Why is K(t):K a valid field extension?

I'm using Stewart's Galois Theory book, which states that a field extension is a pair L:K such that there exists a monomorphism between K and L, and both of them are subfields of $\mathbb{C}$.

I have a problem with considering the set of all rational expressions over K denoted K(t) as a subfield of $\mathbb{C}$, because strictly speaking it is a set of expressions. Then K(t):K would not be a field extension.

Thank you in advance for your help.

• Are you he demands both to be subfields of $\mathbb C$? I'll,bet you mis-read something since that is certainly not a demand. – Ittay Weiss Dec 31 '16 at 17:30
• On the other hand, for example $\Bbb Q(t)$ can be viewed as subfield of $\Bbb C$ by mapping $t\mapsto \pi$ or any other transcendental – Hagen von Eitzen Dec 31 '16 at 17:32
• @IttayWeiss Yes I'm sure, I think it's a typo, but I'm glad it is not a demand, thank you very much. – Santiago Estupiñán Dec 31 '16 at 17:41
• @HagenvonEitzen It is a useful way of visualize it as a subfield. Thank you very much – Santiago Estupiñán Dec 31 '16 at 17:42

The notion of field extension is indeed more general and includes just any pair of fields $K$, $L$ with a (fixed) monomorphism $K\rightarrow L$.
If $K$ is any field and $L$ an extension of degree of transcendence not greater than the degree of transcendence of $K\subset \overline{K}$ (an overline denotes the algebraic closure of the field), the field $L$ can always be seen as a subfield of $\overline K$.
This fact justifies the so called Lefschetz principle which somewhat vaguely and in down-to-earth terms says that to do algebraic geometry in characteristic $0$ it is sufficient to work in the complex field $\Bbb C$.