Field of Fractions

Fraction Rings

The monoid of natural numbers $\mathbb{N}$ defined in [Set Theory] §Natural Numbers and Infinite Sets could (aside from some technical issues) be written in the language of set theory. And $\mathbb{Z}$ 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 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 reciprocals, and since $1/0$ is undefined, if we set $S=\mathbb{Z}\setminus\{0\}$ and consider the monoid of fractions from §Grothendieck Group, ⁋Definition 7, we obtain the multiplicative monoid $\mathbb{Q}$.

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

Theorem 1 Consider a commutative ring $A$ and, in $A$, a subset $S$. 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:

The multiplicative structure on $A_S$ together with this additive structure makes $A_S$ a commutative ring.
$\epsilon$ is a ring homomorphism.

Proof

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

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

to define the quotient monoid $(A\times S')/R$ as $A_S$. Here, with $(\alpha,\gamma)\in A\times S'$ as representative, an element of $A_S$ was denoted by $\alpha/\gamma$, and then $A_S$ still has 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 the two multiplicative monoids was defined by $\alpha\mapsto \alpha/1$, and the fact that this is a monoid homomorphism means that $\epsilon$ is (once we show that $A_S$ is a ring) a function from $A$ to $A_S$ preserving the multiplicative structure.

That is, what we must 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 show 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, for suitable $\alpha,\beta\in A$ and $\gamma,\delta\in S'$, be written as $x=\alpha/\gamma,y=\beta/\delta$. 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 formula above. Then the things to show are as follows.

This definition is independent of the choice of $\alpha,\beta,\gamma,\delta$. That is, suppose they are written in the form $x=\alpha'/\gamma',y=\beta'/\delta'$. We must show that the following equation
\[(\alpha\delta+\beta\gamma)/\delta\gamma=(\alpha'\delta'+\beta'\gamma')/\gamma'\delta'\]
holds in $A_S$. But since $\alpha/\gamma=\alpha'/\gamma',\beta/\delta=\beta'/\delta'$, by definition, satisfying $\alpha\gamma'\zeta=\alpha'\gamma\zeta,\beta\delta'\xi=\beta'\delta\xi$, there exist $\zeta,\xi\in S'$. 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.
The operation $+$ 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 verify that $x_1+(x_2+x_3)$ also has the value on the right-hand side.
The commutativity of $+$ is immediate because addition and multiplication in $A$ are commutative.
$+$ 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.
Inverses with respect to $+$ always exist. This is because for any $x=\alpha/\gamma\in A_S$, $(-\alpha)/\gamma$ satisfies the equation
\[\frac{-\alpha}{\gamma}+\frac{\alpha}{\gamma}=\frac{(-\alpha)\gamma+\alpha\gamma}{\gamma^2}=0\]
$+$ 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, the equality $(x_1+x_2)y=x_1y+x_2y$ can also be shown.

From the above, we see 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, 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 defined by $S$ for $A$, 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 feature that distinguishes them 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 ring $A\neq 0$ 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 $0\neq x\in \mathfrak{a}$. However, in $A$, $x$ has an inverse $x^{-1}$, 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 $Ax\neq 0$. Now, since the only left ideals of $A$ are $0$ or $A$, we must have $Ax=A$, and thus $1\in Ax$. That is, there exists some $\alpha\in A$ such that $\alpha x=1$. Then $\alpha\neq 0$, and by the same logic there exists some $\beta\in A$ 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 is a field is trivial by definition, and this can be extended as follows.

Definition 5 For a ring $A$, if elements $\alpha,\beta$ satisfy $\alpha\beta=0$ but $\alpha\neq 0$ and $\beta\neq 0$, we call $\alpha,\beta$ zerodivisors. That a ring $A$ is an integral domain means that $A$ 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 $A,B$, $A\times B$, from the following equation

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

can never be an integral domain.

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, for $\alpha\in A$ and $\beta\in A\setminus\{0\}$, in the form $\alpha/\beta$. Here, in order 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 by $\Frac(A)$.

The field of fractions has the following universal property.

Proposition 8 Suppose an integral domain $A$, a field $\mathbb{K}$, and an embedding $f:A\hookrightarrow\mathbb{K}$ are given. Then, satisfying $\bar{f}\circ\epsilon=f$, there exists a unique ring homomorphism $\bar{f}:\Frac(A)\rightarrow\mathbb{K}$.

