# Why do we need admissible isomorphisms for differential Galois theory?

Background: In Kaplansky's Introduction to Differential Algebra, an isomorphism between differential fields $K, L$ is defined to be admissible if $K,L$ are contained in a larger differential field $M$. This appears to be an integral part of the development in his book. For instance, Kaplansky proves the theorem that if $K, L$ are contained in a differential field $M$ and $f: K \to L$ is a differential isomorphism, then there is an admissible isomorphism $g: M \to M'$ extending $f$, using various differential analogs of results from commutative algebra (e.g. that a radical ideal is an intersection of prime ideals), which aren't at all necessary in ordinary field theory (as far as I know, anyway).

I find the definition hard to understand partially because it's not arrow-theoretic; it would help to have a more categorical notion than thinking in terms of subsets. Moreover, in the case of Picard-Vessiot extensions (which if I understand correctly is what differential Galois theory focuses on), any admissible isomorphism is an automorphism anyway.

Some googling suggests that one wants to take the (differential?) compositum between a differential field and its image under an admissible isomorphism. Also, I have a suspicion that the absence of an analog of the concept of algebraic closure may be relevant here, but I'm not at all sure.

Question: To what extent is the restriction to admissibility necessary or useful, and if there's no way to avoid it for differential fields, why can one develop ordinary Galois theory without mentioning it?

And if it's indeed unavoidable, is there any way to think of it categorically?

• You can always get such a commutative triangle given an isomorphisn $f: A \to B$ in any category (e.g. using $1_A$ and $f$), so I think Kaplansky has more in mind here. – Akhil Mathew Feb 13 '10 at 16:47
• You're right of course, that was nonsense (therefore removed). Weird. Can you find any statement/proof that uses this embedding into M? – user717 Feb 13 '10 at 17:39
• Kaplansky uses it to prove that any Picard-Vessiot extension of a characteristic zero differential field $K$ whose constant field is algebraically closed is normal (in that any element not in $K$ can be moved by a differential automorphism). It looks like he uses a lemma for Picard-Vessiot extensions to switch between admissible differential isomorphisms (for which he proves existence separately) and automorphisms. – Akhil Mathew Feb 13 '10 at 19:46

## 2 Answers

I don't think I have seen the terminology "admissible isomorphism" being used in differential Galois theory, except for Kaplansky's book. I guess in E. Kolchin's work everything is assumed to lie in a universal differential extension, and therefore he never makes the distinction.

Your observation about Picard-Vessiot extensions is right, and I don't think one needs the notion of admissibility to develop ordinary Picard-Vessiot theory, which is a theory based on the "equation" approach. In fact the efforts done at the time were focused on developing a Galois theory of differential fields that wasn't necessarily associated to differential equations (but it had to be a generalization of PV of course). However, many problems arise when one takes this "extension" approach, in fact finding the right notion of a normal extension is not easy. Classically a field extension $M$ over $K$ is normal if every isomorphism into some extension field of $M$ is an automorphism. However the equivalent statement for differential algebra implies that $M$ is algebraic over $K$ and that is too strong (in fact this is one of the main reasons why one has to allow admissible isomorphisms). Here are two early approaches to normality:

$M$ is weakly normal if $K$ is the fixed field of the set of all differential automorphisms of $M$ over $K$.

Apparently this definition wasn't very fruitful, and not much could be proven. The next step was the following definition:

$M$ is normal over $K$ if it is weakly normal over all intermediate differential fields.

This wasn't bad and Kolchin could prove that the map $L\to Gal(M/L)$ where $K\subset L\subset M$ bijects onto a certain subset of subgroups of $Gal (M/K)$. However the characterization of these subsets was an open question (Kolchin referred to it as a blemish). The property he was missing was already there in the theory of equations, as the existence of a superposition formula (that every solution is some differential rational function of the fundamental solutions and some constants). The relevant section in Kaplansky's book is sec 21. Now an admissible isomorphism of $M$ over $K$ is a differential isomorphism, fixing $K$ element wise, of $M$ onto a subfield of a given larger differential field $N$. Thus, an admissible isomorphism $\sigma$ let's you consider the compositum $M\cdot \sigma(M)$ which is crucial to translating a superposition principle to field extensions. Indeed, if one denotes $C(\sigma)$ to be the field of constants of $M\cdot \sigma(M)$, then Kolchin defined an admissible isomorphism $\sigma$ to be strong if it is the identity on the field of constants of $M$ and satisfies $$M\cdot C(\sigma)=M\cdot \sigma(M)=\sigma (M)\cdot C(\sigma)$$

This was the right interpretation of what was happening in the PV case and so a strongly normal extension $M$ over $K$ was defined as an extension where $M$ is finitely generated over $K$ as a differentiable field, and every admissible isomorphism of $M$ over $K$ is strong. Now the theory became more complete. $Gal(M/K)$ may be identified with an algebraic group and there is a bijection between the intermediate fields and its closed subgroups. Now this incorporates finite normal extensions (when $Gal(M/K)$ is finite), Picard-Vessiot extensions (when $Gal(M/K)$ is linear) or Weierstrass extensions (when $Gal(M/K)$ is isomorphic to an elliptic curve).

For a better exposition of this, see if you can find "Algebraic Groups and Galois Theory in the Work of Ellis R. Kolchin" by Armand Borel.

• It seems to be too late to accept an answer (I placed a bounty on it earlier), but thanks very much! – Akhil Mathew Apr 6 '10 at 22:53

Expanding on Gjergji Zaimi's answer, (1) an hypothesis of admissibility is unnecessary and (2) the right context for differential Galois theories based on properties of extensions of differential fields as opposed to differential equations is somewhat more general than the strongly normal differential Galois theory of Kolchin.

There is an analogue of the notion of an algebraic closure for differential fields, called a constrained closure by Kolchin and his school and a differential closure in the model theoretic literature.

Specializing to characteristic zero, a differential closure $(L,\partial_1,\ldots,\partial_n)$ of a partial differential field $(K,\partial_1,\ldots,\partial_n)$ is a differential field extension having the property that $L$ is differentially closed which means that every finite system of differential equations over $L$ which has a solution in some differential field extension of $L$ already has a solution in $L$ and is universal for differentially closed field extensions of $K$ in the sense that if $K \hookrightarrow M$ is an embedding of $K$ into a differentially closed field, then there is an embedding of $L$ into $M$ over $K$. The differential closure of a differential field $K$ is unique up to isomorphism over $K$, but unlike the algebraic closure it is not minimal. That is, if $L$ is the differential closure of $K$, then there may be a differentially closed field $M$ containing $K$ and properly contained in $L$.

The more general differential Galois theory I mentioned above is explained in the paper Pillay, Anand Differential Galois theory. I. Illinois J. Math. 42 (1998), no. 4, 678--699 at least in the case of ordinary differential fields. Unlike Kolchin's theory, it is possible to have differential algebraic groups (instead of merely algebraic groups) as Galois groups.