§ 9. Fields

1. Fields

Definition 1

A ring $K$ is called a field if it does not consist only of $0$ and every non-zero element of $K$ is invertible.

The set of non-zero elements of the field $K$ , with multiplication, is a group, which is precisely the multiplicative group $K^*$ of the ring $K$ (§ 8, no. 1). The opposite ring of a field is a field. A field is called commutative if its multiplication is commutative; such a field is identified with its opposite. A non-commutative field is sometimes called a skew field.

Example

(1)* We shall define in no. 4 the field of rational numbers; In General Topology the field of real numbers will be defined (General Topology, IV, § 1, no. 3), also the field of complex numbers (General Topology, VIII, § 1, no. 1) and the field of quaternions (General Topology, VIII, § 1, no. 4). These fields are commutative with the exception of the field of quaternions.* (2) The ring $\mathbf{Z}/2\mathbf{Z}$ is obviously a field.

Let $K$ be a field. Every subring $L$ of $K$ which is a field is called a subfield of $K$ and $K$ is then called an extension field of $L$ ; it amounts to the same to say that $L$ is a subring of $K$ and that $x^{-1} \in L$ for every non-zero element $x$ of $L$ . If $(L_i)_{i \in I}$ is a family of subfields of $K$ , then $\bigcap_{i \in I} L_i$ is a subfield of $K$ ; for every subset $X$ of $K$ there thus exists a smallest subfield of $K$ containing $X$ ; it is said to be generated by $X$ .

Proposition 1

Let $K$ be a field. For every subset $X$ of $K$ , the centralizer (§ 8, no. 5, Example 3) $X'$ of $X$ is a subfield of $K$ .

We know (loc. cit.) that $X'$ is a subring of $K$ . On the other hand, if $x \neq 0$ is permutable with $z \in X$ , so is $x^{-1}$ (I, § 2, no. 3, Proposition 5) and hence $X'$ contains the inverse of every non-zero element of $X'$ .

Corollary

The centre of a field $K$ is a (commutative) subfield of $K$ .

Theorem 1

Let $\mathbf{A}$ be a ring. The following conditions are equivalent:

(a) $\mathbf{A}$ is a field; (b) $\mathbf{A}$ is not reduced to 0 and the only left ideals of $\mathbf{A}$ are 0 and $\mathbf{A}$ .

Suppose that $\mathbf{A}$ is a field. Then $\mathbf{A}$ is not reduced to 0. Let $a$ be a left ideal of $\mathbf{A}$ distinct from 0. There exists a non-zero $a$ belonging to $a$ . For all $x \in \mathbf{A}$ , $x = (xa^{-1})a \in a$ ; hence $a = \mathbf{A}$ .

Suppose that $\mathbf{A}$ satisfies condition (b). Let $x \neq 0$ be in $\mathbf{A}$ . We need to prove that $x$ is invertible. The left ideal $Ax$ contains $x$ and hence is not zero, whence $Ax = \mathbf{A}$ . There thus exists $x' \in \mathbf{A}$ such that $x'x = 1$ . Then $x' \neq 0$ since $1 \neq 0$ . Applying the above result to $x'$ , it is seen that there exists $x'' \in \mathbf{A}$ such that $x''x' = 1$ . Then $x'' = x'' . 1 = x''x'x = 1 . x = x$ , hence $xx' = 1$ . Hence $x'$ is the inverse of $x$ .

Remark

In Theorem 1, left ideals may be replaced by right ideals. In Chapter VIII, § 5, no. 2, we shall study non-zero rings $\mathbf{A}$ which have no two-sided ideal distinct from 0 and $\mathbf{A}$ ; such rings (called quasi-simple) are not necessarily fields (for example, the ring $\mathbf{M}_2(\mathbf{Q})$ of square matrices of order 2 with rational coefficients is quasi-simple but is not a field).

