The Hasse derivative, part II: Multivariate partial Hasse derivatives.

Let be a commutative unitary ring, and let $x_i, i \in I$ be a family of indeterminates over . We consider the polynomial ring $S = R[\{ x_i \mid i \in I \}]$ , obtained from by adjoining all indeterminates $x_i, i \in I$ . The two most used examples are $I = \{ \bullet \}$ , where , and $I = \{ 1, \dots, n \}$ , where $S = R[x_1, \dots, x_n]$ . (Note that if we use finitely many polynomials $f_1, \dots, f_r$ , there exist a finite subset $J \subset I$ such that $f_1, \dots, f_r \in R[\{ x_j \mid j \in J \}]$ , 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 $f \in S$ , $i \in I$ and $s \in R^I$ , we can define $f_i(s) = f|_{x_j = s(j) \text{ for } j \neq i}$ , which is an element of . Then $D^{(k)} f_i(s)$ is another element of , whence we can define $D^{(k)}_{x_i} f(s) := (D^{(k)} f_i(s))|_{x_i = s(i)} = (D^{(k)} f_i(s))(s(i))$ . This gives a function

$D^{(k)}_{x_i} f : R^I \to R, \quad s \mapsto D^{(k)}_{x_i} f(s),$

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 $R_i := R[\{ x_j \mid j \in I, j \neq i \}]$ ; here, is another commutative unitary ring. Now is a univariate polynomial ring, whence we have the usual Hasse derivative. Denote the -th Hasse derivative $D^{(k)} : R_i[x_i] \to R_i[x_i]$ by $D^{(k)}_{x_i}$ ; then we obtain an -linear (in fact, -linear) operator $D^{(k)}_{x_i} : S \to S$ . In fact, if we evaluate $D^{(k)} f$ at a point $s \in R^I$ , we obtain the same value as $D^{(k)} f(s)$ defined above. We will prove this in a minute; before that we will prove a more general result on $D^{(k)}_{x_i}$ .

Proposition.

Let $\varphi : R \to R'$ be a homomorphism of unitary commutative rings and (i.e. it additionally satisfies $\varphi(1_R) = 1_{R'}$ ). If is an indeterminate over and , and $\varphi^* : R[x] \to R'[x]$ the natural continuation of $\varphi$ with $\varphi^*(x) = x$ and $\varphi^*(r) = \varphi(r)$ for all $r \in R$ , then for $f \in R[x]$ , we have

$D^{(k)} \varphi^*(f) = \varphi^*(D^{(k)} f).$

Proof.

This follows directly from the definition of the Hasse derivative: if $f = \sum_{i=0}^n a_i x^i \in R[x]$ and $k \in \N$ , we have

$\varphi^*(D^{(k)} f) ={} & \varphi^*\biggl( \sum_{i=k}^n \binom{i}{k} a_i x^{i - k} \biggr) \\ {}={} & \sum_{i=k}^n \binom{i}{k} \varphi(a_i) x^{i - k} = D^{(k)} \varphi^*(f).$

Corollary.

For all $f \in S$ and $s \in R^I$ , we have $(D^{(k)}_{x_i} f)(s) = D^{(k)} f(s)$ .

Proof.

Let $\varphi : R_i \to R$ be the substitution homomorphism $f \mapsto f|_{x_j = s(j) \text{ for } j \neq i}$ . We interpret as an element of ; note that the above defined equals $\varphi^*(f)$ . Then, by the proposition,

$D^{(k) }\varphi^*(f) = \varphi^*(D^{(k)} f) = \varphi^*(D^{(k)}_{x_i} f).$

Now $\varphi^*(D^{(k)}_{x_i} f)|_{x_i = s(i)} = (D^{(k)}_{x_i} f)(s)$ , whence

$(D^{(k)}_{x_i} f)(s) = (D^{(k)} \varphi^*(f))|_{x_i = s(i)} = (D^{(k)} f_i)|_{x_i = s(i)} = D^{(k)}_{x_i} f(s)$

by definition of $D^{(k)}_{x_i} f(s)$ .

Hence, the following definition makes sense:

Definition.

The -linear opeator $D^{(k)}_{x_i} : S \to S$ 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:

Proposition.

Let $i, j \in I$ , $k, \ell \in \N$ and $f \in S$ . Then $D^{(k)}_{x_i} D^{(\ell)}_{x_j} f = D^{(\ell)}_{x_j} D^{(k)}_{x_i} f$ .

For this, consider the notation $R_{i,j} = R[\{ t \in I \mid t \neq i, j \}]$ ; then $S = R_{i,j}[x_i, x_j] = R_{i,j}[x_j][x_i] = R_{i,j}[x_i][x_j]$ .

Proof.