Proof

First we show uniqueness. If such a $\bar{f}$ exists, since any element of $\Frac(A)$ is written in the form $\epsilon(\alpha)\epsilon(\beta)^{-1}$, we must have

\[\bar{f}(\alpha/\beta)=\bar{f}(\epsilon(\alpha)\epsilon(\beta)^{-1})=f(\alpha)f(\beta)^{-1}\]

and therefore $\bar{f}$ is unique if it exists.

To show existence, define $\bar{f}(\alpha/\beta)=f(\alpha)f(\beta)^{-1}$. Since $f$ is injective, $\beta\neq 0$ implies $f(\beta)\neq 0$, and since $\mathbb{K}$ is a field, $f(\beta)$ is invertible. Now let $\alpha/\beta=\alpha'/\beta'$. Since $A$ is an integral domain, the definition of the equivalence relation in Theorem 1 gives $\alpha\beta'=\alpha'\beta$, and thus from $f(\alpha)f(\beta')=f(\alpha')f(\beta)$ we obtain $f(\alpha)f(\beta)^{-1}=f(\alpha')f(\beta')^{-1}$. That is, $\bar{f}$ is well-defined. That it preserves addition and multiplication follows immediately from the definition of addition and multiplication on $\Frac(A)$, and since $\bar{f}(\epsilon(\alpha))=f(\alpha)f(1)^{-1}=f(\alpha)$, it holds that $\bar{f}\circ\epsilon=f$.

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 9 For a commutative ring $A$ and an ideal $\mathfrak{p}\neq A$, the following are all equivalent.

$A/\mathfrak{p}$ is an integral domain.
If $\alpha,\beta\in A\setminus \mathfrak{p}$, then $\alpha\beta\in A\setminus \mathfrak{p}$.
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 that $A/\mathfrak{p}$ 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, for a prime $p$, of the form $p\mathbb{Z}$. Then $(0)$ is a prime ideal but not a maximal ideal.

On the other hand, the following holds.

Proposition 10 For commutative rings $A,B$, a ring homomorphism $\phi:A \rightarrow B$, and a prime ideal of $B$, $\mathfrak{p}$, $\phi^{-1}(\mathfrak{p})$ 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, contrary to the conclusion, that $\alpha\beta\in\phi^{-1}(\mathfrak{p})$ but $\alpha,\beta\not\in\phi^{-1}(\mathfrak{p})$ for some $\alpha,\beta\in A$. 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 9.

On the other hand, by the second equivalence of Proposition 9, 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 denominator of this ring are those of $A\setminus \mathfrak{p}$. This is defined as follows.

Definition 11 For a commutative ring $A$ and a prime ideal $\mathfrak{p}$, the localization of $A$ at $\mathfrak{p}$ is defined to be $(A\setminus \mathfrak{p})^{-1}A$, and is denoted simply by $A_\mathfrak{p}$.

Nilpotents

Definition 12 For a ring $A$, an element $\alpha$ is nilpotent if there exists $n>0$ such that $\alpha^n=0$. If $A$ has no nonzero nilpotent elements, $A$ is called 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 13 For a commutative ring $A$, the collection $\mathfrak{N}$ of nilpotent elements is an ideal.

Proof

If $x\in \mathfrak{N}$, then $x^n=0$ for some $n>0$, 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. Suppose arbitrary $x,y\in \mathfrak{N}$ are given, and let for suitable $m,n>0$ we have $x^m=0$ and $y^n=0$. 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 14 The ideal $\mathfrak{N}$ of Proposition 13 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 a 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 15 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 for some $\mathfrak{p}$ we have $x\not\in \mathfrak{p}$. First, consider the ring of fractions formed from the multiplicative subset $S=\{1,x,x^2,\ldots\}$, namely $A_x=S^{-1}A$. Then in $A_x$, the multiplicative identity $x/x$ must be different from $0/1$, because $x/x=0/1$ means that for some $k\geq 0$ we have $x^{k+1}=0$, 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), $A_x$ necessarily has a maximal ideal $\mathfrak{m}$, and since every maximal ideal is a prime ideal, $\mathfrak{m}$ is a prime ideal of $A_x$. Applying Proposition 10 to $\epsilon:A \rightarrow A_x$, $\epsilon^{-1}(\mathfrak{m})$ 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 $\mathfrak{m}=A_x$, a contradiction.

K

Field of Fractions

Fraction Rings

Fields

Integral Domains

Prime Ideals

Nilpotents

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