Are there non-trivial $\mathbb{Z}$-linear derivations over the real numbers?

Question

Given the algebraic definition of a $\mathbb{Z}$-linear derivation over a commutative ring

$D:R\to R$ with $D(a+b)=D(a)+D(b)$ and $D(a \cdot b)=D(a) \cdot b+a \cdot D(b),$

there is always the trivial derivation where $\forall_{a\in R} \,D(a)=0$.

I wonder if there are non-trivial derivation over the field $\mathbb{R}$ of real numbers.

It is a basic exercise to show that for any derivation $\forall_{r\in\mathbb Q} \, D(r)=0$ and that $D(r^n)=0\implies D(r)=0$.

But I fail to show this for general algebraic numbers and am totally lost on how to tackle transcendental numbers.

Such a map is called a derivation, not derivative. Since $\Bbb R$ has a unit, we have $D(1)=0$ and hence $D=0$. See wikipedia. Also, $D(\lambda x)=\lambda D(x)$ for a linear function $D$. — Dietrich Burde, Feb 10 '25 at 19:15
@Dietrich: I don't follow. The OP does not require $D(\lambda x) = \lambda D(x)$ and this isn't part of the definition. $D(1) = 0$ also does not imply $D = 0$. — Qiaochu Yuan, Feb 10 '25 at 19:24
Is this a question of curiosity for yourself? Or did it come from a text? — Brian Tung, Feb 10 '25 at 19:29
@BrianTung I stumbled over a question integrating $e^f(x)$, from there to the error-function and how to prove $e^x^2$ is not integrable to elementaty functions. — Gyro Gearloose, Feb 10 '25 at 19:31
@QiaochuYuan I took it as $K$-algebra derivation (like in the wikipedia link). I was too quick. — Dietrich Burde, Feb 10 '25 at 19:46
@Dietrich: sure, but you didn't specify $K$. We should take $K = \mathbb{Q}$ here. And again $D(1) = 0$ does not imply $D = 0$ at all. — Qiaochu Yuan, Feb 10 '25 at 19:46
@QiaochuYuan Well, by $K$-linearity, $D(k)=0$ for all $k\in K$ from $D(1)=0$. But this was the mistake, $K=\Bbb Q$, not $K=\Bbb R$. — Dietrich Burde, Feb 10 '25 at 19:48
Based on the comments I have clarified the question. Please check. @DietrichBurde — Martin Brandenburg, Feb 10 '25 at 20:00
@DietrichBurde so you added this for clarity to remove misunderstandings, that's OK. I only asked because of what I had previously tried, everything that that can be constructed from 1 using addition, subtraction, multiplication and division (in other words $\mathbb Q$ has $D(a)=0$). — Gyro Gearloose, Feb 10 '25 at 20:35

Qiaochu Yuan · Accepted Answer · 2025-02-10T20:04:48.287

Yes. You can show it as follows:

A field $K(t)$ which is a purely transcendental extension of another field $K$ of characteristic zero admits a nontrivial $K$-linear derivation $\frac{\partial}{\partial t}$ defined by $\frac{\partial}{\partial t} t = 1$; this gives $\frac{\partial}{\partial t} q(t) = q'(t)$ where $q(t) \in K(t)$ is a rational function and $q'(t)$ is its formal derivative.
If $K$ is a field with a derivation $D$ and $L/K$ is a separable algebraic extension of $K$, then $D$ extends uniquely to a derivation on $L$.

The first claim is straightforward. For the second claim we can reduce to the case of an extension $K[\alpha]/f(\alpha)$ generated by an algebraic element $\alpha$ with separable minimal polynomial $f(\alpha)$. Then if $D$ is our derivation we have $f(\alpha) = 0$ so

$$D f(\alpha) = f'(\alpha) D(\alpha) = 0$$

(this is a "chain rule" for derivations), and since $f'(\alpha) \neq 0$ by separability this forces $D(\alpha) = 0$, and then we can check that this defines an extension of $D$ to $L$.

This second property turns out to be an if-and-only-if characterization of separable extensions; see e.g. Theorem 1.5 in these notes by Keith Conrad.

Corollary: Every field of characteristic zero $K$ containing a transcendental element $t$ admits a derivation $D : K \to K$ such that $D(t) = 1$. If $K$ is an algebraic extension of $\mathbb{Q}$ then it admits no nonzero derivations.

This is because every such field $K$ is a separable (since we are in characteristic zero) extension of a purely transcendental extension of $\mathbb{Q}$, corresponding to a transcendence basis.

So, for example, $\mathbb{R}$ admits a derivation one might call $\frac{\partial}{\partial \pi}$ such that $\frac{\partial}{\partial \pi} \pi = 1$. Whether this is useful for anything is unclear; to construct this derivation requires a transcendence basis of $\mathbb{R}$ over $\mathbb{Q}$ containing $\pi$ which requires the axiom of choice. The construction of the derivation depends on this choice (since you need to specify what the derivation does to other elements of $\mathbb{R}$ and in particular other transcendentals, which can be done arbitrarily on a transcendence basis) and is highly non-unique.

By working a little harder we should be able to show that a derivation on a field $K$ of characteristic zero is completely determined by what it does to a transcendence basis of $K$, and can take arbitrary values on such a basis.

So what you are telling is that $\mathbb Q(\pi)$ provides an example and that we don't have to worry about other transcendental real numbers. Thanks a lot. — Gyro Gearloose, Feb 10 '25 at 19:53
Well, we do have to worry about them in the sense that we need to pick a transcendence basis to decide what our derivation does to other transcendentals. But there is a way to do it, if that's what you mean. — Qiaochu Yuan, Feb 10 '25 at 19:54
I was also wondering about algebraic numbers instead of reals, what do you think: is it worth a separate question, a side note here or too trivial to ask at all? — Gyro Gearloose, Feb 10 '25 at 20:00
@Gyro: this is covered by my answer, in this case there are no nonzero derivations. — Qiaochu Yuan, Feb 10 '25 at 20:05

Are there non-trivial $\mathbb{Z}$-linear derivations over the real numbers?

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