Corollary 1

Let $\mathbf{A}$ be a ring and $a$ a two-sided ideal of $\mathbf{A}$ . For the ring $\mathbf{A}/a$ to be a field, it is necessary and sufficient that $a$ be a maximal left ideal of $\mathbf{A}$ .

The left ideals of $\mathbf{A}/a$ are of the form $b/a$ where $b$ is a left ideal of $\mathbf{A}$ containing $a$ ( $§ 8$ , no. 8, Corollary to Proposition 5). To say that $\mathbf{A}/a \neq 0$ means that $a \neq \mathbf{A}$ . Under this hypothesis, $\mathbf{A}/a$ is a field if and only if the only left ideals of $\mathbf{A}$ containing $a$ are $a$ and $\mathbf{A}$ (Theorem 1), whence the corollary.

Corollary 2

Let $\mathbf{A}$ be a commutative ring which is not 0. There exists a homomorphism of $\mathbf{A}$ onto a commutative field.

By Krull’s Theorem ( $§ 8$ , no. 6, Theorem 1), there exists in $\mathbf{A}$ a maximal ideal $a$ . Then $\mathbf{A}/a$ is a field (Corollary 1).

Corollary 3

Let $a$ be an integer $\geqslant 0$ . For the ring $\mathbf{Z}/a\mathbf{Z}$ to be a field, it is necessary and sufficient that $a$ be prime.

This follows from Corollary 1 and $§ 8$ , no. 11.

For $p$ prime the field $\mathbf{Z}/p\mathbf{Z}$ is denoted by $\mathbf{F}_p$ .

Theorem 2

Let $\mathbf{K}$ be a field and $\mathbf{A}$ a non-zero ring. If $f$ is a homomorphism of $\mathbf{K}$ into $\mathbf{A}$ , then the subring $f(\mathbf{K})$ of $\mathbf{A}$ is a field and $f$ defines an isomorphism of $\mathbf{K}$ onto $f(\mathbf{K})$ .

Let $a$ be the kernel of $f$ . Then $1 \notin a$ for $f(1) = 1 \neq 0$ in $\mathbf{A}$ and, as $a$ is a left ideal of $\mathbf{K}$ , $a = \{0\}$ by Theorem 1. Therefore $f$ is injective and hence an isomorphism of $\mathbf{K}$ onto the subring $f(\mathbf{K})$ of $\mathbf{A}$ ; the latter ring is therefore a field.

2. Integral domains

Definition 2

A ring $\mathbf{A}$ is called an integral domain (or a domain of integrity) if it is commutative, non-zero, and the product of two non-zero elements of $\mathbf{A}$ is non-zero.

The ring $\mathbf{Z}$ of rational integers is an integral domain; it is commutative and non-zero; the product of two integers $> 0$ is non-zero; every non-zero integer is of the form $a$ or $-a$ with $a > 0$ and $(-a)b = -ab,\ (-a)(-b) = ab$ for $a > 0,\ b > 0$ , whence our assertion.

Every commutative field is an integral domain. A subring of an integral domain is an integral domain. In particular, a subring of a commutative field is an integral domain. We shall show that conversely every integral domain $\mathbf{A}$ is isomorphic to a subring of a commutative field. Recall (§ 8, no. 12) that $\mathbf{A}$ has been identified with a subring of its total ring of fractions. Our assertion then follows from the following proposition:

Proposition 2

If $\mathbf{A}$ is an integral domain, the total ring of fractions $\mathbf{K}$ of $\mathbf{A}$ is a commutative ring.

The ring $\mathbf{K}$ is commutative. It is non-zero since $\mathbf{A} \neq \{0\}$ . As $\mathbf{A}$ is an integral domain, every non-zero element of $\mathbf{A}$ is cancellable and $\mathbf{K}$ consists of the fractions $a/b$ with $b \neq 0$ . Now $a/b \neq 0$ implies $a \neq 0$ and the fraction $b/a$ is then inverse of $a/b$ .

