1.2 Fields

Definition 1.1

A field is a set $F$ with two composition laws $+$ and $\cdot$ such that

(a)

$(F,+)$ is a commutative group;
(b)

$(F^{\times},\cdot)$ , where $F^{\times}=F\smallsetminus\{0\}$ , is a commutative group;
(c)

the distributive law holds.

Thus, a field is a nonzero commutative ring such that every nonzero element has an inverse. In particular, it is an integral domain. A field contains at least two distinct elements, $0$ and $1$ . The smallest, and one of the most important, fields is $\mathbb{\mathbb{F}}_{2}=\mathbb{Z}/2\mathbb{Z}=\{0,1\}$ .

A subfield $S$ of a field $F$ is a subring that is closed under passage to the inverse. It inherits the structure of a field from that on $F$ .

Lemma 1.2

A nonzero commutative ring $R$ is a field if and only if it has no ideals other than $(0)$ and $R$ .

Proof.

Suppose that $R$ is a field, and let $I$ be a nonzero ideal in $R$ . If $a$ is a nonzero element of $I$ , then $1=a^{-1}a\in I$ , and so $I=R$ . Conversely, suppose that $R$ is a commutative ring with no proper nonzero ideals. If $a\neq 0$ , then $(a)=R$ , and so there exists a $b$ in $R$ such that $ab=1$ . _□

Example 1.3

The following are fields: $\mathbb{Q}$ , $\mathbb{R}$ , $\mathbb{C}$ , $\mathbb{F}_{p}=\mathbb{Z}/p\mathbb{Z}$ ( $p$ prime).

A homomorphism of fields is simply a homomorphism of rings. Such a homomorphism is always injective, because its kernel is a proper ideal (it doesn’t contain $1$ ), which must therefore be zero.

Let $F$ be a field. An $F$ -algebra (or algebra over $F$ ) is a ring $R$ containing $F$ as a subring (so the inclusion map is a homomorphism). A homomorphism of $F$ -algebras $\alpha\colon R\rightarrow R^{\prime}$ is a homomorphism of rings such that $\alpha(c)=c$ for every $c\in F$ .