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"Quotient field" redirects here. Not to be confused with Quotient ring.

Question for help desk

I use the visual editor almost exclusively, and when I wish to write formulae in LaTeX, the fastest way I know is to type "<math", which opens the editor that hides most of the page. Once a formula has been entered, clicking on it brings up a "Quick Edit" box that doesn't hide the page and shows how the formula will appeared rendered inline. Is there a shortcut to bring up the quick edit box rapidly? Failing that, is there another way to enter TeX quickly short of editing source? ~~~~

In abstract algebra, the field of fractions of an integral domain is the smallest field in which it can be embedded.

The elements of the field of fractions of the integral domain $R$ are equivalence classes (see the construction below) written as

${\frac {a}{b}}$

with

$a$ and $b$ in $R$ and $b\neq 0$.

The field of fractions of $R$ is sometimes denoted by $\operatorname {Frac} (R)$ or $\operatorname {Quot} (R)$.

Mathematicians refer to this construction as the field of fractions, fraction field, field of quotients, or quotient field. All four are in common usage. The expression "quotient field" may sometimes run the risk of confusion with the quotient of a ring by an ideal, which is a quite different concept.

Definition

Given an integral domain and letting $R^{*}=R\setminus \{0\}$, we define an equivalence relation on $R\times R^{*}$ by letting $(n,d)\sim (m,b)$ whenever $nb=md$. We denote the equivalence class of $(n,d)$ by the fraction ${\frac {n}{d}}$. This notion of equivalence is motivated by the rational numbers $\mathbb {Q}$, which have the same property with respect to the underlying ring $\mathbb {Z}$ of integers.

Then the field of fractions is the set ${\text{Frac}}(R)=(R\times R^{*})/\sim$ with addition given by

${\frac {n}{d}}+{\frac {m}{b}}={\frac {nb+md}{db}}$

and multiplication given by

${\frac {n}{d}}\cdot {\frac {m}{b}}={\frac {nm}{db}}$

One may check that these operations are well defined and that, for any integral domain $R$, ${\text{Frac}}(R)$ is indeed a field. In particular, for $n\neq 0$, the multiplicative inverse of ${\frac {n}{d}}$ is as expected: ${\frac {d}{n}}\cdot {\frac {n}{d}}=1$.

Let $R$ be any integral domain,

For $(n,d)\in R\times R^{*}$, we denote by ${\frac {n}{d}}$ the equivalence class of pairs where $(n,d)$ is equivalent to $(m,b)$ if and only if $nb=md$.

(The definition of equivalence is modeled on the property of rational numbers that ${\frac {n}{d}}={\frac {m}{b}}$ if and only if $nb=md$.)

The field of fractions $\operatorname {Frac} (R)$ is defined as the set of all such fractions ${\frac {n}{d}}$ with addition given by

${\frac {n}{d}}+{\frac {m}{b}}={\frac {nb+md}{db}}$

and multiplication given by

${\frac {n}{d}}\cdot {\frac {m}{b}}={\frac {nm}{db}}$.

The embedding of $R$ in $\operatorname {Frac} (R)$ maps each $n$ in $R$ to the fraction ${\frac {en}{e}}$ for any nonzero $e\in R$ (the equivalence class is independent of the choice $e$). This is modeled on the identity ${\frac {n}{1}}=n$.

The field of fractions of $R$ is characterized by the following universal property:

if $h:R\to F$ is an injective ring homomorphism from $R$ into a field $F$, then there exists a unique ring homomorphism $g:\operatorname {Frac} (R)\to F$ which extends $h$.

There is a categorical interpretation of this construction. Let $C$ be the category of integral domains and injective ring maps. The functor from $C$ to the category of fields which takes every integral domain to its fraction field and every homomorphism to the induced map on fields (which exists by the universal property) is the left adjoint of the inclusion functor from the category of fields to $C$. Thus the category of fields (which is a full subcategory) is a reflective subcategory of $C$.

