2 ms·
> And from there, it makes complete sense to look at operators with similar properties in other vector spaces. It does, but this one's on a rig (= ring - negat
by JadeNB 2y ago
> And from there, it makes complete sense to look at operators with similar properties in other vector spaces.
It does, but this one's on a rig (= ring - negatives), not a vector space over any field.
- codeflo 2y agoI don't see how you would define linearity in a ring (I don't mean a module, just a ring). I.e. D(af) = aD(f) doesn't make sense if you don't have scalar multiplication.
- gizmo686 2y agoThe minimum would be to ask for D(a+b) = D(a) + D(b). EDIT: Actually, according to wikipedia, that is exactly what is done in differential algebra https://en.wikipedia.org/wiki/Differential_algebra https://en.wikipedia.org/wiki/Differential_algebra
- JadeNB 2y ago> I don't see how you would define linearity in a ring (I don't mean a module, just a ring). I.e. D(af) = aD(f) doesn't make sense if you don't have scalar multiplication. Every ring is a module over itself. But you wouldn't want the definition you propose; instead, you'd want `D(ab) = aD(b) + D(a)b`. If you really like some sort of linearity to be present, you could observe that this property forces every derivation to be linear as a transformation of `R_0`-modules, where `R_0` is the subring `ker(D)` of "constants".