The total ring of fractions of an integral domain is called its field of fractions. Such a ring is identified with its image in its field of fractions.

Proposition 3

Let $\mathbf{B}$ be a non-zero ring and $\mathbf{A}$ a commutative subring of $\mathbf{B}$ such that every non-zero element of $\mathbf{A}$ is invertible in $\mathbf{B}$ . (a) $\mathbf{A}$ is an integral domain. (b) *Let $\mathbf{A}'$ be the field of fractions $\mathbf{A}$ . The canonical injection of $\mathbf{A}$ into $\mathbf{B}$ can be extended uniquely to an isomorphism $f$ of $\mathbf{A}'$ onto a subfield of $\mathbf{B}$ . (c) *The elements of $f(\mathbf{A}')$ are the $xy^{-1}$ where $x \in \mathbf{A},\ y \in \mathbf{A},\ y \neq 0$ .

Assertion (a) is obvious. The canonical injection of $\mathbf{A}$ into $\mathbf{B}$ extends uniquely to a homomorphism $f$ of $\mathbf{A}'$ into $\mathbf{B}$ (§ 8, no. 12, Theorem 5). Assertion (b) then follows from no. 1, Theorem 2. The elements of $\mathbf{A}'$ are the fractions $x/y$ with $x \in \mathbf{A},\ y \in \mathbf{A},\ y \neq 0$ and $f(x/y) = xy^{-1}$ , whence (c).

3. Prime ideals

Proposition 4

Let $\mathbf{A}$ be a commutative ring and $\mathfrak{p}$ an ideal of $\mathbf{A}$ . The following conditions are equivalent: (a) *the ring $\mathbf{A}/\mathfrak{p}$ is an integral domain; (b) $\mathbf{A} \neq \mathfrak{p}$ and, if $x \in \mathbf{A} - \mathfrak{p}$ and $y \in \mathbf{A} - \mathfrak{p}$ , then $xy \in \mathbf{A} - \mathfrak{p}$ . (c) $\mathfrak{p}$ is the kernel of a homomorphism of $\mathbf{A}$ into a field.

The implications (c) ⇒ (b) ⇒ (a) are obvious. If $A/p$ is an integral domain, let $f$ be the canonical injection of $A/p$ into its field of fractions and $g$ the canonical homomorphism of $A$ onto $A/p$ ; then $p$ is the kernel of $f \circ g$ , whence the implication (a) ⇒ (c).

Definition 3

In a commutative ring $A$ , an ideal $p$ satisfying the conditions of Proposition 4 is called a prime ideal.

Example

(1) Let $A$ be a commutative ring. If $m$ is a maximal ideal of $A$ , $m$ is prime; the ring $A/m$ is a field (no. 1, Corollary 1 to Theorem 1).

(2) If $A$ is an integral domain, the ideal $\{0\}$ of $A$ is prime (but not maximal in general, as the example of the ring $\mathbf{Z}$ proves).

4. The field of rational numbers

Definition 4

The field of fractions of the ring $\mathbf{Z}$ of rational integers is called the field of rational numbers and is denoted by $\mathbf{Q}$ . The elements of $\mathbf{Q}$ are called rational numbers.

Every rational number is thus of the form $a/b$ where $a$ and $b$ are rational integers with $b \neq 0$ (and we may even take $b > 0$ as the relation

a/b = (-a)/(-b)

proves). $\mathbf{Q}_+$ is used to denote the set of rational numbers of the form $a/b$ with $a \in \mathbf{N}$ and $b \in \mathbf{N}^*$ .

