# algebraic extensions of (differential) function fields

Let $K$ be a differential field with algebraically closed constant field $C$ (Think $K=\mathbb{C}(x)$ here). I am looking for an example of a simple algebraic extension $L = K[t]$, such, that $t' \notin K$, i.e., $[ K[t']: K] > 1$. In other words, I am looking for an example of an irreducible polynomial $p(t)$ over the univariate polynomial ring $K[t]$ such, that the derivative $p'(t)$ does not split completely in linear factors over $K$ (the derivative with respect to the canonical extension of the derivative of $K$). I am not sure if such a polynomial even exists, but with $\mathbb{C}(x) = K$ in mind I don't even have a good way to construct examples of irreducible polynomials at all. I would be happy to even see a single example of an irreducible polynomial over $\mathbb{C}(x)$ even if its derivative is not proper algebraic.

Just for some more context: the actual origin of this question is two different definitions I saw for an elementary extension of a differential field. In both they call a differential field extension $L/K$ elementary, if you have a tower of fields $K = K_0 \leq K_1 = K_0[t_1] \leq ... \leq K_n = K_{n-1}[t_n] = L$. The first definition wants $t_{i}$ for $i = 1, ..., k$ to be algebraic over $K_{i-1}$ or for the logarithmic derivative $t_i'/t_i \in K_{i-1}$ to be there, and $t_{k+1}, ..., t_n$ to be integrals of elements in $K$. The second definition just wants each $t_i$ to be either algebraic over $K_{i-1}$, or for the logarithmic derivative to be a derivative already in the field before, i.e. $\exists 0 \neq u \in K_{i-1}$ with $t_i'/t_i = u'$, or for $t_i$ to be the integral of a logarithmic derivative, i.e. $\exists 0 \neq u \in K_{i-1}$ with $t_i' = u_i'/u_i$. Trying to prove these to be equivalent, I came to the thought that the second definition was stronger, and if I find such an extension, it would be of the second class but not of the first one, since the first one does not allow you to have integrals of properly algebraic elements, whereas the second one does, as long as these are derivatives of some element already in the extension.

The questions you are asking are actually different. To see this, consider $t=\sqrt{x}$. The derivative of $t$ is obviously $1/(2\sqrt{x})$ and so does not live in $\mathbb C(x)$, but its minimal polynomial $t^2-x$ has derivative $2t$ which, being linear, certainly splits into linear factors.
So $t=\sqrt{x}$ answers your first question.
For your second question, a sufficiently generic cubic polynomial should do the trick: take $t^3-xt-1$. Its derivative, $3t^2-x$, is irreducible.
For your third question $\sqrt{x}$ is not what you need, since while its derivative is in $K$, its logarithmic derivative is not. In fact, since all polynomials of degree $4$ and less are solvable by radicals, and a $\sqrt[n]{x}=e^{\int x'/nx}$, you need a polynomial equation of degree $5$. Something like $t^5-xt-1$ should do just fine.
• Thanks. Yeah, for some reason I thought the minimal polynomial of the derivative was the derivative of the minimal polynomial, which is just plain wrong. Your example actually also made me realize that it's not so simple for an element to give an algebraic extension: all derivatives have to be algebraic over the base field as well. I actually found out that any proper algebraic element has to anwser the first question (Lemma 1.1 in www.jstor.org/stable/1994827). So actually $K[t,e^t]$ should anwser the second one, for any $t$ such, that $K[t]$ is an algebraic differential field extension. – goens Jun 9 '12 at 10:46