Field of Fractions

edit_note

This post is being revised. The text below is the version as of 2026-08-18, so some statements may be out of date and references to this post from other posts may not line up.

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Fraction Rings

In [Set Theory] §Natural Numbers and Infinite Sets we defined the natural numbers as a monoid \(\mathbb{N}\), which (aside from some technical issues) could be expressed in the language of set theory. Then \(\mathbb{Z}\) was defined as the Grothendieck group of the commutative monoid \(\mathbb{N}\). Thinking back to the number systems learned in middle school, the next object we should define is the set of rational numbers \(\mathbb{Q}\).

If we forget the additive structure of \(\mathbb{Z}\) and remember only the multiplicative structure, then \((\mathbb{Z},\cdot,1)\) is a commutative monoid. What we need to do is adjoin inverses, and since \(1/0\) is undefined, we set \(S=\mathbb{Z}\setminus\{0\}\) and consider the monoid of fractions from §Grothendieck Groups, ⁋Definition 7 to obtain the multiplicative monoid \(\mathbb{Q}\).

In general, this process is made possible by the following theorem.

Theorem 1 Let \(A\) be a commutative ring and \(S\) a subset of \(A\). Regarding \(A\) as a multiplicative monoid, consider the monoid of fractions \(A_S\). Then for the canonical morphism \(\epsilon:A \rightarrow A_S\), there exists a unique additive structure satisfying the following two conditions:

1. The multiplicative structure on \(A_S\) together with this additive structure makes \(A_S\) a commutative ring.
2. \(\epsilon\) is a ring homomorphism.

Proof

Before beginning the proof, let us briefly review the construction from §Grothendieck Groups, ⁋Definition 7. We consider the submonoid \(S'\) of \(A\) generated by \(S\), and define an equivalence relation \(R\) on the monoid \(A\times S'\) by

\[(\alpha,\gamma)\equiv (\beta,\delta)\pmod{R}\iff \alpha\delta\zeta=\beta\gamma\zeta\text{ for some $\zeta\in S'$}\]

and define the quotient monoid \((A\times S')/R\) as \(A_S\). At this point, an element of \(A_S\) having \((\alpha,\gamma)\in A\times S'\) as a representative is denoted \(\alpha/\gamma\), and then \(A_S\) still becomes a multiplicative monoid via the operation

\[\frac{\alpha}{\gamma}\frac{\beta}{\delta}=\frac{\alpha\beta}{\gamma\delta}\]

The canonical morphism \(\epsilon:A \rightarrow A_S\) between two multiplicative monoids was defined by \(\alpha\mapsto \alpha/1\), and the fact that this is a monoid homomorphism means that \(\epsilon\) is a function preserving the multiplicative structure (once we show that \(A_S\) is a ring) from \(A\) to \(A_S\).

That is, what we need to do is endow \(A_S\) with an additive structure satisfying the two conditions of the theorem, show that this defined additive structure makes \(A_S\) a ring, and that \(\epsilon\) actually preserves this additive structure as well.

First, assuming that such an additive structure exists, let us show uniqueness. Any \(x,y\in A_S\) can be written as \(x=\alpha/\gamma,y=\beta/\delta\) for suitable \(\alpha,\beta\in A\) and \(\gamma,\delta\in S'\). Then

\[x=\epsilon(\alpha)\epsilon(\gamma)^{-1}=\epsilon(\alpha\delta)\epsilon(\gamma\delta)^{-1},\qquad y=\epsilon(\beta)\epsilon(\delta)^{-1}=\epsilon(\beta\gamma)\epsilon(\gamma\delta)^{-1}\]

so we must have

\[x+y=(\epsilon(\alpha\delta)+\epsilon(\beta\gamma))\epsilon(\gamma\delta)^{-1}=\frac{\alpha\delta+\beta\gamma}{\gamma\delta}\]

Now, taking a hint from the uniqueness proof, we define the additive structure on \(A_S\) by the above formula. Then what we need to show is as follows.

1. This definition is independent of the choice of \(\alpha,\beta,\gamma,\delta\). That is, suppose \(x,y\) are written in the form \(x=\alpha'/\gamma',y=\beta'/\delta'\). We need to show that the equality

   \[(\alpha\delta+\beta\gamma)/\delta\gamma=(\alpha'\delta'+\beta'\gamma')/\gamma'\delta'\]

   holds in \(A_S\). But since \(\alpha/\gamma=\alpha'/\gamma'\) and \(\beta/\delta=\beta'/\delta'\), by definition there exist \(\zeta,\xi\in S'\) satisfying \(\alpha\gamma'\zeta=\alpha'\gamma\zeta\) and \(\beta\delta'\xi=\beta'\delta\xi\). From this we can verify

   \[(\alpha\delta+\beta\gamma)(\gamma'\delta')(\zeta\xi)=(\alpha'\delta'+\beta'\gamma')(\gamma\delta)(\zeta\xi)\]

   so the desired equality holds.

