The field of constants of a differential ring. Derivative of real and complex numbers.

Let $D$ be a derivation operator over a ring $R$: $$D(a + b) = D(a) + D(b) \\ D(ab) = D(a)b + aD(b)$$ for all elements $a,b\in R$.

If the ring is the field $\mathbb{Q}$, all derivatives should be equal to zero since $$D(0)=D(0+0)=2D(0)=0 \\ D(1)=D(1\times1)=2D(1)=0\\ D(n\in\mathbb N)=D(\underbrace{1+...+1}_n)=nD(1)=0\\ D\left(\frac{m}{n}\right)=\frac{D(m)n-mD(n)}{n^2}=0$$

But I saw the statement saying it can be proved that the derivative of any real or complex number must be zero.

I have made several attempts trying to constrain or approximate (by sequences) the real number by rationals, but I need the fact that if $a\leq b$ then $D(a)\leq D(b)$ for any real number that I have not been able to establish (and I'm not sure if it is possible to do).

Solution 1:

As written, the statement you've described is not true. To see this, let $X\sqcup\{x_0\}$ be a transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$, so that $\mathbb{R}$ is algebraic over $\mathbb{Q}(X)(x_0)$. By the argument here, to exhibit a non-trivial derivation on $\mathbb{R}$ it thus suffices to exhibit one on $\mathbb{Q}(X)(x_0)$. Furthermore, derivations on an integral domain extend to its fraction field via the 'quotient rule', so to exhibit a non-trivial derivation on $\mathbb{Q}(X)(x_0)$ it suffices to exhibit one on $\mathbb{Q}(X)[x_0]$. But this is just a polynomial ring over a field, so we can define $D:\mathbb{Q}(X)[x_0]\to \mathbb{Q}(X)[x_0]$ in the usual way, by taking $$D(\lambda_0+\dots+\lambda_nx_0^n)=\lambda_1+2\lambda_2x+\dots+n\lambda_nx_0^{n-1}$$ for all $\lambda_i\in \mathbb{Q}(X)$. An essentially identical argument can also give a non-trivial derivation on $\mathbb{C}$.

Now, if you were to additionally require that $D$ be continuous, then it would necessarily be identically zero, since $\mathbb{Q}$ is a dense subset of $\mathbb{R}$. Is it possible this is what the reference you were looking at was referring to?