Let be a commutative unitary ring, and let be a family of indeterminates over . We consider the polynomial ring , obtained from by adjoining all indeterminates . The two most used examples are , where , and , where . (Note that if we use finitely many polynomials , there exist a finite subset such that , i.e. it suffices to consider finitely many variables.)

In classical analysis, one defines the partial derivative by fixing all variables but one, and then using the classical one-dimensional derivative. This can be done similarly in the case of Hasse derivatives. Given , and , we can define , which is an element of . Then is another element of , whence we can define . This gives a function

which can be considered as the *-th partial Hasse derivative of with respect to *. Now we would like the derivative to be another element of . In classical analysis, polynomial functions are in bijection to polynomials, and one can show using certain rules that the partial derivative of a polynomial is again a polynomial. In case is an arbitrary ring, one has in general no longer the bijection between polynomials and polynomial functions.

In the case of derivatives over arbitrary rings, we have the advantage that we can naturally identify with , where ; here, is another commutative unitary ring. Now is a univariate polynomial ring, whence we have the usual Hasse derivative. Denote the -th Hasse derivative by ; then we obtain an -linear (in fact, -linear) operator . In fact, if we evaluate at a point , we obtain the same value as defined above. We will prove this in a minute; before that we will prove a more general result on .

Let be a homomorphism of unitary commutative rings and (i.e. it additionally satisfies ). If is an indeterminate over and , and the natural continuation of with and for all , then for , we have

This follows directly from the definition of the Hasse derivative: if and , we have

For all and , we have .

Let be the substitution homomorphism . We interpret as an element of ; note that the above defined equals . Then, by the proposition,

Now , whence

by definition of .

Hence, the following definition makes sense:

The -linear opeator is called the *-th partial Hasse derivative with respect to *.

As in the case of usual partial derivatives (in the case of “nice” functions, i.e. the partial derivatives are continuous in a neighborhood of the point we are interested in), the Hasse derivatives commute:

Let , and . Then .

For this, consider the notation ; then .

By -linearity of and , it suffices to show the result for with . Now

Now, let us define the following notation. We let be the set of functions which are zero for all but finitely many elements of . In case is finite, in the usual sense. The set will be the set of multiindices we use. For , define by , where is the support of . By the above proposition, is independent of the order of , and as , we can also include elements outside the support. We sometimes use the more suggestive notation

which is hence also justified. Moreover, for and , define . Again, this is well-defined. Using this notation, we obtain Taylor's formula:

For and , we have

By the -linearity of the equality, it suffices to consider with pairwise distinct and . Define . If satisfies for , then . Hence,

moreover, one can restrict to the subring , as every appearing object lies in that ring. Hence, it suffices to show the result for , i.e. for a finite index set . We can assume .

But now we can prove this by induction on . If we write with , and with , for and , then

Hence,

Now, by induction hypothesis ( with the base ring ),

whence

But by the classical univariate case which we already proved, this equals itself.

Again, we get the Identity Theorem as a direct corollary:

Let and . If for all , then .

In a similar manner as Taylor's formula, we can also adapt other one-dimensional results to the multivariate case.

For example, for , let us define if both and have no non-zero values outside . Note that . We now obtain the following:

For , we have . In particular, .

Let denote the ring , where and have no non-zero values outside ; then we have reduced to the ring , i.e. to the case of being finite, say . Write and . As and commute for , we obtain

Using the corresponding univariate result, we have . Hence,

We also obtain the Leibniz rule:

For and , we have

here, the sum ranges over all with . As a special case, for , we have

Using the standard argument, we reduce to the case and we use the standard decomposition . Hence,

where the second sum ranges over all and the third sum ranges over all . Now

and applying the univariate result to the base ring , we obtain

By induction hypothesis, this equals

## Comments.

Hello! Thank you for this excellent article! I think it will be useful for many peoples.

But I note that in the first proposition it is necessary to add one of the following constraints on the ring homomorphism: or . If both requirements does not hold its easy to make a counterexample for situation when proposition is false (, ).

Ah, I think this is just a problem with notation. When I wrote, “

homomorphism of unitary commutative rings”, I assumed that the homomorphism satisfies . :-) I think I have to make it a bit more clearly, to make it less confusing... Thanks for the hint!