4 ms·

It is pretty common to talk about derivatives in abstract algebra in contexts where the calculus definition makes no sense. Broadly, there is a class of functi

It is pretty common to talk about derivatives in abstract algebra in contexts where the calculus definition makes no sense. Broadly, there is a class of functions refered to as "derivations" that can be viewed as a generalization of the derivative. In particular, a derivation satisfies 2 properties: 1) It is linear 2) It satisfies the product rule. Any function that satisfies these rules is often called a "derivative". Notably, the function discussed in the article fails the linearity test, which is a pretty big problem for calling it a derivative.

codeflo 2y ago
Also, the abstract algebra definition is an actual generalization, because the (analysis) derivative acts as a linear operator in the vector space of (suitably smooth) funtions. It blew my mind when I first encountered this way of looking at derivatives. And from there, it makes complete sense to look at operators with similar properties in other vector spaces. The definition presented here is a loose analogy to derivatives rather than an actual generalization, which doesn't fully justify using the name IMO.
- deleted 2y ago
  [deleted]
- 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 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.
    - gizmo686 2y ago
      The 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".
magicalhippo 2y ago
> It satisfies the product rule. It's been many years since I had calculus at uni, and never any abstract math. Why is the product rule picked as the "interesting" attribute of derivatives, ie to serve as the basis for the generalization? Is there some deeper connection of the product rule in ordinary derivatives that singles out the product rule over the other properties a derivative has? For me, a key aspect of derivatives is that it allows for something like Taylor expansion or integrals to exist. Are there any equivalent things to these product-rule-generalized derivatives?
- gizmo686 2y ago
  I actually think it is better to not think of derivations as a generalization of derivatives; but to think of derivatives as just one of many examples of a derivation. That the name 'derivative' comes from calculus and became the name for this abstraction is just a quirk of history. You could also consider the notion of a linear function to be a generalization of derivatives. Or even the notion of homomorphisms in general. Having said that, there are many usages of other derivations where the calculus inspiration is clear, even if the geometric meaning that motivated the calculus is lost. For instance, we often talk about polynomials over arbitrary fields. In general, there is no way to graph such polynomials. There is no notion of tangent lines, slope, "continuous", or even "less than". There is, however, still the notion of roots and the multiplicity of roots. These notions turn out to be quite important. When working with any polynomial, you can define the "formal derivative" as a derivation that also satisfies D(x) = 1, D(a) = 0 (where a is an element of the underlying field). This operator behaves as you would naively expect a derivative to behave over polynomials. In Galois theory, it is important to distinguish between polynomials that have repeated roots, and those that do not. If you have a polynomial f(x), you can determine this by taking its formal derivative f'(x). Then, you can easily compute their greatest common divisor [0]. If this is a constant, then you know f has no repeated roots. [0] Using Euclid's algorithm, this is a purely mechanical process that can be done without needing to factor either f or f'. A similar trick has actually been used to attack real word cryptography. If there are secret primes p and q, and a public number pq, many cryptosystems assume that it is infeasible to determine what p and q are. However, if there is a bad random number generator, you might get 2 different keys that share a prime, so have the public numbers pq and ps, then you can easily determine that p is a common factor, from which you can easily recover q and s. This means that you can look for a large collection of public keys and try this attack on possible pair of them.
bjornsing 2y ago
> 1) It is linear This ”derivative” is not linear though, and that was sort of what motivated my question.

codeflo 2y ago

Also, the abstract algebra definition is an actual generalization, because the (analysis) derivative acts as a linear operator in the vector space of (suitably smooth) funtions. It blew my mind when I first encountered this way of looking at derivatives. And from there, it makes complete sense to look at operators with similar properties in other vector spaces. The definition presented here is a loose analogy to derivatives rather than an actual generalization, which doesn't fully justify using the name IMO.

deleted 2y ago
[deleted]
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 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.
  - gizmo686 2y ago
    The 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".

magicalhippo 2y ago

> It satisfies the product rule. It's been many years since I had calculus at uni, and never any abstract math. Why is the product rule picked as the "interesting" attribute of derivatives, ie to serve as the basis for the generalization? Is there some deeper connection of the product rule in ordinary derivatives that singles out the product rule over the other properties a derivative has? For me, a key aspect of derivatives is that it allows for something like Taylor expansion or integrals to exist. Are there any equivalent things to these product-rule-generalized derivatives?

gizmo686 2y ago
I actually think it is better to not think of derivations as a generalization of derivatives; but to think of derivatives as just one of many examples of a derivation. That the name 'derivative' comes from calculus and became the name for this abstraction is just a quirk of history. You could also consider the notion of a linear function to be a generalization of derivatives. Or even the notion of homomorphisms in general. Having said that, there are many usages of other derivations where the calculus inspiration is clear, even if the geometric meaning that motivated the calculus is lost. For instance, we often talk about polynomials over arbitrary fields. In general, there is no way to graph such polynomials. There is no notion of tangent lines, slope, "continuous", or even "less than". There is, however, still the notion of roots and the multiplicity of roots. These notions turn out to be quite important. When working with any polynomial, you can define the "formal derivative" as a derivation that also satisfies D(x) = 1, D(a) = 0 (where a is an element of the underlying field). This operator behaves as you would naively expect a derivative to behave over polynomials. In Galois theory, it is important to distinguish between polynomials that have repeated roots, and those that do not. If you have a polynomial f(x), you can determine this by taking its formal derivative f'(x). Then, you can easily compute their greatest common divisor [0]. If this is a constant, then you know f has no repeated roots. [0] Using Euclid's algorithm, this is a purely mechanical process that can be done without needing to factor either f or f'. A similar trick has actually been used to attack real word cryptography. If there are secret primes p and q, and a public number pq, many cryptosystems assume that it is infeasible to determine what p and q are. However, if there is a bad random number generator, you might get 2 different keys that share a prime, so have the public numbers pq and ps, then you can easily determine that p is a common factor, from which you can easily recover q and s. This means that you can look for a large collection of public keys and try this attack on possible pair of them.

bjornsing 2y ago

> 1) It is linear This ”derivative” is not linear though, and that was sort of what motivated my question.