Inabstract algebra, thefield of fractions of anintegral domain is the smallestfield in which it can beembedded. The construction of the field of fractions is modeled on the relationship between the integral domain ofintegers and the field ofrational numbers. Intuitively, it consists of ratios between integral domain elements.

The field of fractions of an integral domain${\displaystyle R}$ is sometimes denoted by${\displaystyle \mathrm{Frac}(R)}$ or${\displaystyle \mathrm{Quot}(R)}$, and the construction is sometimes also called thefraction field,field of quotients, orquotient field of${\displaystyle R}$. All four are in common usage, but are not to be confused with thequotient of a ring by an ideal, which is a quite different concept. For acommutative ring that is not an integral domain, the analogous construction is called thelocalization or ring of quotients.

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

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

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

and multiplication given by

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

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

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

The field of fractions of${\displaystyle R}$ is characterized by the followinguniversal property:

if${\displaystyle h:R\to F}$ is aninjectivering homomorphism from${\displaystyle R}$ into a field${\displaystyle F}$, then there exists a unique ring homomorphism${\displaystyle g:\mathrm{Frac}(R)\to F}$ that extends${\displaystyle h}$.

There is acategorical interpretation of this construction. Let${\displaystyle \mathbf{C}}$ be thecategory of integral domains and injective ring maps. Thefunctor from${\displaystyle \mathbf{C}}$ to thecategory of fields that takes every integral domain to its fraction field and every homomorphism to the induced map on fields (which exists by the universal property) is theleft adjoint of theinclusion functor from the category of fields to${\displaystyle \mathbf{C}}$. Thus the category of fields (which is a full subcategory) is areflective subcategory of${\displaystyle \mathbf{C}}$.

Amultiplicative identity is not required for the role of the integral domain; this construction can be applied to anynonzero commutativerng${\displaystyle R}$ with no nonzerozero divisors. The embedding is given by${\displaystyle r\mapsto \frac{rs}{s}}$ for any nonzero${\displaystyle s\in R}$.^[1]

• The field of fractions of the ring ofintegers is the field ofrationals:${\displaystyle \mathbb{Q}=\mathrm{Frac}(\mathbb{Z})}$.
• Let${\displaystyle R:=\{a+b\mathrm{i}\mid a,b\in \mathbb{Z}\}}$ be the ring ofGaussian integers. Then${\displaystyle \mathrm{Frac}(R)=\{c+d\mathrm{i}\mid c,d\in \mathbb{Q}\}}$, the field ofGaussian rationals.
• The field of fractions of a field is canonicallyisomorphic to the field itself.
• Given a field${\displaystyle K}$, the field of fractions of thepolynomial ring in one indeterminate${\displaystyle K[X]}$ (which is an integral domain), is called thefield of rational functions,field of rational fractions, orfield of rational expressions^[2][3][4][5] and is denoted${\displaystyle K(X)}$.
• The field of fractions of theconvolution ring of half-line functions yields a space of operators, including theDirac delta function,differential operator, andintegral operator. This construction gives an alternate representation of theLaplace transform that does not depend explicitly on an integral transform.^[6]

For anycommutative ring${\displaystyle R}$ and anymultiplicative set${\displaystyle S}$ in${\displaystyle R}$, thelocalization${\displaystyle S^{-1}R}$ is thecommutative ring consisting offractions

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

with${\displaystyle r\in R}$ and${\displaystyle s\in S}$, where now${\displaystyle (r,s)}$ is equivalent to${\displaystyle (r^{\prime },s^{\prime })}$ if and only if there exists${\displaystyle t\in S}$ such that${\displaystyle t(r{s}^{\prime }-r^{\prime }s)=0}$.

Two special cases of this are notable:

• If${\displaystyle S}$ is the complement of aprime ideal${\displaystyle P}$, then${\displaystyle S^{-1}R}$ is also denoted${\displaystyle R_P}$.
  When${\displaystyle R}$ is anintegral domain and${\displaystyle P}$ is the zero ideal,${\displaystyle R_P}$ is the field of fractions of${\displaystyle R}$.
• If${\displaystyle S}$ is the set of non-zero-divisors in${\displaystyle R}$, then${\displaystyle S^{-1}R}$ is called thetotal quotient ring.
  Thetotal quotient ring of anintegral domain is its field of fractions, but thetotal quotient ring is defined for anycommutative ring.

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

Thesemifield of fractions of acommutative semiring in which every nonzero element is (multiplicatively) cancellative is the smallestsemifield in which it can beembedded. (Note that, unlike the case of rings, a semiring with nozero divisors can still have nonzero elements that are not cancellative. For example, let${\displaystyle \mathbb{T}}$ denote thetropical semiring and let${\displaystyle R=\mathbb{T}[X]}$ be thepolynomial semiring over${\displaystyle \mathbb{T}}$. Then${\displaystyle R}$ has no zero divisors, but the element${\displaystyle 1+X}$ is not cancellative because${\displaystyle (1+X)(1+X+X^2)=1+X+X^2+X^3=(1+X)(1+X^2)}$).

The elements of the semifield of fractions of the commutativesemiring${\displaystyle R}$ areequivalence classes written as

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

with${\displaystyle a}$ and${\displaystyle b}$ in${\displaystyle R}$ and${\displaystyle b\ne 0}$.

• Ore condition; condition related to constructing fractions in the noncommutative case.
• Total ring of fractions

• Field (mathematics)
• Commutative algebra

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