Short Exact Sequence] A short exact sequence of vector spaces over a field $k$ is a diagram of linear maps \begin{align*} 0\to U\xrightarrow{i} V\xrightarrow{q} W\to 0 \end{align*} such that $i$ is injective, $q$ is surjective, and \begin{align*} \operatorname{Range}(i)=\ker(q). \end{align*} [/definition] Exactness packages quotient data without naming cosets. If $U$ sits inside $V$ as the kernel of a surjection onto $W$, then $W$ ought to be the quotient of $V$ by that hidden copy of $U$. The next theorem makes that interpretation canonical, so a short exact sequence can be read as a quotient construction. [quotetheorem:9494] [example: Imposing a Single Linear Relation] Let $V=k^3$, and let $T:k^3\to k$ be defined by \begin{align*} T(x_1,x_2,x_3)=x_1+x_2+x_3. \end{align*} This map is linear because for $x=(x_1,x_2,x_3)$, $y=(y_1,y_2,y_3)$, and $\lambda\in k$, \begin{align*} T(x+\lambda y)=T(x_1+\lambda y_1,x_2+\lambda y_2,x_3+\lambda y_3). \end{align*} Expanding the definition of $T$ gives \begin{align*} T(x+\lambda y)=(x_1+\lambda y_1)+(x_2+\lambda y_2)+(x_3+\lambda y_3). \end{align*} Rearranging terms in the field $k$ gives \begin{align*} T(x+\lambda y)=(x_1+x_2+x_3)+\lambda(y_1+y_2+y_3)=T(x)+\lambda T(y). \end{align*} The kernel is exactly the plane of triples whose coordinates sum to zero: \begin{align*} \ker(T)=\{(x_1,x_2,x_3)\in k^3:T(x_1,x_2,x_3)=0\}. \end{align*} Substituting the formula for $T$ gives \begin{align*} \ker(T)=\{(x_1,x_2,x_3)\in k^3:x_1+x_2+x_3=0\}. \end{align*} Now define \begin{align*} \Phi:k^3/\ker(T)\to k,\qquad \Phi((x_1,x_2,x_3)+\ker(T))=x_1+x_2+x_3. \end{align*} This rule is well-defined. If two representatives determine the same coset, then their difference lies in $\ker(T)$, so for $x=(x_1,x_2,x_3)$ and $y=(y_1,y_2,y_3)$ we have \begin{align*} x-y=(x_1-y_1,x_2-y_2,x_3-y_3)\in\ker(T). \end{align*} By the description of the kernel, \begin{align*} (x_1-y_1)+(x_2-y_2)+(x_3-y_3)=0. \end{align*} Rearranging gives \begin{align*} (x_1+x_2+x_3)-(y_1+y_2+y_3)=0. \end{align*} Hence \begin{align*} x_1+x_2+x_3=y_1+y_2+y_3, \end{align*} so $\Phi$ gives the same value on both representatives. The map $\Phi$ is linear because for $x,y\in k^3$ and $\lambda\in k$, \begin{align*} \Phi((x+\ker(T))+\lambda(y+\ker(T)))=\Phi((x+\lambda y)+\ker(T)). \end{align*} Using the definition of $\Phi$ and the linearity of $T$ computed above, \begin{align*} \Phi((x+\lambda y)+\ker(T))=T(x+\lambda y)=T(x)+\lambda T(y). \end{align*} Therefore \begin{align*} \Phi((x+\ker(T))+\lambda(y+\ker(T)))=\Phi(x+\ker(T))+\lambda\Phi(y+\ker(T)). \end{align*} The map is injective: if $\Phi(x+\ker(T))=0$, then $T(x)=0$, so $x\in\ker(T)$. Therefore \begin{align*} x+\ker(T)=\ker(T), \end{align*} which is the zero coset. It is surjective because for every $c\in k$, \begin{align*} \Phi((c,0,0)+\ker(T))=c+0+0=c. \end{align*} Thus $k^3/\ker(T)$ is isomorphic to $k$. The quotient collapses precisely the triples with coordinate sum $0$ and keeps exactly the single value $x_1+x_2+x_3$. [/example] ## Quotients Beyond Vector Spaces The same idea appears throughout algebra, but each category has its own condition for when operations descend. For vector spaces, every subspace works. For groups, rings, and modules, the subobject must be compatible with the operations that survive in the quotient. [definition: Quotient Module] Let $R$ be a ring, let $M$ be a left $R$-module, and let $N\subset M$ be a submodule. The quotient module $M/N$ is the set \begin{align*} M/N:=\{m+N:m\in M\}, \end{align*} equipped with addition \begin{align*} +:(M/N)\times(M/N)\to M/N,\qquad (m+N,m'+N)\mapsto (m+m')+N, \end{align*} and scalar multiplication \begin{align*} \cdot:R\times(M/N)\to M/N,\qquad (r,m+N)\mapsto (rm)+N \end{align*} for $m,m'\in M$ and $r\in R$. [/definition] Modules