1.6 Extensions

Let $F$ be a field. A field containing $F$ is called an extension of $F$ .⁵⁵ 5 This is the usual definition of “extension” (Wikipedia: field extension), but “overfield” would be a better term because Bourbaki, for example, uses “extension” to mean a field $E$ together with a homomorphism from $F$ to $E$ . In other words, an extension is an $F$ -algebra whose underlying ring is a field. An extension $E$ of $F$ is, in particular, an $F$ -vector space, whose dimension is called the degree of $E$ over $F$ . It is denoted by $[E\colon F]$ . An extension is said to finite if its degree is finite, and quadratic, cubic, etc. if it is of degree $2$ , $3$ , etc.

When $E$ and $E^{\prime}$ are extensions of $F$ , an $F$ -homomorphism $E\rightarrow E^{\prime}$ is a homomorphism $\varphi\colon E\rightarrow E^{\prime}$ such that $\varphi(c)=c$ for all $c\in F$ .

Example 1.19

(a) The field of complex numbers $\mathbb{C}$ has degree $2$ over $\mathbb{R}$ (basis $\{1,i\}).$

(b) The field of real numbers $\mathbb{R}$ has infinite degree over $\mathbb{Q}$ : the field $\mathbb{Q}{}$ is countable, and so every finite-dimensional $\mathbb{Q}{}$ -vector space is also countable, but a famous argument of Cantor shows that $\mathbb{R}{}$ is not countable.

\mathbb{Q}(i)\overset{\smash{\lower 1.31395pt\hbox{{def}}}}{=}\{a+bi\in\mathbb% {C}\mid a,b\in\mathbb{Q}\}

has degree $2$ over $\mathbb{Q}$ (basis $\{1,i\}$ ).

(d) The field $F(X)$ has infinite degree over $F$ ; in fact, even its subspace $F[X]$ has infinite dimension over $F$ (basis $1,X,X^{2},\ldots$ ).

Proposition 1.20 (multiplicativity of degrees)

Consider fields $L\supset E\supset F$ . Then $L/F$ is of finite degree if and only if $L/E$ and $E/F$ are both of finite degree, in which case

[L\colon F]=[L\colon E][E\colon F].

Proof.

If $L$ is finite over $F$ , then it is certainly finite over $E$ ; moreover, $E$ , being a subspace of a finite-dimensional $F$ -vector space, is also finite-dimensional.

Thus, assume that $L/E$ and $E/F$ are of finite degree, and let $(e_{i})_{1\leq i\leq m}$ be a basis for $E$ as an $F$ -vector space and let $(l_{j})_{1\leq j\leq n}$ be a basis for $L$ as an $E$ -vector space. To complete the proof of the proposition, it suffices to show that $(e_{i}l_{j})_{1\leq i\leq m,1\leq j\leq n}$ is a basis for $L$ over $F$ , because then $L$ will be finite over $F$ of the predicted degree.

First, $(e_{i}l_{j})_{i,j}$ spans $L$ . Let $\gamma\in L$ . Then, because $(l_{j})_{j}$ spans $L$ as an $E$ -vector space,

\gamma={\textstyle\sum\nolimits_{j}}\alpha_{j}l_{j},\qquad\text{some }\alpha_{% j}\in E,

and because $(e_{i})_{i}$ spans $E$ as an $F$ -vector space,

\alpha_{j}={\textstyle\sum\nolimits_{i}}a_{ij}e_{i},\qquad\text{some $a_{ij}% \in F$}.

On putting these together, we find that

\gamma={\textstyle\sum\nolimits_{i,j}}a_{ij}e_{i}l_{j}.

Second, $(e_{i}l_{j})_{i,j}$ is linearly independent. A linear relation $\sum a_{ij}e_{i}l_{j}=0$ , $a_{ij}\in F$ , can be rewritten $\sum_{j}(\sum_{i}a_{ij}e_{i})l_{j}=0$ . The linear independence of the $l_{j}$ ’s now shows that $\sum_{i}a_{ij}e_{i}=0$ for each $j$ , and the linear independence of the $e_{i}$ ’s shows that each $a_{ij}=0$ . _□