A multiplicative identity is not required for the role of the integral domain; this construction can be applied to any nonzero commutative rng $R$ with no nonzero zero divisors. The embedding is given by $r\mapsto {\frac {rs}{s}}$ for any nonzero $s\in R$.

Examples

• The field of fractions of the ring of integers is the field of rationals, $\mathbb {Q} =\operatorname {Frac} (\mathbb {Z} )$.
• Let $R:=\{a+b\mathrm {i} \mid a,b\in \mathbb {Z} \}$ be the ring of Gaussian integers. Then $\operatorname {Frac} (R)=\{c+d\mathrm {i} \mid c,d\in \mathbb {Q} \}$, the field of Gaussian rationals.
• The field of fractions of a field is canonically isomorphic to the field itself.
• Given a field $K$, the field of fractions of the polynomial ring in one indeterminate $K[X]$ (which is an integral domain), is called the field of rational functions or field of rational fractions and is denoted $K(X)$.

Generalizations

Localization

Main article: Localization (commutative algebra)

For any commutative ring $R$ and any multiplicative set $S$ in $R$, the localization $S^{-1}R$ is the commutative ring consisting of fractions

${\frac {r}{s}}$

with $r\in R$ and $s\in S$, where now $(r,s)$ is equivalent to $(r',s')$ if and only if there exists $t\in S$ such that $t(rs'-r's)=0$.

Two special cases of this are notable:

• If $S$ is the complement of a prime ideal $P$, then $S^{-1}R$ is also denoted $R_{P}$. When $R$ is an integral domain and $P$ is the zero ideal, $R_{P}$ is the field of fractions of $R$.
• If $S$ is the set of non-zero-divisors in $R$, then $S^{-1}R$ is called the total quotient ring. The total quotient ring of an integral domain is its field of fractions, but the total quotient ring is defined for any commutative ring.

Note that it is permitted for $S$ to contain 0, but in that case $S^{-1}R$ will be the trivial ring.

Semifield of fractions

The semifield of fractions of a commutative semiring with no zero divisors is the smallest semifield in which it can be embedded.

The elements of the semifield of fractions of the commutative semiring $R$ are equivalence classes written as

${\frac {a}{b}}$

with $a$ and $b$ in $R$.

See also

• Ore condition; condition related to constructing fractions in the noncommutative case.
• Projective line over a ring; alternative structure not limited to integral domains.

Exact sequence

Vector calculus in $\mathbb {R} ^{3}$ provides another example, which may offer some sense of the utility of exact sequences. Consider the space of smooth scalar functions $F=C^{\infty }(\mathbb {R} ^{3})$ and the space of smooth vector fields $V=C^{\infty }(\mathbb {R} ^{3})^{3}$. Three of the basic operators in vector calculus are taking the gradient of a function, and taking the curl or divergence of a vector field:$$\nabla :C^{\infty }(\mathbb {R} ^{3})\to C^{\infty }(\mathbb {R} ^{3})^{3}\quad \quad \nabla \times :C^{\infty }(\mathbb {R} ^{3})^{3}\to C^{\infty }(\mathbb {R} ^{3})^{3}\quad \quad \nabla \cdot :C^{\infty }(\mathbb {R} ^{3})^{3}\to C^{\infty }(\mathbb {R} ^{3})$$In order to construct an exact sequence, the exactness requirements must be met, that ${\text{im}}(\nabla )=\ker(\nabla \times )$ and ${\text{im}}(\nabla \times )=\ker(\nabla \cdot )$. The inclusion of the images in the kernels is easily verified for any domain, but the reverse inclusion only holds for simply connected subdomains of $\mathbb {R} ^{3}$ (including the whole space).The results that ${\text{im}}\nabla =\ker \nabla \times$ and that ${\text{im}}\nabla \times =\ker \nabla \cdot$ in $\mathbb {R} ^{3}$ are sometimes called Poincare's lemma for gradients or curls respectively. Therefore we have the exact sequence

