$\Bbb Q(x) \supseteq \Bbb Q$ - How to make sense of the indeterminate $x$ I am currently studying field theory (as a beginner),
and am always uncertain about how to make sense of the indeterminate x in field extensions.
As a simple example, $\Bbb Q(\sqrt 2)$ as an extension field of the field of rational numbers $\Bbb Q$, seems to make sense to me, as a field containing all the rational numbers and all possible combinations (if that is a way to express it) of it
But then, for example, if $K=F(x)$, the field of rational functions over a field F, am I to understand the "$x$" as a number, just like I did for $\sqrt 2$? Is it a "variable"? Although outwardly the field $K=F(x)$ looks like a "simple extension", is it safe to call it a simple extension?
Furthermore, the fact that $\Bbb Q(x) \supseteq \Bbb Q$ is not an algebraic extension is also quite difficult for me to grasp, perhaps because I am uncomfortable with the notion of variable "$x$" being there.
 A: Before studying this in field-theory, it would be a good idea to make sense of $\Bbb F[x]$ (or even $R[x]$ for a general ring) as a ring. In short, you don't interpret it as a number. You do interpret it as a variable (or an indeterminate). This means that you think of polynomials as formal expressions of the form $$a_0 + a_1x + \cdots + a_nx^n,$$
where $a_i \in \Bbb F$.
If you wish to go super-formal, think of these as sequences in $\Bbb F$ which are eventually $0$. $$a_0 + a_1x + \cdots + a_nx^n \leftrightarrow (a_0, a_1, \ldots, a_n, 0 , 0, \cdots).$$
With this identification, $x^i$ is simply a placeholder (or a "bookmark") to tell you which entry of the sequence you're looking at. The thing to remember is that $+$ is defined coordinate wise but $\cdot$ is not.
You can now check that $\Bbb F[x]$ is a ring. Moreover, it is actually an integral domain. Thus, you can talk about its field of fractions. This is simply all expressions of the form $p(x)/q(x)$ where $q(x) \neq 0$ and scaling the numerator and denominator by the same non-zero polynomial does not change anything.

Finally, yes, $\Bbb F(x)$ is indeed a simple extension since $x$ acts as a primitive element. However, keep in mind that as opposed to $\Bbb Q(\sqrt 2)$, this is not an algebraic extension but a transcendental one. In particular, $\Bbb F(x)$ is infinite-dimensional as an $\Bbb F$ vector space.

An aside, to elaborate more on the last part: Given fields $\Bbb F \subset \Bbb E \subset \Bbb K$ and $\alpha \in \Bbb K$, when do we say that $\Bbb E = \Bbb F(\alpha)$?
That happens precisely when, $\Bbb E$ is the smallest (w.r.t. $\subset$) subfield of $\Bbb K$ containing both $\Bbb F$ and $\alpha$. (This implies that $\alpha$ must necessarily be in $\Bbb E$.)
Thus, you must argue that the smallest subfield of $\Bbb F(x)$ containing both $\Bbb F$ and $x$ indeed all of $\Bbb F(x)$.
(This is not too tough; if a subfield $\Bbb E \subset \Bbb F(x)$ contains $x$ and $\Bbb F$, then it must contain all polynomials. (Why?)
Being a subfield, it must have multiplicative inverses and so...)
