K

Grad. student in mathematics

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

Ring of Fractions

In [Set Theory] §Natural Numbers and Infinite Sets, we defined the monoid \(\mathbb{N}\) of natural numbers, which (apart from some technical issues) could be written 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 we learned in middle school, the next object we need to define is the set of rational numbers \(\mathbb{Q}\).

If we forget the additive structure of \(\mathbb{Z}\) and remember only its 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 group \(\mathbb{Q}\).

In general, this process is possible through the following theorem.

Theorem 1 Consider a commutative ring \(A\) and a subset \(S\) of \(A\). Viewing \(A\) 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 \(A_S\). We denote the element of \(A_S\) with representative \((\alpha,\gamma)\in A\times S'\) by \(\alpha/\gamma\); 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).

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 verify 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 we can write

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

and therefore 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 \(A_S\). Since \(\alpha/\gamma=\alpha'/\gamma'\) and \(\beta/\delta=\beta'/\delta'\), by definition there exist \(\zeta,\xi\in S'\) satisfying \(a p' s = a' p s\) and \(b q' t = b' q t\). From this we can verify

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

    and thus the desired equality holds.

  2. The \(+\) defined this way satisfies associativity. 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. Commutativity of \(+\) is obvious 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 \(+\). For any \(x=\alpha/\gamma\in A_S\), \((-\alpha)/\gamma\) satisfies

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

  6. \(+\) satisfies distributivity over 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 \(1,\gamma\in S'\) one can check that these two expressions have the same value. Similarly the equality \((x_1+x_2)y=x_1y+x_2y\) can be shown.

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 by \(S^{-1}A\).

If \(S\) was the set of cancellable elements of \(A\), then it is obvious that \(\epsilon\) is an injection, and thus we can think of \(A\) 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 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 \(0\neq x\in \mathfrak{a}\). Since the inverse \(x^{-1}\) of \(x\) exists in \(A\),

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

and therefore \(\mathfrak{a}=A\). 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 therefore \(1\in Ax\). 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 therefore \(\alpha\) is the multiplicative inverse of \(x\).

Integral Domains

\(\mathbb{Q}\) is by its definition the total ring of fractions of \(\mathbb{Z}\). That this is a field is obvious by 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 \(\beta\neq 0\), then \(\alpha,\beta\) are called 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 obvious 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 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 \(\beta/\alpha\in K\) is also well-defined, and then \(\beta/\alpha\) becomes the inverse of \(\alpha/\beta\).

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

Prime Ideals

From the fourth isomorphism theorem for ring homomorphisms, we know that for any ring \(A\neq 0\) and maximal left ideal \(\mathfrak{m}\), the only left ideals of \(A/\mathfrak{m}\) are \(0\) and \(A/\mathfrak{m}\) itself. Therefore 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

Conditions 2 and 3 are contrapositives of each other, so it is sufficient to show equivalence with condition 1. First suppose \(A/\mathfrak{p}\) 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.

Meanwhile, 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 \(B\), \(\phi^{-1}(\mathfrak{p})\) is a prime ideal of \(A\).

Proof

Suppose 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.

Meanwhile, by the second equivalence in Proposition 8, if we view 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 \(A\). Therefore 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 write it simply as \(A_\mathfrak{p}\).

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. Therefore 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 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 \((\alpha x)^n=\alpha^n x^n=0\), so we can show that \(\alpha x\in \mathfrak{N}\).

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

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

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

By definition, \(A\) being reduced is equivalent to the nilradical of \(A\) being \(0\). Meanwhile, 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 is sufficient to show that \(x\not\in \mathfrak{p}\) for some \(\mathfrak{p}\). First consider the ring of fractions \(A_x=S^{-1}A\) with multiplicative subset \(S=\{1,x,x^2,\ldots\}\). Then we can check that the multiplicative identity \(x/x\) of \(A_x\) is necessarily different from \(0/1\), and in particular \(A_x\neq 0\). Now from [Definition of Rings] §…, ⁋Theorem 9, a maximal ideal \(\mathfrak{m}\) of \(A_x\) must exist, 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 \(x/1\in \mathfrak{p}\); but \(x/1\) is invertible in \(A_x\), so we would have \(\mathfrak{p}=A_x\), a contradiction.

References

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

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