K

Grad. student in mathematics

Ring of Fractions

The monoid \(\mathbb{N}\) of natural numbers defined in [Set Theory] §Natural Numbers and Infinite Sets could be written in the language of set theory (with some minor technical issues). Then \(\mathbb{Z}\) was defined as the Grothendieck group of the commutative monoid \(\mathbb{N}\). Recalling the number systems learned in middle school, the next object to 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 add inverses, and since \(1/0\) is undefined, we set \(S=\mathbb{Z}\setminus\{0\}\) and consider the monoid of fractions as in §Grothendieck Groups, ⁋Definition 7, obtaining the multiplicative group \(\mathbb{Q}\).

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 \(A_S\) as the quotient monoid \((A\times S')/R\). Here, an element of \(A_S\) having \((\alpha,\gamma)\in A\times S'\) as a representative is denoted by \(\alpha/\gamma\), and \(A_S\) carries a multiplicative monoid structure 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 this being a monoid homomorphism means that \(\epsilon\) is a function from \(A\) to \(A_S\) that preserves the multiplicative structure (after we show that \(A_S\) is a ring).

Thus, what we need to do is to give \(A_S\) an additive structure satisfying the two conditions of the theorem, show that this additive structure makes \(A_S\) a ring, and verify that \(\epsilon\) indeed preserves this additive structure as well.

First, assuming such an additive structure exists, let us prove uniqueness. Any \(x,y\in A_S\) can be written as \(x=\alpha/\gamma,y=\beta/\delta\) for some \(\alpha,\beta\in A\) 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. What we need to show is:

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

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

    holds in \(A_S\). Since \(\alpha/\gamma=\alpha'/\gamma',\beta/\delta=\beta'/\delta'\), 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 satisfies the associative law. 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 verify that \(x_1+(x_2+x_3)\) has the same value as the right-hand side.

  3. The commutative law for \(+\) is clear since the addition and multiplication of \(A\) are commutative.

  4. \(+\) has the 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 \(x\in \alpha/\gamma\in A_S\), the element \((-\alpha)/\gamma\) satisfies

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

  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 are equal. Similarly, the equality \((x_1+x_2)y=x_1y+x_2y\) can be shown.

From the above, we conclude that \(A_S\) has a commutative ring structure. Finally, to show that \(\epsilon\) is a ring homomorphism, it suffices to show that \(\epsilon\) preserves addition, which 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 by \(S^{-1}A\).

If \(S\) is the collection of cancellable elements of \(A\), then it is clear that \(\epsilon\) is an injection, so we can regard \(A\) as a subring of \(S^{-1}A\). In this case, \(S^{-1}A\) is called the total ring of fractions of \(A\).

Fields

The rational numbers \(\mathbb{Q}\) have the following distinguishing characteristic 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 A nonzero ring \(A\) is a division ring if and only if 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 \(0\neq x\in \mathfrak{a}\). Since in \(A\) the inverse \(x^{-1}\) of \(x\) exists,

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

so \(\mathfrak{a}=A\). 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 \(A\). Since \(0\neq x\in Ax\), we have \(Ax\neq 0\). Since the only left ideals of \(A\) are \(0\) or \(A\), we must have \(Ax=A\), so \(1\in Ax\). That is, there exists \(\alpha\in A\) such that \(\alpha x=1\). Then \(\alpha\neq 0\), and by the same argument, there exists \(\beta\in A\) such that \(\beta\alpha=1\). Now

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

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

Integral Domains

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

Definition 5 Elements \(\alpha,\beta\) of a ring \(A\) are called zero divisors if \(\alpha\beta=0\) but \(\alpha\neq 0\) and \(\beta\neq 0\). A ring \(A\) is called an integral domain if \(A\) is commutative, \(0\neq 1\), and \(A\) has no zero divisors.

By definition, it is clear 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 since

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

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 have \(S=A\setminus\{0\}\). Thus, any element of \(S^{-1}A\) can be written as \(\alpha/\beta\) for some \(\alpha\in A\) and \(\beta\in A\setminus\{0\}\). Here, \(\alpha/\beta\neq 0\) requires \(\alpha\neq 0\), so \(\beta/\alpha\in K\) is well-defined, and \(\beta/\alpha\) becomes the inverse of \(\alpha/\beta\).

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

Prime Ideals

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

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

The second and third conditions are contrapositives of each other, so it suffices to show equivalence with the first. 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 verify that the prime ideals of \(\mathbb{Z}\) are only \((0)\) and those of the form \(p\mathbb{Z}\) for prime numbers \(p\). Thus \((0)\) 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 \(A,B\) and a prime ideal \(\mathfrak{p}\) of \(B\), the preimage \(\phi^{-1}(\mathfrak{p})\) is a prime ideal of \(A\).

Proof

Assume for contradiction that there exist \(\alpha,\beta\in A\) such that \(\alpha\beta\in\phi^{-1}(\mathfrak{p})\) 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 in Proposition 8.

On the other hand, by condition 2 of the equivalence in Proposition 8, if we regard a commutative ring \(A\) as a multiplicative monoid, then for any prime ideal \(\mathfrak{p}\), the complement \(A\setminus\mathfrak{p}\) can be viewed as a submonoid of \(A\). Thus the ring of fractions \((A\setminus \mathfrak{p})^{-1}A\) is well-defined, and only elements of \(A\setminus \mathfrak{p}\) appear in the denominators of this ring. We make the following definition.

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

Nilpotent Elements

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

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

Proposition 12 For a commutative ring \(A\), the collection \(\mathfrak{N}\) of nilpotent elements forms 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 \((\alpha x)^n=\alpha^nx^n=0\), showing that \(\alpha x\in \mathfrak{N}\).

Now we need to show that \(\mathfrak{N}\) is closed under addition. Let \(x,y\in \mathfrak{N}\) be given, and suppose \(x^m=0\) and \(y^n=0\) for some \(m,n>0\). Then

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

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

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

By definition, \(A\) is reduced if and only if the nilradical of \(A\) is \(0\). On the other hand, if \(x\in \mathfrak{N}\), then from the equation \(x^n=0\) and the definition of prime ideals, we see that \(x\in \mathfrak{p}\) for all prime ideals \(\mathfrak{p}\). 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 there exists \(\mathfrak{p}\) such that \(x\not\in \mathfrak{p}\). First, consider the ring of fractions \(A_x=S^{-1}A\) where \(S=\{1,x,x^2,\ldots\}\). The multiplicative identity \(x/x\) of \(A_x\) is necessarily different from \(0/1\), so in particular \(A_x\neq 0\). By §Definition of Rings, ⁋Theorem 9, \(A_x\) has a maximal ideal \(\mathfrak{m}\), and since every maximal ideal is a prime ideal, \(A_x\) has a prime ideal. Now applying Proposition 9 to \(\epsilon:A \rightarrow A_x\), we see that \(\epsilon^{-1}(\mathfrak{p})\) is a prime ideal of \(A\), and if \(x\in\epsilon^{-1}(\mathfrak{p})\), then \(x/1\in \mathfrak{p}\), and since \(x/1\) is invertible in \(A_x\), we have \(\mathfrak{p}=A_x\), a contradiction.

References

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

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