$C^{\infty }(\mathbb {R} ^{3})\xrightarrow {\nabla } C^{\infty }(\mathbb {R} ^{3})^{3}\xrightarrow {\operatorname {\nabla \times } } C^{\infty }(\mathbb {R} ^{3})^{3}\xrightarrow {\nabla \cdot } C^{\infty }(\mathbb {R} ^{3})$

This sequence can be extended to a short exact sequence, although a proof is slightly more involved. The observation that these operators form an exact sequence for simply connected subdomains of $\mathbb {R} ^{3}$, and form a cochain complex when the domain is not simply connected, is the basis of the theory of de Rahm cohomology, which investigates similar relationships on general manifolds.

A sequence in which the image of each map is contained in the kernel of the next, but not necessarily equal, is called a (co)chain complex.

Another example can be derived from differential geometry, especially relevant for work on the Maxwell equations.

Consider the Hilbert space $L^{2}$ of scalar-valued square-integrable functions on three dimensions $\left\lbrace f:\mathbb {R} ^{3}\to \mathbb {R} \right\rbrace$. Taking the gradient of a function $f\in \mathbb {H} _{1}$ moves us to a subset of $\mathbb {H} _{3}$, the space of vector valued, still square-integrable functions on the same domain $\left\lbrace f:\mathbb {R} ^{3}\to \mathbb {R} ^{3}\right\rbrace$ -- specifically, the set of such functions that represent conservative vector fields. (The generalized Stokes' theorem has preserved integrability.)

First, note the curl of all such fields is zero -- since

${\begin{aligned}\operatorname {curl} (\operatorname {grad} f)&\equiv \nabla \times (\nabla f)=0\end{aligned}}$

for all such f. However, this only proves that the image of the gradient is a subset of the kernel of the curl. To prove that they are in fact the same set, prove the converse: that if the curl of a vector field ${\vec {F}}$ is 0, then ${\vec {F}}$ is the gradient of some scalar function. This follows almost immediately from Stokes' theorem (see the proof at conservative force.) The image of the gradient is then precisely the kernel of the curl, and so we can then take the curl to be our next morphism, taking us again to a (different) subset of $\mathbb {H} _{3}$.

Similarly, we note that

${\begin{aligned}\operatorname {div} (\operatorname {curl} {\vec {v}})&\equiv \nabla \cdot \nabla \times {\vec {v}}=0,\end{aligned}}$

so the image of the curl is a subset of the kernel of the divergence. The converse is somewhat involved:

Proof that $\operatorname {div} {\vec {F}}$ = 0 implies ${\vec {F}}=\operatorname {curl} {\vec {A}}$ for some ${\vec {A}}$