By $R_{i,j}$ -linearity of $D^{(k)}_{x_i}$ and $D^{(\ell)}_{x_j}$ , it suffices to show the result for with $s, t \in \N$ . Now

$D^{(k)}_{x_i} D^{(\ell)}_{x_j} f ={} & D^{(k)}_{x_i} \binom{t}{\ell} x_i^s x_j^{t - \ell} = \binom{s}{k} \binom{t}{\ell} x_i^{s - k} x_j^{t - \ell} \\ {}={} & D^{(\ell)}_{x_j} \binom{s}{k} x_i^{s - k} x_j^t = D^{(\ell)}_{x_j} D^{(k)}_{x_i} f.$

Now, let us define the following notation. We let $\N^{(I)}$ be the set of functions $I \to \N$ which are zero for all but finitely many elements of . In case is finite, $\N^{(I)} = \N^I$ in the usual sense. The set $\N^{(I)}$ will be the set of multiindices we use. For $s \in \N^{(I)}$ , define $D^{(s)} : S \to S$ by $D^{(s)} := D^{(s(i_1))}_{x_{i_1}} \circ \dots \circ D^{(s(i_t))}_{x_{i_t}}$ , where $\{ i_1, \dots, i_t \} \subseteq I$ is the support of . By the above proposition, $D^{(s)}$ is independent of the order of $i_1, \dots, i_t$ , and as $D^{(0)}_{x_i} = \id_S$ , we can also include elements outside the support. We sometimes use the more suggestive notation

$D^{(s)} = \prod_{i \in I} D^{(s(i))}_{x_i},$

which is hence also justified. Moreover, for $s \in \N^{(I)}$ and $\lambda \in R^I$ , define $(x - \lambda)^s := \prod_{j=1}^t (x_j - \lambda(j))^{s(j)} \in S$ . Again, this is well-defined. Using this notation, we obtain Taylor's formula:

Theorem (Taylor formula).

For $f \in S$ and $\lambda \in R^I$ , we have

$f = \sum_{s \in \N^{(I)}} (D^{(s)} f)(\lambda) (x - \lambda)^s.$

Proof.

By the -linearity of the equality, it suffices to consider $f = x_{i_1}^{e_1} \cdots x_{i_t}^{e_t}$ with $i_1, \dots, i_t \in I$ pairwise distinct and $e_1, \dots, e_t \in \N$ . Define $I' := \{ i_1, \dots, i_t \}$ . If $s \in \N^{(I)}$ satisfies $s(i) \neq 0$ for $i \not\in I'$ , then $D^{(s)} f = 0$ . Hence,

$\sum_{s \in \N^{(I)}} (D^{(s)} f)(\lambda) (x - \lambda)^s = \sum_{s \in \N^{(I')}} (D^{(s)} f)(\lambda) (x - \lambda)^s;$

moreover, one can restrict to the subring $S' = R[x_{i_1}, \dots, x_{i_t}]$ , 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 $I = \{ 1, \dots, n \}$ .

But now we can prove this by induction on . If we write with $s' \in \N^{n-1}$ , $s_n \in \N$ and $\lambda = (\lambda', \lambda_n)$ with $\lambda' \in R^{n-1}$ , $\lambda_n \in R$ for $s \in \N^n$ and $\lambda \in R^n$ , then

$D^{(s)} f = D^{(s_n)} D^{(s')} f \quad \text{and} \quad (x - \lambda)^s = (x - \lambda')^{s'} (x_n - \lambda_n)^{s_n}.$

Hence,

$& \sum_{s \in \N^n} (D^{(s)} f)(\lambda) (x - \lambda)^s \\ {}={} & \sum_{s \in \N^n} (D^{(s_n)} D^{(s')} f)(\lambda) (x - \lambda')^{s'} (x_n - \lambda_n)^{s_n} \\ {}={} & \sum_{s_n \in \N} \sum_{s' \in \N^{n-1}} (D^{(s_n)} D^{(s')} f)(\lambda) (x - \lambda')^{s'} (x_n - \lambda_n)^{s_n} \\ {}={} & \sum_{s_n \in \N} \sum_{s' \in \N^{n-1}} \bigl(D^{(s_n)} (D^{(s')} f)(\lambda') \bigr)(\lambda_n) (x - \lambda')^{s'} (x_n - \lambda_n)^{s_n} \ {}={} & \sum_{s_n \in \N} D^{(s_n)} \biggl( \sum_{s' \in \N^{n-1}} (D^{(s')} f)(\lambda') (x - \lambda')^{s'} \biggr)(\lambda_n) (x_n - \lambda_n)^{s_n}.$

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

$\sum_{s' \in \N^{n-1}} (D^{(s')} f)(\lambda') (x - \lambda')^{s'} = f,$

whence

$\sum_{s \in \N^n} (D^{(s)} f)(\lambda) (x - \lambda)^s = \sum_{s_n \in \N} D^{(s_n)} f(\lambda_n) (x_n - \lambda_n)^{s_n}.$

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

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

Corollary (Identity Theorem).

Let $f \in S$ and $\lambda \in R^I$ . If $D^{(s)} f(\lambda) = 0$ for all $s \in \N^{(I)}$ , then .

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

For example, for $s, t \in \N^{(I)}$ , let us define $\binom{s + t}{s} := \prod_{j=1}^u \binom{s(i_j) + t(i_j)}{s(i_j)}$ if both and have no non-zero values outside $\{ i_1, \dots, i_u \}$ . Note that $\binom{s + t}{s} = \binom{s + t}{t}$ . We now obtain the following:

Proposition.

