Differentiating by Prime Numbers

It is likely a fair assumption that you, the reader, are not only familiar with but even quite adept at differentiating by $x$. What about differentiating by 13? That certainly didn’t come up in my calculus class! From a calculus perspective, this is ridiculous: are we supposed to take a limit as 13 changes?

One notion of differentiating by 13, or any other prime number, is the notion of $p$-derivation discovered independently by Joyal Joy85 and Buium Bui96. $p$-derivations have been put to use in a range of applications in algebra, number theory, and arithmetic geometry. Despite the wide range of sophisticated applications, and the fundamentally counterintuitive nature of the idea of differentiating by a number, $p$-derivations are elementary to define and inviting for exploration.

In this article, we will introduce $p$-derivations and give a few basic ways in which they really do act like derivatives by numbers; our hope is that you will be inspired and consider adding $p$-derivations to your own toolkit!

$p$-Derivations on ${\mathbb{Z}}$

First we want to discuss differentiating one number $n$, by another, $p$; i.e., what we will call $p$-derivations on $\mathbb{Z}$. Before we succeed, we need to abandon the notion of derivative as a limit as the input varies by a small amount: the thing $p$ that we are differentiating by does not vary, and the thing $n$ that we are differentiating does not even have an input! Instead, we take a little inspiration from elementary number theory.

Let $p$ be a prime number. By Fermat’s little theorem, for any integer $n$, we have

$$\begin{equation*} n \equiv n^p \mod \ p, \end{equation*}$$

so we can divide the difference $n-n^p$ by $p$. The starting point of our journey is that not only can we divide by $p$ here, but we should. The $p$-derivation on $\mathbb{Z}$ is the result of this process. Namely:

Definition 1.

For a prime number $p$, the $p$-derivation on $\mathbb{Z}$ is defined as the function ${\delta _p:\mathbb{Z}\to \mathbb{Z}}$ given by the formula

$$\begin{equation*} \delta _p(n) = \frac{n-n^p}{p}. \end{equation*}$$

So, in particular, there is the $2$-derivation on $\mathbb{Z}$ and the $13$-derivation on $\mathbb{Z}$ given respectively by

$$\begin{equation*} \delta _2(n) = \frac{n-n^2}{2} \ \ \text{and} \ \ \delta _{13}(n) = \frac{n-n^{13}}{13}. \end{equation*}$$

Let’s plug in a few values:

$$\begin{equation*} \begin{array}{ c | c c c } n & \delta _2(n) & \delta _3(n) & \delta _5(n) \\\hline \vdots &\vdots &\vdots & \vdots \\-4 & -10 & 20 & 204 \\-3 & -6 & 8 & 48 \\-2 & -3 & 2 & 6 \\-1 & -1 & 0 & 0 \\0 & 0 & 0 & 0 \\1 & 0 & 0 & 0 \\2 & -1 & -2 & -6 \\3 & -3 & -8 &-48 \\4 & -6 & -20 &-204 \\5 & -10 & -40 &-624 \\6 & -15 & -70 &-1554 \\\vdots &\vdots &\vdots &\vdots \end{array} \end{equation*}$$

A quick look at this table suggests a few observations, easily verified from the definition:

•

Numbers are no longer “constants” in the sense of having derivative zero, but at least $0$ and $1$ are.

•

These functions are neither additive nor multiplicative, e.g.:$$\begin{equation*} \delta _p(1) + \delta _p(1) \neq \delta _p(2), \end{equation*}$$

$$\begin{equation*} \delta _p(1) \delta _p(2) \neq \delta _p(2). \end{equation*}$$

•

$\delta _p$ is an odd function, at least for $p\neq 2$.

•

The outputs of $\delta _2$ are just the negatives of the triangular numbers.

We might also note that the outputs are very large in absolute value, and think that this operation is simply making a mess of our numbers. However, something more informative occurs if we think about largeness of the outputs from the point of view of $p$, namely, the $p$-adic order of $n$—the number of copies of $p$ in its prime factorization. Writing $n=p^a m$ with $\gcd (m,p)=1$, if $a>0$, we get

$$\begin{equation*} \delta _p(p^am) = \frac{p^a m - (p^a m)^p}{p} = p^{a-1} m (1-p^{ap-a}m^{p-1}). \end{equation*}$$

Since $p\geq 2$ and $a\geq 1$, we must have $ap-a\geq 1$, so $p$ does not divide $1-p^{ap-a}m^{p-1}$. In particular, the $p$-derivation decreases the $p$-adic order of a multiple of $p$ by exactly one. This leads to our first comparison with old-fashioned $\frac{d}{dx}$:

Comparison 1 (Order-decreasing property).