2. The \(+\) defined this way is associative. For any \(x_1=\alpha_1/\gamma_1,x_2=\alpha_2/\gamma_2,x_3=\alpha_3/\gamma_3\),

   \[(x_1+x_2)+x_3=\frac{\alpha_1\gamma_2+\alpha_2\gamma_1}{\gamma_1\gamma_2}+\frac{\alpha_3}{\gamma_3}=\frac{(\alpha_1\gamma_2+\alpha_2\gamma_1)\gamma_3+\alpha_3(\gamma_1\gamma_2)}{\gamma_1\gamma_2\gamma_3}=\frac{\alpha_1\gamma_2\gamma_3+\gamma_1\alpha_2\gamma_3+\gamma_1\gamma_2\alpha_3}{\gamma_1\gamma_2\gamma_3}\]

   and similarly one can check that \(x_1+(x_2+x_3)\) also has the value on the right-hand side.

3. The commutativity of \(+\) is immediate because addition and multiplication in \(A\) are commutative.

4. \(+\) has an additive identity \(0/1\). This is because for any \(x=\alpha/\gamma\in A_S\),

   \[\frac{0}{1}+\frac{\alpha}{\gamma}=\frac{\alpha}{\gamma}\]

   holds.

5. Additive inverses always exist. For any \(x=\alpha/\gamma\in A_S\), \((-\alpha)/\gamma\) satisfies the equality

   \[\frac{-\alpha}{\gamma}+\frac{\alpha}{\gamma}=\frac{(-\alpha)\gamma+\alpha\gamma}{\gamma^2}=0\]

6. \(+\) satisfies the distributive law with respect to multiplication. For any \(x=\alpha/\gamma,y_1=\beta_1/\delta_1,y_2=\beta_2/\delta_2\),

   \[x(y_1+y_2)=\frac{\alpha}{\gamma}\left(\frac{\beta_1}{\delta_1}+\frac{\beta_2}{\delta_2}\right)=\frac{\alpha}{\gamma}\frac{\beta_1\delta_2+\delta_1\beta_2}{\delta_1\delta_2}=\frac{\alpha\beta_1\delta_2+\alpha\delta_1\beta_2}{\gamma\delta_1\delta_2}\]

   and

   \[xy_1+xy_2=\frac{\alpha\beta_1}{\gamma\delta_1}+\frac{\alpha\beta_2}{\gamma\delta_2}=\frac{\alpha\beta_1\gamma\delta_2+\alpha\beta_2\gamma\delta_1}{\gamma^2\delta_1\delta_2}\]

   and using \(1,\gamma\in S'\) we can verify that these two expressions have the same value. Similarly one can show the equality \((x_1+x_2)y=x_1y+x_2y\).

From the above we see that \(A_S\) has a commutative ring structure. Finally, that \(\epsilon\) is a ring homomorphism is sufficient to show that \(\epsilon\) preserves addition, and this follows from

\[\epsilon(\alpha+\beta)=(\alpha+\beta)/1=\alpha/1+\beta/1=\epsilon(\alpha)+\epsilon(\beta)\]

Definition 2 The ring obtained as above is called the ring of fractions of \(A\) defined by \(S\), and is denoted \(S^{-1}A\).

If \(S\) was the set of cancellable elements of \(A\), then it is immediate that \(\epsilon\) is injective, and thus we may regard \(A\) as a subring of \(S^{-1}A\). In this case we call \(S^{-1}A\) the total ring of fractions of \(A\).

Fields

The rational numbers \(\mathbb{Q}\) have the following distinguishing feature from general rings.

Definition 3 A ring \(A\) is called a division ring if \(A\neq0\) and every nonzero element of \(A\) has a multiplicative inverse. A commutative division ring is called a field.

Proposition 4 For a ring \(A\neq 0\), the following are equivalent: \(A\) is a division ring, and the only left ideals of \(A\) are \(0\) and \(A\).

Proof

First suppose \(A\) is a division ring. If a left ideal \(\mathfrak{a}\neq 0\) is given, then there exists \(0\neq x\in \mathfrak{a}\). But in \(A\) the inverse \(x^{-1}\) of \(x\) exists, so

\[1=x^{-1}x\in A\mathfrak{a}=\mathfrak{a}\]

and thus \(\mathfrak{a}=A\). Conversely, suppose the only left ideals of \(A\) are \(0\) and \(A\). For any \(0\neq x\in A\), consider the left ideal \(Ax\) of \(A\); then \(0\neq x\in Ax\) so \(Ax\neq 0\). Since the left ideals of \(A\) are only \(0\) or \(A\), we must have \(Ax=A\), and thus \(1\in Ax\). That is, there exists suitable \(\alpha\in A\) such that \(\alpha x=1\). Then \(\alpha\neq 0\), and by the same logic there exists suitable \(\beta\in A\) such that \(\beta\alpha=1\). Now from

\[\beta=\beta1=\beta\alpha x=x\]

we see that \(\beta=x\), and thus \(\alpha\) is the multiplicative inverse of \(x\).