For $s, t \in \N^{(I)}$ , we have $D^{(s)} \circ D^{(t)} = \binom{s + t}{s} D^{(s + t)}$ . In particular, $D^{(s)} \circ D^{(t)} = D^{(t)} \circ D^{(s)}$ .

Proof.

Let denote the ring $R[x_j \mid j \neq i_1, \dots, i_u\}]$ , where and have no non-zero values outside $\{ i_1, \dots, i_u \}$ ; then we have reduced to the ring $R'[x_{i_1}, \dots, x_{i_u}]$ , i.e. to the case of being finite, say $I = \{ 1, \dots, n \}$ . Write $s = (s_1, \dots, s_n)$ and $t = (t_1, \dots, t_n)$ . As $D^{(k)}_{x_i}$ and $D^{(\ell)}_{x_j}$ commute for $i \neq j$ , we obtain

$D^{(s)} \circ D^{(t)} = \prod_{i=1}^n (D^{(s_i)}_{x_i} \circ D^{(t_i)}_{x_i}).$

Using the corresponding univariate result, we have $D^{(s_i)}_{x_i} \circ D^{(t_i)}_{x_i} = \binom{s_i + t_i}{s_i} D^{(s_i + t_i)}_{x_i}$ . Hence,

$D^{(s)} \circ D^{(t)} = \prod_{i=1}^n \biggl( \binom{s_i + t_i}{s_i} D^{(s_i + t_i)}_{x_i} \biggr) = \binom{s + t}{s} D^{(s + t)}.$

We also obtain the Leibniz rule:

Proposition (Generalized Leibniz Rule).

For $f_1, \dots, f_m \in S$ and $s \in \N^{(I)}$ , we have

$D^{(s)} \prod_{i=1}^m f_i = \sum_{s_1 + \dots + s_m = s} \prod_{i=1}^m D^{(s_i)} f_i;$

here, the sum ranges over all $(s_1, \dots, s_m) \in (\N^{(I)})^m$ with $s_1 + \dots + s_m = s$ . As a special case, for $f, g \in S$ , we have

$D^{(s)}(f g) = \sum_{s' + s'' = s} D^{(s')}(f) D^{(s'')}(g).$

Proof.

Using the standard argument, we reduce to the case $I = \{ 1, \dots, n \}$ and we use the standard decomposition $\N^n = \N^{n-1} \times \N$ . Hence,

$\sum_{s_1 + \dots + s_m = s} \prod_{i=1}^m D^{(s_i)} f_i = \sum_{s'_1 + \dots + s_m' = s'} \sum_{s_{1,n} + \dots + s_{m,n} = s_n} \prod_{i=1}^m D^{(s_i')} D^{(s_{i,n})} f_i,$

where the second sum ranges over all $(s'_1, \dots, s_ m') \in (\N^{n-1})^m$ and the third sum ranges over all $(s_{1,n}, \dots, s_{m,n}) \in \N^m$ . Now

$& \sum_{s'_1 + \dots + s_m' = s'} \sum_{s_{1,n} + \dots + s_{m,n} = s_n} \prod_{i=1}^m D^{(s_i')} D^{(s_{i,n})} f_i \ {}={} & \sum_{s'_1 + \dots + s_m' = s'} D^{(s_i')} \sum_{s_{1,n} + \dots + s_{m,n} = s_n} \prod_{i=1}^m D^{(s_{i,n})} f_i,$

and applying the univariate result to the base ring $R[x_1, \dots, x_{n-1}]$ , we obtain

$\sum_{s_1 + \dots + s_m = s} \prod_{i=1}^m D^{(s_i)} f_i ={} & \sum_{s'_1 + \dots + s_m' = s'} D^{(s_i')} D^{(s_n)}_{x_n} \prod_{i=1}^m f_i \ {}={} & D^{(s_n)}_{x_n} \sum_{s'_1 + \dots + s_m' = s'} D^{(s_i')} \prod_{i=1}^m f_i.$

By induction hypothesis, this equals

$D^{(s_n)}_{x_n} D^{(s')} \prod_{i=1}^m f_i = D^{(s)} \prod_{i=1}^m f_i.$

Comments.

Alexey Maevskiy wrote on October 5, 2009:

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: $\varphi(1_R)=1_{R'}$ or $\varphi^*(x) = \varphi(1_R)x$ . If both requirements does not hold its easy to make a counterexample for situation when proposition is false ( $\varphi:Z_{10}\rightarrow Z_{10}$ , $\varphi(a)=6a$ ).

Felix Fontein wrote on October 5, 2009:

Ah, I think this is just a problem with notation. When I wrote, “homomorphism of unitary commutative rings”, I assumed that the homomorphism satisfies $\varphi(1_R) = 1_{R'}$ . :-) I think I have to make it a bit more clearly, to make it less confusing... Thanks for the hint!

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The Hasse derivative, part II: Multivariate partial Hasse derivatives.

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The Hasse derivative, part II: Multivariate partial Hasse derivatives.

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