•

If $f\in \mathbb{R}[x]$ is a polynomial and $x=r$ is a root of $f$ of multiplicity $a>0$, then $x=r$ is a root of the polynomial $\frac{d}{dx}(f(x))$ of multiplicity $a-1$.

•

If $n$ is an integer and $p$ is a prime factor of $n$ of multiplicity $a>0$, then $p$ is a prime factor of the integer $\delta _p(n)$ of multiplicity $a-1$.

In particular, if $r$ is a simple root or $p$ is a simple factor, then it is no longer a root or factor of $\frac{d}{dx}(f(x))$ or $\delta _p(n)$ respectively.

Let’s check this against our table: the numbers $-2,2$, and $6$ that were divisible by $2$ but not $4$ result in odd numbers when we apply $\delta _2$, whereas $\pm 4$ returned even numbers no longer divisible by $4$. Note that this order-decreasing property says nothing about what happens when you apply $\delta _2$ to an odd number, and indeed, based on the table we observe that even and odd numbers can result. You can convince yourself that

$$\begin{equation*} \delta _2(n) \ \text{is} \ \begin{cases} \text{even} \ \text{if} \ n\equiv 0,1 \mod 4 \\ \text{odd} \ \text{if} \ n\equiv 2,3 \mod 4.\end{cases} \end{equation*}$$

We’ve observed already that these $p$-derivations on $\mathbb{Z}$ are not additive. This can be a bit unsettling for those of us (like myself) who are usually accustomed to the luxury of additive operators. However, any function satisfying the order-decreasing property of $\delta _p$ above must not be additive, since an additive function has to take multiples of $p$ to multiples of $p$. However, the error term can be made concrete:

$$\begin{equation*} \begin{aligned} \delta _p(m+n) -( \delta _p(m) + \delta _p(n) ) &= \frac{m^p + n^p -(m+n)^p}{p} \\ &= -\sum _{i=1}^{p-1} \frac{\binom{p}{i}}{p} m^i n^{p-i}.\end{aligned} \end{equation*}$$

All of the binomial coefficients $\binom{p}{i}$ appearing above are multiples of $p$, so this expression is, given a particular value of $p$, a particular polynomial in $m$ and $n$ with integer coefficients; let’s call it $C_p(m,n)$ for convenience. This gives us the following “sum rule” for $\delta _p$:

$$\begin{equation} \delta _p(m+n) = \delta _p(m) + \delta _p(n) + C_p(m,n). {\tag{$+$}} \end{equation}$$

Products satisfy a rule with a similar flavor:

$$\begin{equation} \delta _p(mn) = m^p \delta _p(n) + n^p \delta _p(m) + p \delta _p(m) \delta _p(n). {\tag{$\times $}} \end{equation}$$

The fact that we have rules to break things down into sums and products gives the basis for another comparison with old-fashioned $\frac{d}{dx}$:

Comparison 2 (Sum and product rules).

•

For polynomials $f(x), g(x)$, one can compute each of $\frac{d}{dx}(f+g)$ and $\frac{d}{dx}(fg)$ as a fixed polynomial expression in the inputs $f,g,\frac{d}{dx}(f),\frac{d}{dx}(g)$, namely $\frac{d}{dx}(f+g)= \frac{d}{dx}(f)+\frac{d}{dx}(g)$ and $\frac{d}{dx}(fg) = f \frac{d}{dx}(g) + g \frac{d}{dx}(f)$.

•

For integers $m,n$, one can compute each of ${\delta _p(m+n)}$ and $\delta _p(mn)$ as a fixed polynomial expression in the inputs $m,n,\delta _p(m),\delta _p(n)$, namely $+$ and $\times$.

We might pause to ask whether we could have hoped for a simpler way to differentiate by 13. If we want Comparsion 2 to hold, then the following theorem of Buium provides a definitive answer.

Theorem 1 (Buium Bui97).