We shall proceed by construction: given a vector field ${\vec {F}}=F_{x}{\hat {i}}+F_{y}{\hat {j}}+F_{z}{\hat {k}}$ such that $\nabla \cdot {\vec {F}}=0$, we produce a field ${\vec {A}}$ such that ${\begin{aligned}\nabla \times {\vec {A}}&=\left(\partial _{y}A_{z}-\partial _{z}A_{y}\right){\hat {i}}+\left(\partial _{z}A_{x}-\partial _{x}A_{z}\right){\hat {j}}+\left(\partial _{x}A_{y}-\partial _{y}A_{x}\right){\hat {k}}\\&=F_{x}{\hat {i}}+F_{y}{\hat {j}}+F_{z}{\hat {k}}={\vec {F}}\end{aligned}}$ First, note that since as proved above $\operatorname {curl} (\operatorname {grad} f)=0$, we can add the gradient of any scalar function to ${\vec {A}}$ without changing the curl. We can use this gauge freedom to set any one component of ${\vec {A}}$ to zero without changing its curl; choosing arbitrarily the z-component, we thus require simply that $-\partial _{z}A_{y}{\hat {i}}+\partial _{z}A_{x}{\hat {j}}+\left(\partial _{x}A_{y}-\partial _{y}A_{x}\right){\hat {k}}=F_{x}{\hat {i}}+F_{y}{\hat {j}}+F_{z}{\hat {k}}$ Then by simply integrating the first two components, and noting that the 'constant' of integration may still depend on any variable not integrated over, we find that ${\begin{aligned}A_{x}&=\int F_{y}dz+{\vec {C_{1}}}(x,y)\\A_{y}&=-\int F_{x}dz+{\vec {C_{2}}}(x,y)\end{aligned}}$ Note that since the two integration terms ${\vec {C_{1}}},{\vec {C_{2}}}$ both depend only on x and y and not on z, then we can add another gradient of some function $f(x,y)$ that also does not depend on z. This permits us to eliminate either of the terms in favor of the other, without spoiling our earlier work that set $A_{z}$ to zero. Choosing to eliminate ${\vec {C_{2}}}$ and applying the last component as a constraint, we have ${\begin{aligned}-\partial _{x}\int F_{x}dz+\partial _{x}{\vec {C_{1}}}(x,y)-\partial _{y}\int F_{y}dz&=F_{z}\\-\int \left(\partial _{x}F_{x}+\partial _{y}F_{y}\right)dz+\partial _{x}{\vec {C_{1}}}(x,y)&=F_{z}\end{aligned}}$ By assumption, $\operatorname {div} {\vec {F}}=\partial _{x}F_{x}+\partial _{y}F_{y}+\partial _{z}F_{z}=0$, and so $\int \partial _{z}F_{z}dz+\partial _{x}{\vec {C_{1}}}(x,y)=F_{z}$ Since the fundamental theorem of calculus requires that the first term above be precisely $F_{z}$ plus a constant in z, a solution to the above system of equations is guaranteed to exist.

Having thus proved that the image of the curl is precisely the kernel of the divergence, this morphism in turn takes us back to the space we started from $L^{2}$. Since definitionally we have landed on a space of integrable functions, any such function can (at least formally) be integrated in order to produce a vector field which divergence is that function -- so the image of the divergence is the entirety of $L^{2}$, and we can complete our sequence:

$0\to L^{2}\;\;{\xrightarrow {\operatorname {grad} }}\;\;\mathbb {H} _{3}\;\;{\xrightarrow {\operatorname {curl} }}\;\;\mathbb {H} _{3}\;\;{\xrightarrow {\operatorname {div} }}\;\;L^{2}\to 0$

Equivalently, we could have reasoned in reverse: in a simply connected space, a curl-free vector field (a field in the kernel of the curl) can always be written as a gradient of a scalar function (and thus is in the image of the gradient). Similarly, a divergenceless field can be written as a curl of another field. (Reasoning in this direction thus makes use of the fact that 3-dimensional space is topologically trivial.)

This short exact sequence also permits a much shorter proof of the validity of the Helmholtz decomposition that does not rely on brute-force vector calculus. Consider the subsequence

$0\to L^{2}\;\;{\xrightarrow {\operatorname {grad} }}\;\;\mathbb {H} _{3}\;\;{\xrightarrow {\operatorname {curl} }}\;\;\operatorname {im} (\operatorname {curl} )\to 0.$

Since the divergence of the gradient is the Laplacian, and since the Hilbert space of square-integrable functions can be spanned by the eigenfunctions of the Laplacian, we already see that some inverse mapping $\nabla ^{-1}:\mathbb {H} _{3}\to L^{2}$ must exist. To explicitly construct such an inverse, we can start from the definition of the vector Laplacian

$\nabla ^{2}{\vec {A}}=\nabla (\nabla \cdot {\vec {A}})+\nabla \times (\nabla \times {\vec {A}})$

Since we are trying to construct an identity mapping by composing some function with the gradient, we know that in our case $\nabla \times {\vec {A}}=\operatorname {curl} ({\vec {A}})=0$. Then if we take the divergence of both sides

