2

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).

Kubrick
  • 362
  • 2
  • 11
  • “I saw the statement….” Where? Are you sure it was the same notion of derivative? – Thomas Andrews Dec 02 '21 at 02:40
  • @ThomasAndrews "François Boulier. A Differential Algebra Introduction For Tropical Differential Geometry (https://hal.archives-ouvertes.fr/hal-02378197/document)" that refers to "Joseph Fels Ritt - Differential Algebra". – Kubrick Dec 02 '21 at 02:45

1 Answers1

3

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?

Atticus Stonestrom
  • 11,659
  • 2
  • 18
  • 45
  • Thank's for your answer. Let me ask you several questions. Am I right that "$D$" here is "$d$" from the link because (I suppose) $\mathbb{Q}(X)(x_0)\subseteq \mathbb{R}$? And why do $\lambda_i\in \mathbb{Q}(X)$ and not in $\mathbb{Q}$? Why do you single out a separate element $x_0$ from the basis? – Kubrick Dec 04 '21 at 00:35
  • And no, there weren't any additional assumptions (hal.archives-ouvertes.fr/hal-02378197/document), so I think it's some kind of mistake. – Kubrick Dec 04 '21 at 00:47
  • hi @Kubrick; sorry for the late response! I have been traveling. I will respond to your questions shortly :) – Atticus Stonestrom Dec 06 '21 at 00:27
  • No problem at all) – Kubrick Dec 06 '21 at 17:14
  • By definition of the transcendental basis $A=X\sqcup{x_0}$ for $\mathbb{R}/\mathbb{Q}$ $\mathbb{R}$ is algebraic over $\mathbb{Q}(A=X\sqcup{x_0})$ that is the fraction field of $\mathbb{Q}[A]$. Do you use here some isomorphism to consider not the fraction field of $\mathbb{Q}[A=X\sqcup{x_0}]$, but the fraction field of $\mathbb{Q}(X)[x_0]$? – Kubrick Dec 06 '21 at 17:56
  • hi @Kubrick; alright, let me try to address your questions! regarding the first; I am indeed using $D$ to denote a derivation, following the notation in your post. – Atticus Stonestrom Dec 07 '21 at 05:41
  • regarding the second and third, here is an overview of what I do in the post; it is a general fact that, if $R$ is an integral domain and $K\supseteq R$ is a field algebraic over the fraction field of $R$, then any derivation on $R$ extends to $K$. in particular, if we want to find a non-trivial derivation on $K$, it suffices to find one on $R$ – Atticus Stonestrom Dec 07 '21 at 05:44
  • in this case, I am taking $K=\mathbb{R}$ and $R=\mathbb{Q}(X)[x_0]$. the reason that the $\lambda_i$ lie in $\mathbb{Q}(X)$ is that elements of $R$ can all be written uniquely in the form $$\lambda_0+\dots+\lambda_n x_0^n$$ for some $\lambda_i\in\mathbb{Q}(X)$; in other words, $R$ is isomorphic to the polynomial ring in one variable over $\mathbb{Q}(X)$. (this is the nature of transcendence bases) – Atticus Stonestrom Dec 07 '21 at 05:45
  • the reason that I single out $x_0$ is the following: for any field $L$, there always exists a non-trivial derivation on the polynomial ring $L[t]$, given by $$D(\lambda_0+\dots+\lambda_n t^n)=\lambda_1+\dots+ n\lambda_n t^{n-1}$$ for all $\lambda_i\in L$. since the ring $R$ mentioned in the previous comment is a polynomial ring in the variable $t=x_0$ over field $L=\mathbb{Q}(X)$, this is precisely what we exploit; singling out $x_0$ is just a notational convenience to make this fact easier to express – Atticus Stonestrom Dec 07 '21 at 05:49
  • 1
    for your final question, we have the following fact; suppose $T$ is an integral domain and $F\supseteq T$ is the fraction field of $T$. then, for any ring $T'$ with $T\subseteq T'\subseteq F$, we have that $F$ is also the fraction field of $T'$. so, in this case, take $T=\mathbb{Q}[A]$ and $F=\mathbb{Q}(A)$. you are absolutely right that $F$ is the fraction field of $T$; but, in this case, we take $T'=\mathbb{Q}X$. by the remark in this paragraph, $F$ is still the fraction field of $T'$, so that's all we need – Atticus Stonestrom Dec 07 '21 at 05:55
  • does that make sense? please let me know if anything is still unclear or if you have other questions :) ... sorry once again for the delayed response! – Atticus Stonestrom Dec 07 '21 at 05:55
  • 1
    Thanks for all your answers. The last comment cleared everything up! – Kubrick Dec 07 '21 at 05:59
  • 1
    @Kubrick my pleasure, very happy it helped! :) – Atticus Stonestrom Dec 07 '21 at 06:01