I am a bit confused with the Notation $K(X)$ or $K(x)$. In my notes I found two definitions:
- $K(X)$ is the field of rational functions with indeterminate $X$.
- $K(x)$ is the smallest field containing $K$ and $x$. For example $\mathbb{R}(i) = \mathbb{C}$
Now I don't quite see how I should be able to distinguish these two uses of $K(x)$ or I don't see that these two concepts are the same.
If the two are the same, wouldn't that mean that $K(X)$ is an extension of $K$? But how can an element of $K$ be an element of $K(X)$, which contains functions?
I tried to search about this, but haven't found any sources which deal with both these fields at the same time.
It's somewhat of an abuse of notation, but an acceptable one, because there is a trivial embedding of $K$ into $K(X)$, in the same way that $K$ embeds into the polynomial ring $K[X]$ with indeterminate $X$. In both cases the embedding is just the map from an element in $K$ to the corresponding constant function.
More precisely, let $φ(a) = ( X \mapsto a )$ for any $a \in K$. Then $φ$ is clearly a ring or field homomorphism from $K$ to $K[X]$ or $K(X)$ respectively.
Also, it is true that even if you interpret $K(X)$ as a field whose elements are (true) functions, it is still indeed isomorphic to $K(x)$ for any $x$ in a field extension of $K$ such that $x$ is not algebraic over $K$. In other words, extending a field by adjoining an indeterminate is essentially the same as adjoining a transcendental element (in a larger containing field). That is probably why you don't see many sources talk about both at the same time, as they are no different in structure.
Finally, I should mention that you must not forget what exactly you mean when you talk about common concepts like $i = \sqrt{-1}$. Technically speaking there is no such thing until you have shown that there is really a field containing $\mathbb{R}$ that has an element $i$ such that $i^2 = -1$. If you have not, then you can't just talk about $\mathbb{R}(i)$ since you must already have such a larger field to begin with, before you can talk about the smallest field containing $\mathbb{R}$ as well as $i$.