Example of how to classify a function's field in Liouville's theorem?

In building off of this question Can anyone show an example of going through Liouville's differential algebra theorem?

I'm slowly starting to understand more components of Liouville's theorem after organizing what I do and don't know, writing notes and looking back through the theorem again.

But, just when I thought I'd finally figured something out about it, I hit a roadblock.

Let's say I start with the field of rational functions $C(x)$ and I want to work with this to build my starting-field $\mathbb{F}.$

Now, I'm given an element $a \in \mathbb{F}$ where $a = e^{\sqrt{\ln(x)-1}}$ and I suppose $\int{a}$ belongs to some elementary field extension $\mathbb{G}$ of $\mathbb{F}.$

My confusion and question is, what field extensions were necessary to construct $a$ so that I can begin applying Liouville's theorem and adding extensions to check $\int a$?

The reason I'm confused is, rational functions are always the singular go-to example for this theorem. So, I figure that if $a$ belongs to a differential field, it must be $C(x,\ln(x),e^x,\sqrt{x})$ but I'm worried this is incorrect! The reason being, these are functions rational in the listed variables, $\frac{P(x,\ln(x),e^x,\sqrt{x})}{Q(x,\ln(x),e^x,\sqrt{x})}$. This says nothing about composites of these functions in of itself! So if I'm given a function like this, what field extension can I work with other than that of $a$ itself?

And, if I have to add finite field extensions, how could I ever see ahead and know whether it's possible or impossible to add such extensions? Supposedly this is what Liouville's theorem addresses but I'm not sure.


The extensions to build $e^\sqrt{\ln(x) - 1}$ from $F_0 = \mathbb{C}(x)$ are:

  1. Make a logarithmic extension $\ln(x)$ to $F_0$ to get $F_1 = \mathbb{C}(x, \ln(x))$.
  2. Now $\ln(x) - 1 \in F_1$; make an algebraic extension $\sqrt{\ln(x) - 1}$ to $F_1$ get a field $F_2$.
  3. Now $\sqrt{\ln(x) - 1} \in F_2$; make an exponential extension $e^\sqrt{\ln(x) - 1}$ to $F_2$ to get a field $F_3$.

Now $F_3$ is your starting field containing $e^\sqrt{\ln(x) - 1}$.

Note that at each stage you can adjoin an exponential, logarithm, or radical (or more generally, any algebraic extension) using any elements you had in the previous stage. You don't go back to $\mathbb{C}(x)$.

The above only describes field extensions, not differential field extensions, which is what you actually need. To make a differential field extension you also need to specify the derivatives of the new elements you are adjoining. This is done with the usual rules of calculus.

For example, when you adjoin $\ln(x)$ to $\mathbb{C}(x)$, you specify that its derivative is $1/x$, and then all other functions in $F_1$ are differentiated by using the usual rules of calculus. Similarly when you adjoin the $\ln$ of some other element, say $\ln u$ for $u \in F$ then you specify that $(\ln u)' = u'/u$.

Regarding the last comment, yes you are correct that field extensions here do not give composites. They only give rational functions in the new adjoined element, as is always the case for all field extensions (ignoring derivatives).