behave closest to vector spaces: submodules are exactly the subsets for which the quotient operations are well-defined. Rings and groups need stronger compatibility because multiplication or conjugation must also respect representatives. [definition: Quotient Ring] Let $R$ be a ring, and let $I\trianglelefteq R$ be a two-sided ideal. The [quotient ring](/page/Quotient%20Ring) $R/I$ is the set \begin{align*} R/I:=\{r+I:r\in R\}, \end{align*} equipped with addition \begin{align*} +:(R/I)\times(R/I)\to R/I,\qquad (r+I,s+I)\mapsto (r+s)+I, \end{align*} and multiplication \begin{align*} \cdot:(R/I)\times(R/I)\to R/I,\qquad (r+I,s+I)\mapsto rs+I. \end{align*} [/definition] The ideal condition is exactly what prevents multiplication from depending on representatives. This is why quotient rings encode algebraic equations so effectively. [example: A Polynomial Relation Quotient Ring] Let $k$ be a field, and write $I=(x^2)\subset k[x]$. In the quotient ring $k[x]/I$, the class of $x^2$ is zero because $x^2\in I$, so \begin{align*} x^2+I=I. \end{align*} Every polynomial $p\in k[x]$ can be written as \begin{align*} p=a+bx+x^2q \end{align*} for some $a,b\in k$ and $q\in k[x]$: take $a$ to be the constant coefficient of $p$, take $b$ to be the coefficient of $x$, and put all terms of degree at least $2$ into $x^2q$. Therefore \begin{align*} p+I=(a+bx+x^2q)+I=(a+bx)+I \end{align*} because $x^2q\in I$. Thus every class has a representative of the form $a+bx$. Now multiply two such representatives. Expanding in $k[x]$ gives \begin{align*} (a+bx)(c+dx)=a(c+dx)+bx(c+dx) \end{align*} and hence \begin{align*} (a+bx)(c+dx)=ac+adx+bcx+bdx^2. \end{align*} Combining the two $x$-terms gives \begin{align*} (a+bx)(c+dx)=ac+(ad+bc)x+bdx^2. \end{align*} Since $bdx^2\in (x^2)=I$, this becomes in the quotient \begin{align*} ((a+bx)+I)((c+dx)+I)=(ac+(ad+bc)x)+I. \end{align*} The classes $1+I$ and $x+I$ span $k[x]/I$ by the representative calculation above. They are linearly independent: if \begin{align*} a(1+I)+b(x+I)=I, \end{align*} then \begin{align*} (a+bx)+I=I, \end{align*} so $a+bx\in I$. Thus $a+bx=x^2q$ for some $q\in k[x]$, and the constant and $x$ coefficients of $x^2q$ are both $0$, so $a=0$ and $b=0$. Hence $k[x]/(x^2)$ is a two-dimensional $k$-vector space with basis $(1+I,x+I)$, while its multiplication remembers the imposed nilpotent relation $(x+I)^2=I$. [/example] For groups, the obstruction is even sharper: left and right multiplication must be compatible with the chosen subgroup. Normality is the condition that makes the [quotient group](/theorems/790) possible. [definition: Quotient Group] Let $G$ be a group, and let $N\trianglelefteq G$ be a [normal subgroup](/page/Normal%20Subgroup). The quotient group $G/N$ is the set \begin{align*} G/N:=\{gN:g\in G\}, \end{align*} equipped with multiplication \begin{align*} \cdot:(G/N)\times(G/N)\to G/N,\qquad (gN,hN)\mapsto ghN. \end{align*} [/definition] If the subgroup is not normal, representative choices can change the product. This failure is the group-theoretic analogue of a representative-dependent formula on a vector-space quotient. [example: Non-Normal Subgroup Failure] Let $G=S_3$, and let $H=\{e,(12)\}$. We show that the proposed product of left cosets, \begin{align*} (gH)(hH)=ghH, \end{align*} depends on the representative chosen from the first coset. First, \begin{align*} eH=\{e,(12)\}=H. \end{align*} Also, \begin{align*} (12)H=\{(12)e,(12)(12)\}=\{(12),e\}=H. \end{align*} Hence $eH=(12)H$. Now multiply this same left coset by $(23)H$. Using the representative $e$ gives \begin{align*} e(23)H=(23)H. \end{align*} The coset on the right is \begin{align*} (23)H=\{(23)e,(23)(12)\}=\{(23),(132)\}. \end{align*} Using the other representative $(12)$ gives \begin{align*} (12)(23)H=(123)H, \end{align*} because $(12)(23)=(123)$ under the usual right-to-left convention for composing permutations. This second coset is \begin{align*} (123)H=\{(123)e,(123)(12)\}=\{(123),(13)\}. \end{align*} The two resulting cosets are different, since \begin{align*} (23)H=\{(23),(132)\} \end{align*} while \begin{align*} (123)H=\{(123),(13)\}. \end{align*} Thus the single equality $eH=(12)H$ would force two different products with $(23)H$, so the rule $(gH)(hH)=ghH$ is not well-defined for this subgroup. The obstruction is exactly that $H$ is not normal in $S_3$. [/example] ## Quotient Topologies and Norms ### Collapsing Points Continuously Algebraic quotients collapse structure; topological quotients collapse points while preserving the ability to test continuity. The guiding question changes from “are the operations well-defined?” to “which topology makes maps out of the quotient continuous exactly when their pullbacks are continuous?” If we merely identify points of a space and then choose an arbitrary topology on the set of equivalence classes, continuity becomes disconnected from the original space. The [quotient topology](/page/Quotient%20Topology) solves this by making the collapse map the source of all open-set information, so the new space remembers precisely the continuity data that survives identification. [definition: Quotient Topology] Let $X$ be a [topological space](/page/Topological%20Space), let $Y$ be a set, and let $q:X\to Y$ be a surjective map. The quotient topology on $Y$ induced by $q$ is the topology in which a subset $O\subset Y$ is open exactly when $q^{-1}(O)$ is open in $X$. [/definition] The definition is engineered for a universal property: continuity out of the quotient can be checked before collapse. This is the topological counterpart of the linear universal property. [quotetheorem:1031] [example: Gluing the Endpoints of an Interval] Let $Y$ be the set obtained from $[0,1]$ by declaring $0\sim 1$ and leaving every other point equivalent only to itself. Write \begin{align*} q:[0,1]\to Y,\qquad q(t)=[t]. \end{align*} Thus $q(0)=q(1)$, while $q(s)=q(t)$ for $s,t\in(0,1)$ forces $s=t$. The quotient topology on $Y$ is defined by the rule that $O\subset Y$ is open exactly when $q^{-1}(O)$ is open in the [subspace topology](/page/Subspace%20Topology) on $[0,1]$. To see the circle explicitly, define \begin{align*} F:[0,1]\to S^1,\qquad F(t)=(\cos(2\pi t),\sin(2\pi t)). \end{align*} The endpoints have the same image because \begin{align*} F(0)=(\cos 0,\sin 0)=(1,0) \end{align*} and \begin{align*} F(1)=(\cos 2\pi,\sin 2\pi)=(1,0). \end{align*} So $F$ is constant on the only nontrivial equivalence class, and therefore defines a map \begin{align*} \overline F:Y\to S^1,\qquad \overline F([t])=F(t). \end{align*} This rule is well-defined: if $[s]=[t]$, then either $s=t$, which gives $F(s)=F(t)$, or $\{s,t\}=\{0,1\}$, which gives $F(s)=F(t)=(1,0)$. The map $\overline F$ is bijective. It is surjective because every point of $S^1$ has the form $(\cos\theta,\sin\theta)$ for some $\theta\in[0,2\pi]$, and then with $t=\theta/(2\pi)$ we have \begin{align*} \overline F([t])=(\cos\theta,\sin\theta). \end{align*} It is injective because if $\overline F([s])=\overline F([t])$, then \begin{align*} (\cos(2\pi s),\sin(2\pi s))=(\cos(2\pi t),\sin(2\pi t)). \end{align*} For $s,t\in[0,1]$, this means either $s=t$ or $\{s,t\}=\{0,1\}$, and in both cases $[s]=[t]$. Continuity of $\overline F$ follows from the quotient topology because \begin{align*} F=\overline F\circ q, \end{align*} and $F$ is continuous as a coordinatewise trigonometric map. Since $[0,1]$ is compact, $Y=q([0,1])$ is compact; since $S^1\subset\mathbb R^2$ is Hausdorff, the standard *compact-to-Hausdorff homeomorphism criterion* makes the continuous bijection $\overline F:Y\to S^1$ a homeomorphism. Thus gluing the two endpoints of the interval turns the quotient space into the usual topological circle. [/example] ### Measuring Quotient Classes Normed spaces add one more layer: after collapsing a subspace, the norm of a class should measure the shortest distance from the representative to the collapsed subspace. This is not the norm of a chosen representative, because representatives in the same class may have different lengths. The only quantity attached to the class itself is the best distance from that class to the subspace being collapsed. [definition: Quotient Norm] Let $X$ be a [normed vector space](/page/Normed%20Vector%20Space) over $k$, and let $Y\subset X$ be a linear subspace. The quotient seminorm on $X/Y$ is \begin{align*} \|x+Y\|_{X/Y}:=\inf_{y\in Y}\|x-y\|_X. \end{align*} If $Y$ is closed, this seminorm is called the quotient norm. [/definition] Closedness matters because non-closed subspaces may contain vectors arbitrarily close to points outside the subspace, forcing nonzero classes to have zero seminorm. Once the collapsed subspace is closed, the quotient norm separates points; the next question is whether completeness survives the collapse. [quotetheorem:2631] This theorem is the analytic payoff of the quotient norm. It says that when $X$ is complete and $Y$ is closed, passing from $X$ to $X/Y$ does not lose the Banach-space setting: Cauchy sequences of classes can still be represented and controlled by vectors in $X$. Closedness is the separating hypothesis that prevents distinct classes from collapsing to distance zero, while completeness of $X$ is the supply of limits needed to prove completeness in the quotient. The accompanying open mapping statement is equally useful in practice: the canonical projection $q:X\to X/Y$ sends open balls in $X$ to open balls in the quotient, so estimates and neighborhoods can be pushed down to the quotient without adding an artificial topology by hand. Thus closed quotient spaces are not merely algebraic objects with a convenient notation. They are Banach spaces in their own right, and this is what allows quotient constructions to be used throughout functional analysis, for example when one identifies spaces modulo null functions, kernels, or closed constraint subspaces. ## Beyond and Connected Topics Quotient spaces are the linear prototype for much of modern algebra. In [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules), the first isomorphism theorem says that homomorphisms factor through quotients by kernels; the same slogan reappears for groups, rings, modules, and many categorical settings. The linear version sits naturally alongside [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra), where bases, dimension, kernels, and images provide the computational language for these identifications. Exact sequences turn quotients into a structural language. Instead of saying “$W$ is $V$ modulo $U$,” one records the short exact sequence $0\to U\to V\to W\to 0$. This is the natural bridge to homological algebra, module presentations, and extension problems. Topology and analysis add compatibility conditions. Quotient topologies describe gluing spaces, while quotient norms describe Banach spaces modulo closed subspaces. These ideas continue into [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology), where quotient constructions interact with continuity, compactness, and connectedness. The algebraic idea remains the same, but continuity, openness, and completeness become part of the construction. ## References Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Sheldon Axler, *Linear Algebra Done Right* (2015). Serge Lang, *Algebra* (2002). Michael Artin, *Algebra* (2011). Atiyah and Macdonald, *Introduction to Commutative Algebra* (1969).