Any function ${\delta :\mathbb{Z}\to \mathbb{Z}}$ that satisfies

•

a sum rule $\delta (m+n)=S(m,n,\delta (m),\delta (n))$ for some polynomial $S$ with integer coefficients

•

a product rule $\delta (mn)=P(m,n,\delta (m),\delta (n))$ for some polynomial $P$ with integer coefficients

is of the form

$$\begin{equation*} \delta (n) = \pm \frac{n-n^{p^e}}{p} + f(n) \end{equation*}$$

for some prime integer $p$, positive integer $e$, and polynomial $f$ with integer coefficients.

That is, any function satisfying a sum rule and a product rule is a mild variation on a $p$-derivation.

With the properties of $p$-derivations we have so far, we can recreate analogues of some familiar aspects of calculus. For example, from the product rule $\times$ and a straightforward induction, we obtain a power rule:

$$\begin{equation*} \delta _p(n^a) = \sum _{i=1}^a \binom{a}{i} p^{i-1} \delta _p(n)^{i} n^{(a-i)p}. \end{equation*}$$

Note that the $i=1$ term in the sum above, $a n^{(a-1)p} \delta _p(n)$, looks a bit like the power rule for usual derivatives. If we allow ourselves to extend $\delta _p$ to a map on $\mathbb{Q}$, then we get an analogue of the quotient rule:

$$\begin{equation*} \delta _p\left(\frac{m}{n}\right) = \frac{n^p\delta _p(m) -m^p \delta _p(n)}{n^{2p}+pn^p\delta _p(n)}. \end{equation*}$$

Of the main cast of characters in a first class on derivatives, perhaps the most conspicuous one missing at this point is the chain rule. Since there is no way to compose a number with a number, we will need a notion of $p$-derivations for functions to state a sensible analogue of the chain rule.

$p$-Derivations for General Commutative Rings

One can define $p$-derivations for commutative rings with $1$.

Definition 2.

Let $R$ be a commutative ring with $1$ and $p$ a prime integer. A $p$-derivation on $R$ is a function $\delta :R\to R$ such that $\delta (0)=\delta (1)=0$ and $\delta$ satisfies the sum rule $(+)$ and the product rule $(\times )$ above; i.e., for all $r,s\in R$,

$$\begin{equation} \delta _p(r+s) = \delta _p(r) + \delta _p(s) - \sum _{i=1}^{p-1} \frac{\binom{p}{i}}{p} r^i s^{p-i} {\tag{$+$}} \end{equation}$$

and

$$\begin{equation} \delta _p(rs) = r^p \delta _p(s) + r^p \delta _p(s) + p \delta _p(r) \delta _p(s). {\tag{$\times $}} \end{equation}$$

Evidently, the functions $\delta _p$ we defined on $\mathbb{Z}$ above are $p$-derivations. In fact, for a fixed $p$, a simple induction and the sum rule show that for any $p$-derivation $\delta$ on a ring $R$ and any $n$ in the prime subring (image of $\mathbb{Z}$) of $R$, $\delta (n)=\delta _p(n)$.

The other basic example is as follows. Take the ring of polynomials in $n$ variables with integer coefficients, $R=\mathbb{Z}[x_1,\dots ,x_n]$. For any polynomial $f(x_1,\dots ,x_n)$, we can consider its $p$th power $f(x_1,\dots ,x_n)^p$, or we can plug in $p$th powers of the variables as inputs to get $f(x_1^p,\dots ,x_n^p)$. These are different, but they agree modulo $p$ as a consequence of the “freshman’s dream.” Namely, in the quotient ring $R/pR\cong \mathbb{Z}/p\mathbb{Z}[x_1,\dots ,x_n]$,

$$\begin{equation*} (f+g)^p=f^p + \sum _{i=1}^{p-1} \binom{p}{i} f^i g^{p-i} + g^p = f^p + g^p, \end{equation*}$$

since each $\binom{p}{i}$ is a multiple of $p$, and

$$\begin{equation*} (fg)^p = f^p g^p \end{equation*}$$

as a consequence of commutativity, so the map ${f\mapsto f^p}$ is a ring homomorphism in $R/pR$, called the Frobenius map. Thus, in $R/pR$, taking $p$th powers before doing polynomial operations is just as good as after. So, back in $R$ we can divide the difference by $p$, and we will! Namely, we can define the function

$$\begin{equation*} \delta (f(x_1,\dots ,x_n))= \frac{f(x_1^p,\dots ,x_n^p) - f(x_1,\dots ,x_n)^p}{p}, \end{equation*}$$

and this function is a $p$-derivation. Just so we can refer to this function later, let’s call this the standard $p$-derivation on $\mathbb{Z}[x_1,\dots ,x_n]$ and denote it by $\delta _{\mathrm{st},p}$ (though this notation is not at all standard).