${\begin{aligned}\nabla \cdot \nabla ^{2}{\vec {A}}&=\nabla \cdot \nabla (\nabla \cdot {\vec {A}})\\&=\nabla ^{2}(\nabla \cdot {\vec {A}})\\\end{aligned}}$

we see that if a function is an eigenfunction of the vector Laplacian, its divergence must be an eigenfunction of the scalar Laplacian with the same eigenvalue. Then we can build our inverse function $\nabla ^{-1}$ simply by breaking any function in $\mathbb {H} _{3}$ into the vector-Laplacian eigenbasis, scaling each by the inverse of their eigenvalue, and taking the divergence; the action of $\nabla ^{-1}\circ \nabla$ is thus clearly the identity. Thus by the splitting lemma,

$\mathbb {H} _{3}\cong L^{2}\oplus \operatorname {im} (\operatorname {curl} )$,

or equivalently, any square-integrable vector field on $\mathbb {R} ^{3}$ can be broken into the sum of a gradient and a curl -- which is what we set out to prove.

Example: the circle

For example, $[0,1]/\{0,1\}$ is homeomorphic to the circle $S^{1}$.

A simple example is the construction of the circle $S^{1}$ as a quotient space of the unit interval $[0,1]$. Consider the unit interval with the subspace topology from $\mathbb {R}$, e.g. $\tau _{[0,1]}=\{U\cap [0,1]:U\in \tau _{\mathbb {R} }\}$. Introduce an equivalence relation $\sim$ by defining $a\sim b$ if and only if $|a-b|=1$, so that $0\sim 1$ but no other point is equivalent to any other. Then the quotient space $[0,1]/\sim$ is homeomorphic to $S^{1}$

Examples

A section $s$ of a bundle $p\colon E\to B$. A section $s$ allows the base space $B$ to be identified with a subspace $s(B)$ of $E$.

A vector field on $\mathbb {R} ^{2}$. A section of a tangent vector bundle is a vector field.

In the mathematical field of topology, a section (or cross section) of a fiber bundle $E$ is a continuous right inverse of the projection function $\pi$. In other words, if $E$ is a fiber bundle over a base space, $B$:

$\pi \colon E\to B$

then a section of that fiber bundle is a continuous map,

$\sigma \colon B\to E$

such that

$\pi (\sigma (x))=x$ for all $x\in B$.

A section is an abstract characterization of what it means to be a graph. The graph of a function $g\colon B\to Y$ can be identified with a function taking its values in the Cartesian product $E=B\times Y$, of $B$ and $Y$:

$\sigma (x)=(x,g(x))\in E,\ \sigma :B\to E$

Let $\pi \colon E\to B$ be the projection onto the first factor: $\pi (x,y)=x$. Then a graph is any function $\sigma$ for which $\pi (\sigma (x))=x$.

The language of fibre bundles allows this notion of a section to be generalized to the case when $E$ is not necessarily a Cartesian product. If $\pi \colon E\to B$ is a fibre bundle, then a section is a choice of point $\sigma (x)$ in each of the fibres. The condition $\pi (\sigma (x))=x$ simply means that the section at a point $x$ must lie over $x$. (See image.)

For example, when $\pi :E\to M$, often denoted just $E$, is a vector bundle over a smooth manifold $M$, a section $\sigma$of $E$ is a map from $M$to $E$such that, $\forall p\in M$, $\sigma _{p}:=\sigma (p)\in \pi ^{-1}(p)$, or equivalently that $\pi \circ \sigma (p)=p$. The set of all (global) sections of $E$is denoted $\Gamma (E)$. In particular, a vector field $X$on a smooth manifold $M$ is a choice of tangent vector $X_{p}\in T_{p}M$at each point $p\in M$: this is a section of the tangent bundleIn particular$TM:=\cup _{p\in M}T_{p}M$. Likewise, a 1-form $\tau$ on $M$is a section of the cotangent bundle $T^{*}M$, and higher k-forms are elements of $\Gamma (\wedge ^{k}T^{*}M)$, sections of the k^th exterior power of the cotangent bundle.