We have the relations:

(1)

\mathbf{Q}_+ + \mathbf{Q}_+ = \mathbf{Q}_+

(2)

\mathbf{Q}_+ \cdot \mathbf{Q}_+ = \mathbf{Q}_+

(3)

\mathbf{Q}_+ \cap (-\mathbf{Q}_+) = \{0\}

(4)

\mathbf{Q}_+ \cup (-\mathbf{Q}_+) = \mathbf{Q}

(5)

\mathbf{Q}_+ \cap \mathbf{Z} = \mathbf{N}

The first two follow from the formulae $a/b + a'/b' = (ab' + a'b)/bb'$ , $(a/b)(a'/b') = aa'/bb'$ , $0 \in \mathbf{Q}_+$ , $1 \in \mathbf{Q}_+$ and the fact that $\mathbf{N}$ is stable under addition and multiplication and $\mathbf{N}^*$ under multiplication. Obviously $0 \in \mathbf{Q}_+$ , whence $0 \in (-\mathbf{Q}_+)$ ; let $x$ be in $\mathbf{Q}_+ \cap (-\mathbf{Q}_+)$ ; then there exist positive integers $a, b, a', b'$ with $b \neq 0,\ b' \neq 0$ and $x = a/b = -a'/b'$ ; then $ab' + ba' = 0$ , whence $ab' = 0$ (for $ab' \geq 0$ and $ba' \geq 0$ ) and therefore $a = 0$ since $b' \neq 0$ ; in other words, $x = 0$ . Finally, obviously $\mathbf{N} \subset \mathbf{Z}$ and $\mathbf{N} \subset \mathbf{Q}_+$ . Conversely, if $x$ belongs to $\mathbf{Z} \cap \mathbf{Q}_+$ , it is a rational integer; there exist two rational integers $a$ and $b$ with $a \geq 0,\ b > 0$ and $x = a/b$ , whence $a = bx$ ; if $x \notin \mathbf{N}$ , then $-x > 0$ , whence $-a = b(-x) > 0$ and consequently $a < 0$ contrary to the hypothesis.

Given two rational numbers $x$ and $y$ , we write $x \leq y$ if $y - x \in \mathbf{Q}_+$ . It is easily deduced from formulae (1), (3) and (4) that $x \leq y$ is a total ordering on

Q, from (5) that this relation induces the usual order relation on Z. Finally, from (1) it follows that the relations $x \leq y$ and $x' \leq y'$ imply $x + x' \leq y + y'$ and from (2) that the relation $x \leq y$ implies $xz \leq yz$ for all $z \geq 0$ and $xz \geq yz$ for $z \leq 0$ (cf. VI, § 2, no. 1).

Let $x$ be a rational number. $x$ is said to be positive if $x \geq 0$ , strictly positive if $x > 0$ , negative if $x \leq 0$ and strictly negative if $x < 0$ .† The set of positive rational numbers is $\mathbf{Q}_+$ and that of negative numbers is $-\mathbf{Q}_+$ . If $\mathbf{Q}^*$ denotes the set of non-zero rational numbers, the set $\mathbf{Q}_+^*$ of strictly positive numbers is equal to $\mathbf{Q}^* \cap \mathbf{Q}_+$ and $-\mathbf{Q}_+^*$ is the set of strictly negative numbers.

The sets $\mathbf{Q}_+^*$ and $\{1, -1\}$ are subgroups of the multiplicative group $\mathbf{Q}^*$ . Every rational number $x \neq 0$ can be expressed in one and only one way in the form $1.y, (-1).y$ , where $y > 0$ ; hence the multiplicative group $\mathbf{Q}^*$ is the product of the subgroups $\mathbf{Q}_+^*$ and $\{1, -1\}$ , the component of $x$ in $\mathbf{Q}_+^*$ is called the absolute value of $x$ and is denoted by $|x|$ ; the component of $x$ in $\{-1, 1\}$ (equal to 1 if $x > 0$ , to $-1$ if $x < 0$ ) is called the sign of $x$ and denoted by $\operatorname{sgn}\, x$ .

Usually these two functions are extended to the whole of $\mathbf{Q}$ by setting $|0| = 0$ and $\operatorname{sgn}\, 0 = 0$ .

Exercises

See the exercises for § 9.