Let $f\in R[x]$ (where $R$ is a unital ring), we can define a map $D:R[x]\to R[x]$ called the "Hasse Derivative" as:
$$D(f) = \sum_{i=1}^n i a_i x^{i-1}$$
In this case, we can show that there is "chain rule" for the Hasse derivative:$$D(f\circ g) = D(f)\circ g \cdot D(g)$$
The Hasse derivative can then be extended to the Multivariate Partial Hasse Derivative (MPHD) which defines the following map:
$$D_i:R[x_1,\cdots,x_n]\to R[x_1,\cdots,x_n]$$
$$D_i(f)= D(f^{(i)})$$
Where $f:R[x_1,\cdots,x_n]\to R[x_1,\cdots,x_n]$, and $f^{(i)}$ is the "univariate part" polynomial in $x_i$ having fixed the other $x_j$ to some value in $R^{n-1}$.
Now suppose, we had a function $F: R^n\to R^m$ given by $m$ coordinate functions $f_i: R^n\to R$. It is then possible to define a "Hasse Jacobian" $J_F$ given by the matrix:
$$ J_F = \left( \begin{matrix} D_1(f_1) & D_2(f_1) & \cdots & D_n(f_1) \\ D_1(f_2) & D_2(f_2) & \cdots & D_n(f_2) \\ \vdots & \vdots & \ddots & \vdots \\ D_1(f_m) & D_2(f_m) & \cdots & D_n(f_m) \end{matrix} \right) $$
My question is, then, is there a chain rule for the Hasse Jacobian, ie. some relationship equating $J_{F\circ G}$ to expression with $J_F, J_G$ and $G$?
Best Answer
Yes, it's the same as the usual chain rule and it's much easier to establish in this algebraic setting. (I'll take $R$ to be commutative for safety, though.) Adjoin an infinitesimal $\varepsilon$ satisfying $\varepsilon^2 = 0$; then $D = \frac{d}{dx}$ can be defined via the identity
$$f(x + \varepsilon) = f(x) + \varepsilon D(f(x))$$
and similarly for the partial derivatives. (This gives quick and easy proofs that $D$ is linear and satisfies the product rule; a good exercise if you haven't seen this before.) For a polynomial function $F : \mathbb{A}^n \to \mathbb{A}^n$ we can similarly define the Jacobian via
$$F(v + \varepsilon w) = F(v) + \varepsilon J_F(v) w$$
which allows for a very straightforward and intuitive proof and derivation of the chain rule:
$$\begin{align*} (F \circ G)(v + \varepsilon w) &= F \left( G(v) + \varepsilon J_G(v) w \right) \\ &= F(G(v)) + \varepsilon J_F(G(v)) J_G(v) w \\ &= (F \circ G)(v) + \varepsilon J_{F \circ G}(v) w \end{align*}.$$
so we get $J_{F \circ G}(v) = (J_F \circ G)(v) J_G(v)$ as expected. Similar proofs are available in calculus using asymptotic notation but here they are particularly simple; we really get to consider first-order expansions only and ignore higher-order terms by fiat.
I am not familiar with the term "Hasse derivative" for this operation; Wikipedia uses this term for a slightly different map, and the term I've seen is formal derivative (although that Wikipedia article is a bit strange). But see the Wikipedia article on derivations for more.