Integral Domains

By its definition, \(\mathbb{Q}\) is the total ring of fractions of \(\mathbb{Z}\). That this becomes a field is immediate from the definition, and this can be extended as follows.

Definition 5 If elements \(\alpha,\beta\) of a ring \(A\) satisfy \(\alpha\beta=0\) but \(\alpha\neq 0\) and \(\beta\neq 0\), we call \(\alpha,\beta\) zerodivisors. A ring \(A\) is called an integral domain if \(A\) is commutative, \(0\neq 1\), and \(A\) has no zerodivisors.

By definition, it is immediate that a subring of an integral domain is an integral domain. For any nonzero rings \(A,B\), the product \(A\times B\) can never be an integral domain, as seen from the equality

\[(1,0)(0,1)=(0,0)\]

Proposition 6 The total ring of fractions of an integral domain \(A\) is a field.

Proof

From the assumption that \(A\) is an integral domain, we know that \(S=A\setminus\{0\}\). That is, any element of \(S^{-1}A\) can be written in the form \(\alpha/\beta\) for \(\alpha\in A\), \(\beta\in A\setminus\{0\}\). Here, for \(\alpha/\beta\neq 0\) we must have \(\alpha\neq 0\), so \(\beta/\alpha\in S^{-1}A\) is also well defined and then \(\beta/\alpha\) becomes the inverse of \(\alpha/\beta\). On the other hand, since \(A\) is an integral domain we have \(1/1\neq 0/1\), and in particular \(S^{-1}A\neq 0\).

Definition 7 The field \(S^{-1}A\) obtained from Proposition 6 above is called the field of fractions of \(A\), and is denoted \(\Frac(A)\).

Prime Ideals

From the fourth isomorphism theorem for ring homomorphisms, we know that for any commutative ring \(A\neq 0\) and maximal ideal \(\mathfrak{m}\), the only ideals of \(A/\mathfrak{m}\) are \(0\) and \(A/\mathfrak{m}\) itself. Thus by Proposition 4, \(A/\mathfrak{m}\) is a division ring, that is, a field. Integral domains can also be characterized in a similar manner.

Proposition 8 For a commutative ring \(A\) and an ideal \(\mathfrak{p}\neq A\), the following are all equivalent.

1. \(A/\mathfrak{p}\) is an integral domain.
2. If \(\alpha,\beta\in A\setminus \mathfrak{p}\), then \(\alpha\beta\in A\setminus \mathfrak{p}\).
3. If \(\alpha\beta\in \mathfrak{p}\), then \(\alpha\in \mathfrak{p}\) or \(\beta\in \mathfrak{p}\).

Proof

Conditions 2 and 3 are contrapositives of each other, so it suffices to show equivalence with condition 1. First assume \(A/\mathfrak{p}\) is an integral domain. That is, if

\[(\alpha+\mathfrak{p})(\beta+\mathfrak{p})=0+\mathfrak{p}\]

then necessarily \(\alpha+\mathfrak{p}=0+\mathfrak{p}\) or \(\beta+\mathfrak{p}=0+\mathfrak{p}\). From this we see that if condition 1 holds then condition 3 holds. This argument also works in the reverse direction.

An ideal \(\mathfrak{p}\) satisfying the above equivalent conditions is called a prime ideal. Since every field is an integral domain, every maximal ideal is a prime ideal. The converse does not hold; for example, one can easily check that the prime ideals of \(\mathbb{Z}\) are only \((0)\) and those of the form \(p\mathbb{Z}\) for a prime number \(p\). Then \((0)\) is a prime ideal but not a maximal ideal.

On the other hand, the following holds.

Proposition 9 For a ring homomorphism \(\phi:A \rightarrow B\) between commutative rings and a prime ideal \(\mathfrak{p}\) of \(B\), the preimage \(\phi^{-1}(\mathfrak{p})\) is a prime ideal of \(A\).

Proof

First, \(\phi^{-1}(\mathfrak{p})\) is a subgroup under addition, and for any \(\alpha\in A\) and \(x\in\phi^{-1}(\mathfrak{p})\) we have \(\phi(\alpha x)=\phi(\alpha)\phi(x)\in\mathfrak{p}\), so it is an ideal of \(A\). Also since \(\mathfrak{p}\neq B\) we have \(1\not\in\mathfrak{p}\), and from \(\phi(1)=1\) we obtain \(1\not\in\phi^{-1}(\mathfrak{p})\), that is, \(\phi^{-1}(\mathfrak{p})\neq A\). Now suppose for contradiction that there exist \(\alpha,\beta\in A\) with \(\alpha\beta\in\phi^{-1}(\mathfrak{p})\) but \(\alpha,\beta\not\in\phi^{-1}(\mathfrak{p})\); then \(\phi(\alpha)\phi(\beta)=\phi(\alpha\beta)\in \mathfrak{p}\) but \(\phi(\alpha),\phi(\beta)\not\in \mathfrak{p}\), contradicting the equivalence of Proposition 8.