For example,

$$\begin{equation*} \begin{aligned} \delta _{\mathrm{st},2}(x^3 + 5x) &= \frac{(x^2)^3 + 5(x^2) - (x^3+5x)^2}{2} \\ &= -5x^4 - 10x^2.\end{aligned} \end{equation*}$$

As this operator $\delta _{\mathrm{st},p}$ measures the failure of the freshman’s dream, one might think of this as a freshman’s nightmare. In fact, in large generality, $p$-derivations all arise from some freshman’s nightmare. Let’s make this precise. Given a ring $R$, we say that a map $\Phi :R\to R$ is a lift of Frobenius if it is a ring homomorphism and the induced map from $R/pR \to R/pR$ is just the Frobenius map, i.e., $\Phi (r) \equiv r^p \mod pR$ for all $R$. Given a $p$-derivation $\delta :R\to R$, the map $\Phi :R\to R$ given by

$$\begin{equation*} \Phi (r)=r^p + p \delta (r) \end{equation*}$$

is a lift of Frobenius. Indeed, the congruence condition is automatic, and the sum rule and product rule on $\delta$ translate exactly to the conditions that $\Phi$ respects addition and multiplication. Conversely, if $p$ is a nonzero divisor on $R$, and $\Phi$ is a lift of Frobenius, then the map $\delta (r) = \frac{\Phi (r) - r^p}{p}$ is a $p$-derivation: the freshman’s nightmare associated to the lift of Frobenius $\Phi$.

It is worth noting that not every ring admits a $p$-derivation. For a quick example, no ring $R$ of characteristic $p$ admits a $p$-derivation, since we would have

$$\begin{equation*} 0=\delta (0)=\delta (p)=\delta _p(p) = 1-p^{p-1} = 1 \end{equation*}$$

in $R$. Much more subtle obstructions exist, and it is an interesting question to determine which rings admit $p$-derivations; see AWZ21 for some recent work on related questions.

Note that the power rule from before follows for any $p$-derivation on any ring, since we just used the product rule to see it. The order-decreasing property holds in general, too, at least if $p$ is a nonzero divisor on $R$—this follows from writing $s=p^a r$ with $p\nmid r$ and applying the product rule:

$$\begin{equation*} \begin{aligned} \delta &(p^a r) = p^{ap} \delta (r) + r^p \delta (p^a) + p \delta (p^a) \delta (r) \\ &= p^{ap} \delta (r) + (p^{a-1} - p^{ap-1}) r^p + p (p^{a-1} - p^{ap-1}) \delta (r) \\ &= p^{a-1} r^p + p^a ( \delta (r) - p^{a(p-1)-1} r^p - p^{a(p-1)}\delta (r)). \end{aligned} \end{equation*}$$

Let’s wrap up our cliffhanger from the previous section. Now that we have $p$-derivations of polynomials, we have the ingredients needed for a chain rule: given a polynomial $f(x)$ and a number $n$, we will think of the number $f(n)$ as the composition of the function $f$ and the number $n$, and we can try to compare $\delta _p(f(n))$ with $\delta _\mathrm{st}(f)$ and $\delta _p(n)$. Here’s the chain rule:

$$\begin{equation*} \delta _p(f(n)) = \delta _{\mathrm{st},p}(f)(n) + \sum _{j=1}^{\deg (f)} {p^{j-1}} \frac{d^j f}{j! \,dx^j}(n^p) \delta _p(n)^j. \end{equation*}$$

This is a bit more complicated than the original, but let’s notice in passing that the $j=1$ term in the sum, $\frac{df}{dx}(n^p) \delta _p(n)$, looks pretty close to the classic chain rule, besides the $p$th power on $n$. The curious reader is encouraged⁠¹ to prove the formula above.

