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In the context of Synthetic Differential geometry as in the first few sections of Kock's book, so that $R$ is ring satisfying Axiom 1. Let ${f : R \rightarrow R}$ be such that ${f' = 0}$. How do you prove that $f$ is constant?

Boogie
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1 Answers1

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In most developments of classical analysis, one proves this theorem early on.

However, in synthetic differential geometry, the proof requires the use of the more complicated order and integration axioms, so is usually left for later.

Hadamard's lemma (Proposition 13.1 on page 50 in my edition of Kock's book) gives the following:

For any $a \leq b \in $ and function $f: [a,b] \rightarrow R$, and any $x,y \in [a,b]$, we have that $$ f(y) - f(x) = (y - x) \int_{0}^{1} f'(x + t(y-x)) \:dt$$

Therefore, if $f:R \rightarrow R$ satisfies $f'(z) = 0$ for all $z \in [a,b]$, we immediately get that $$f(y) - f(x) = (y-x) \int_{0}^{1} 0 \:dt = 0$$ and so $f(y) = f(x)$. Since $a,b,x,y$ were arbitrary, it follows that $f$ is constant everywhere.

Z. A. K.
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  • Thanks for your reply @z-a-k . I clicked on the arrow up but I did not click the tick because I really meant just Axiom 1 in the question. Do you know of a model of SDG where "zero-derivative implies constant" does not hold? – Boogie Mar 03 '22 at 18:53
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    There are definitely such models of Axiom 1, but they are nasty (in particular, by the argument above they cannot satisfy the integration axioms). I think that if you start with a real-closed field $\mathbb{F}$ of cardinality strictly above $2^{\aleph_0}$, then the "free cocompletion of the opposite category of finitely presented $\mathbb{F}$-algebras" will model Axiom 1 and will contain non-constant functions that have zero derivative everywhere. But for obvious reasons, I'm unwilling to check the details :) – Z. A. K. Mar 04 '22 at 20:53
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    But whether this particular construction works or not, the idea behind constructing such a model is to make the "real line object" of your model so long that it becomes disconnected. Since Axiom 1 is purely local, you'll be able to construct a function that takes a constant value $x$ around $0$, and a different constant value $y$ in some segment disconnected from $0$. Since derivatives are purely local, they'll be zero at every point, but the function won't be globally constant. – Z. A. K. Mar 04 '22 at 21:02
  • @z-a-k , thanks for the further thoughts. The intuition about connectedness makes me wonder if, on top of axiom 1, we further require that $R$ is local, then "zero-derivative implies constant". – Boogie Mar 04 '22 at 23:05