D-Modules = Derivatives + Vector Space (That’s it)

Let's recap the 12 rules for how to build a vector space is:

1. You have a set (M) of elements, a set of "vectors"
2. You have a field (K) of elements, called "scalars" (ex: real numbers, complex numbers, etc.)
3. The addition operation (+) where vectors A, B in M add up to still be inside M: A + B = C still in M
4. The multiplication operation (*) where c in K and A in M where c * A = L still in M
5. The addition operation is associative for three vectors A, B, C in M. (A + B) + C = A + (B + C)
6. The addition operation commutes: A + B = B + A for vectors A and B in M
7. There is a zero vector, ZV, in M where ZV + A = A for any vector A in M
8. Every vector H has an additive inverse AH where AH + H = ZV
9. Multiplication respects an associative-like rule a * (b * A) = (a * b) * A where a and b are in K and A is in M
10. If U is the identity element of K, then U * A = A for all A in M
11. Distributive property holds for adding vectors: c * (A + B) = c * A + c * B where c in K and A, B in M
12. Distributive property holds for adding scalars: (g + h) * A = g * A + h * A for g, h in K and M

Now, you can replace the field K (or "enlarge" it) with a ring Q. This ring is made up of all linear combinations of x^n and d^n/dx^n. (Linear combinations using complex numbers). Note that Q contains all complex numbers, so if K was the complex field, then we are "enlarging" the field K to the ring Q.

Q is called the Weyl algebra, and if you did replace K with Q, then the previous vector space is now a D-module. Also called "a module 'over' the linear differential operators of one variable." The D historically stands for differential operator.

We say that M is "acted on" by Q. We could also have said M is acted on by K but the "action" of numbers (the elements of a number field, K in this case) is not that interesting.

If you replace a vector space's field with a ring, in general you get a module. Some of the 12 rules will be broken, can you tell which ones? For example, rings are not requires to be commutative.