Difference/similarities: Field of rational functions in X - smallest field containing K and x? 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.
 A: 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 $i$.