¹

For a hint, consider the lift of Frobenius on $\mathbb{Z}[x]$ that sends $x\mapsto x^p + p \delta _p(n)$, and use Taylor expansion to rewrite the associated $p$-derivation in terms of the standard $p$-derivation and derivatives of $f$.

✖

We have collected a decent set of analogues for the basics of differential calculus for $p$-derivations. One can ask how far this story goes, and the short answer is very far. Buium has developed extensive theories of arithmetic differential equations and arithmetic differential geometry, building analogues of the classical (nonarithmetic) versions of these respective theories with $p$-derivations playing the role of usual derivatives. The reader is encouraged to check out Bui05Bui17 to learn more about these beautiful theories, though our story now diverges from these. Instead, we will turn our attention towards using $p$-derivations to give some algebraic results with geometric flavors.

A Jacobian Criterion

One natural geometric consideration is whether, and where, a shape has singularities: points that locally fail to look flat, due to some sort of crossing or crinkle (or some harder-to-imagine higher-dimensional analogue of a crossing or crinkle). For example, the double cone cut out by the equation $z^2-x^2-y^2=0$ has a singularity at the origin where the two cones meet, but any other point on the cone is not a singularity, see Figure 1.

Figure 1.

The cone of solutions of $z^2-x^2-y^2=0$. The origin is the unique singular point.

We are going to consider shapes like this that are cut out by polynomial equations, though to state the classical Jacobian criterion, we will consider their solution sets over the complex numbers.

Since it is difficult to envision higher-dimensional shapes (and impossible to envision what we’re doing next!), it will be useful to give a somewhat more algebraic heuristic definition of singularity. We will say that a point $x$ is a nonsingular point in $X$ if within $X$ one can locally cut out $x$ by exactly $d=\dim (X)$-many equations without taking roots, and singular otherwise. For example, the point $(1,0,1)$ in the cone is nonsingular, and I claim that the two equations $y=0$, $z-1=0$ “work” for our definition: with these two equations and the equation for $X$, we get

$$\begin{equation*} y=z-1=z^2-x^2-y^2=0. \end{equation*}$$

Substituting in, we get $0 = x^2-1 = (x-1)(x+1)$, and “near $(1,0,1)$,” $x+1$ is nonzero, so we can divide out and get $x-1=0$, so $x=1,y=0, z=1$. On the other hand, $(0,0,0)$ is singular, and the two equations $y=0$, $z=0$ don’t “work” for our definition: we have

$$\begin{equation*} y=z=z^2-x^2-y^2=0, \end{equation*}$$

so $x^2=0$, but we need to take a root to get $x=0$.

The classical Jacobian criterion gives a recipe for the locus of all singularities of a shape cut out by complex polynomials.

Theorem 2 (Jacobian criterion).

Let $X\subseteq \mathbb{C}^n$ be the solution set of the system of polynomial equations

$$\begin{equation*} f_1 = \cdots = f_m=0. \end{equation*}$$

If the dimension of $X$ is $d=n-h$, and $f_1,\dots ,f_m$ generate a prime ideal⁠² in the polynomial ring $\mathbb{C}[x_1,\dots ,x_n]$, then the set of singular points is the solution set within $X$ of the system of polynomial equations

²

We recall that an ideal $I$ is prime if it is proper and $gh\in I$ implies $g\in I$ or $h\in I$. Experts will recognize this condition as overkill, but something needs to be done to avoid examples like $(x-y)^2=0$ whose solution set $X$ is the (complex) line $\{(a,a)\}$, which is nonsingular, but $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} =0$ for every point in $X$, or $x^2-x=xyz=0$ whose solution set $Y$ is two-dimensional but has a crossing singularity at $(0,0,1)$ that is not a solution of the $1\times 1$ minors of the Jacobian matrix.

✖

$$\begin{equation*} \text{all $h \times h$ minors of} \ \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_m}{\partial x_1} & \cdots & \frac{\partial f_m}{\partial x_n}\end{bmatrix} =0. \end{equation*}$$

In particular, if $f$ is irreducible, the set of singular points of the solution set $f=0$ is the solution set of

$$\begin{equation*} f= \frac{\partial f}{\partial x_1}= \cdots = \frac{\partial f}{\partial x_n} =0. \end{equation*}$$

For example, for the Whitney umbrella cut out by the polynomial