I'm reading through this article on Liouville's differential algebra theorem which is very tedious and advanced with respect to my background. I'm looking at Propositions 1.13 and 4.2 trying to break down what they mean to make sure I understand. Unfortunately it says "(an algebraic result assumed familiar to readers$^5$)" and almost none of this is familiar to me
First, 1.13 uses the term $L$ \ $K.$ Quotient groups hadn't made much sense to me, but now reflecting on them in a quotient field, if that is what this means, probably isn't a much bigger leap from that. What I hope $L$ \ $K$ means for simplicity, is merely the set of elements $L$ and $K$ have in common, which is basically the specific, newly added elements of $L$ that $K$ didn't have, since $K \subset L.$ A common notation I see in other instances of abstract algebra and linear algebra is something like $\mathbb{Q}$ \ $\{ 0 \}$ where the division symbol here means to take away the element $0,$ and presumably in the general case, $\mathbb{Q}$ \ $\mathbb{E}$ means to take some set of elements $\mathbb{E}$ away from the set $\mathbb{Q}.$ If $K$ relates to a field of rational functions, then probably some condition about equivalence classes is important so that expressions like $x/x=1$ at $x=0,$ and also because when you integrate a function with countably many discontinuities, the result is still a continuous function.
Next, this article uses two different notations, $K(\ell)$ and $K[\ell],$ rectangular brackets and curved brackets. I don't know what the difference is between those two yet, but in looking at this other article, I might be inclined to infer that $K[\ell]$ is a differential field with elements that are polynomial in $\ell,$ and that $K(\ell)$ is a field extension of $K$. What were the original elements of $K$ before such a field extension? I don't know, the article doesn't seem to specify.
So, the statement of the theorem might be that, if $\ell$ is one of the newly added elements of the field extension $L = K(\ell),$ then the derivative of $\ell,$ $\ell ',$ is an element of $K$ that is polynomial in $\ell.$
That doesn't quite add up to me because it should be more general than just polynomials, but I wouldn't know because polynomials are a big deal in the world of abstract algebra. There's also the subtly in the difference between a ring and a field, so I don't quite understand what the theorem is saying, but it might be saying that if you extend your differential field, the derivatives of those new elements belong to that field but you lose the guarantee of an inverse or commutability of some kind.
This is also important to Proposition 4.2 where both the curved and square brackets occur, I don't know what the difference is.