On the other hand, by the second equivalence of Proposition 8, if we regard a commutative ring \(A\) as a multiplicative monoid, then for its prime ideal \(\mathfrak{p}\) the set \(A\setminus\mathfrak{p}\) can be viewed as a submonoid of \(A\). Thus the ring of fractions \((A\setminus \mathfrak{p})^{-1}A\) is well defined, and the only elements appearing in the denominators of this ring are those of \(A\setminus \mathfrak{p}\). This is defined as follows.

Definition 10 For a commutative ring \(A\) and a prime ideal \(\mathfrak{p}\), we define the localization of \(A\) at \(\mathfrak{p}\) as \((A\setminus \mathfrak{p})^{-1}A\), and denote it simply by \(A_\mathfrak{p}\).

Nilpotents

Definition 11 An element \(\alpha\) of a ring \(A\) is nilpotent if there exists \(n>0\) such that \(\alpha^n=0\). If \(A\) has no nonzero nilpotent elements, we call \(A\) reduced.

By definition, a nonzero nilpotent element is a zerodivisor. Thus every integral domain is a (commutative) reduced ring. Moreover, if we restrict to commutative rings, we obtain the following.

Proposition 12 For a commutative ring \(A\), the set \(\mathfrak{N}\) of nilpotent elements is an ideal.

Proof

If \(x\in \mathfrak{N}\), then there exists \(n>0\) such that \(x^n=0\), and for any \(\alpha\in A\) we have \((\alpha x)^n=\alpha^nx^n=0\), so we can show that \(\alpha x\in \mathfrak{N}\).

Now we must show that \(\mathfrak{N}\) is closed under addition. Given arbitrary \(x,y\in \mathfrak{N}\), suppose \(x^m=0\) and \(y^n=0\) for suitable \(m,n>0\). Then

\[(x+y)^{m+n}=x^{m+n}+\binom{m+n}{1}x^{m+n-1}y+\cdots+\binom{m+n}{n}x^my^n+\binom{m+n}{n+1}x^{m-1}y^{n+1}+\cdots+y^{m+n}\]

and we can see that all terms on the right-hand side are \(0\). From the above, \(x+y\in \mathfrak{N}\).

Definition 13 The ideal \(\mathfrak{N}\) of Proposition 12 is called the nilradical of \(A\).

By definition, \(A\) being reduced is equivalent to the nilradical of \(A\) being \(0\). On the other hand, if \(x\in \mathfrak{N}\), then from the equation \(x^n=0\) and the definition of prime ideal we see that \(x\in \mathfrak{p}\) holds for every prime ideal \(\mathfrak{p}\). That is, the inclusion

\[\mathfrak{N}\subseteq\bigcap_\text{\scriptsize$\mathfrak{p}$: prime} \mathfrak{p}\]

holds.

Proposition 14 For a commutative ring \(A\) and its nilradical \(\mathfrak{N}\),

\[\mathfrak{N}=\bigcap_\text{\scriptsize$\mathfrak{p}$: prime} \mathfrak{p}\]

holds.

Proof

If \(x\not\in \mathfrak{N}\), it suffices to show that \(x\not\in \mathfrak{p}\) for some suitable \(\mathfrak{p}\). First consider the ring of fractions \(A_x=S^{-1}A\) formed from the multiplicative subset \(S=\{1,x,x^2,\ldots\}\). Then the multiplicative identity \(x/x\) of \(A_x\) must be different from \(0/1\), because \(x/x=0/1\) is equivalent to \(x^{k+1}=0\) for some \(k\geq 0\), which contradicts the assumption that \(x\not\in\mathfrak{N}\). In particular \(A_x\neq 0\). Now from §Definition of a Ring, ⁋Theorem 10 (Krull), a maximal ideal \(\mathfrak{m}\) of \(A_x\) must exist, and since every maximal ideal is a prime ideal, \(\mathfrak{m}\) is a prime ideal of \(A_x\). Applying Proposition 9 to \(\epsilon:A \rightarrow A_x\), we see that \(\epsilon^{-1}(\mathfrak{m})\) is a prime ideal of \(A\), and if \(x\in\epsilon^{-1}(\mathfrak{m})\) then \(x/1\in \mathfrak{m}\) and \(x/1\) is invertible in \(A_x\), so we would have \(\mathfrak{m}=A_x\), a contradiction.

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.

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