Sections, particularly of principal bundles and vector bundles, are also very important tools in differential geometry. In this setting, the base space $B$ is a smooth manifold $M$, and $E$ is assumed to be a smooth fiber bundle, often the tangent or cotangent bundle, such that $E$ is a smooth manifold with induced charts and topology, and $\pi \colon E\to M$ is a smooth map. The set of smooth sections of the tangent bundle $TM$(smooth vector fields) is often denoted ${\mathfrak {X}}(M)$, and the set of smooth sections of $\wedge ^{k}T^{*}M$(the smooth k-forms) is denoted $\Omega ^{k}(M)$In this case, one considers . It is also useful in geometric analysis to consider spaces of sections with intermediate regularity, such as $C^{k}$(k-times differentiable) sections, or sections with regularity in the sense of Hölder conditions or Sobolev spaces.

Local and global sections

Fiber bundles do not in general have such global sections (consider, for example, the fiber bundle over $S^{1}$ with fiber $F=\mathbb {R} \setminus \{0\}$ obtained by taking the Möbius bundle and removing the zero section), so it is also useful to define sections only locally. A local section of a fiber bundle is a continuous map $s\colon U\to E$ where $U$ is an open set in $B$ and $\pi (s(x))=x$ for all $x$ in $U$. If $(U,\varphi )$ is a local trivialization of $E$, where $\varphi$ is a homeomorphism from $\pi ^{-1}(U)$ to $U\times F$ (where $F$ is the fiber), then local sections always exist over $U$ in bijective correspondence with continuous maps from $U$ to $F$. The (local) sections form a sheaf over $B$ called the sheaf of sections of $E$.

The space of continuous sections of a fiber bundle $E$ over $U$ is sometimes denoted $C(U,E)$, while the space of global sections of $E$ is often denoted $\Gamma (E)$ or $\Gamma (B,E)$.

Extending to global sections

Sections are studied in homotopy theory and algebraic topology, where one of the main goals is to account for the existence or non-existence of global sections. An obstruction denies the existence of global sections since the space is too "twisted". More precisely, obstructions "obstruct" the possibility of extending a local section to a global section due to the space's "twistedness". Obstructions are indicated by particular characteristic classes, which are cohomological classes. For example, a principal bundle has a global section if and only if it is trivial. On the other hand, a vector bundle always has a global section, namely the zero section. However, it only admits a nowhere vanishing section if its Euler class is zero.

Generalizations

Obstructions to extending local sections may be generalized in the following manner: take a topological space and form a category whose objects are open subsets, and morphisms are inclusions. Thus we use a category to generalize a topological space. We generalize the notion of a "local section" using sheaves of abelian groups, which assigns to each object an abelian group (analogous to local sections).

There is an important distinction here: intuitively, local sections are like "vector fields" on an open subset of a topological space. So at each point, an element of a fixed vector space is assigned. However, sheaves can "continuously change" the vector space (or more generally abelian group).

This entire process is really the global section functor, which assigns to each sheaf its global section. Then sheaf cohomology enables us to consider a similar extension problem while "continuously varying" the abelian group. The theory of characteristic classes generalizes the idea of obstructions to our extensions.

See also

• Fibration
• Gauge theory
• Principal bundle
• Pullback bundle
• Vector bundle

References

• Norman Steenrod, The Topology of Fibre Bundles, Princeton University Press (1951). ISBN 0-691-00548-6.
• David Bleecker, Gauge Theory and Variational Principles, Addison-Wesley publishing, Reading, Mass (1981). ISBN 0-201-10096-7.
• Husemöller, Dale (1994), Fibre Bundles, Springer Verlag, ISBN 0-387-94087-1

External links

• Fiber Bundle, PlanetMath
• Weisstein, Eric W. "Fiber Bundle". MathWorld.