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I am confused by properties of $R$, the geometric line in synthetic differential geometry. In the book Synthetic Differential Geometry by Kock, he assumes $R$ is only a commutative ring. However, in these notes by Mike Shulman, it is said that $R$ is a field in the following constructive sense: $$x\neq 0\implies x\text{ is invertible}.$$

This seems to be an assumption which Kock does not make: he explicitly states that we must sometimes assume $R$ is a $\mathbb Q$-algebra, for instance in order to prove $D$ is not an ideal.

Why should or shouldn't one make this assumption?

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  • $\begingroup$ Apparently he's substituting Kock-Lawvere axiom by this one. Mike claims too apparently that this one is stronger, but codifies the essential part. I think that it means that being invertible makes every map $D \rightarrow R$ extends canonically, but this seems kind of forced. $\endgroup$ – user40276 Feb 29 '16 at 10:46
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The number line $R$ is a ring. It must be in order to accommodate for the interval $D=[x:x^2=0]$ which elements are nilpotent, and not invertible: if $ x\in D\rightarrow \nexists~x^{-1}$. Then $R$ is taken as an algebra over the rationals in order to take care of further proofs as you indicate, but let's go back to what Shulman says. Applying the law of contrapositive on the statement $x\neq 0\implies x$ is invertible, we get $$\forall x\in R: ~\nexists~x^{-1}\implies x~\neg (\neq 0)$$ Notice how $ x~\neg (\neq 0)$ does not imply $x=0$ since we can not use the principle of excluded middle. Therefore for these $x: x~\neg (\neq 0)\implies x\in D$. I think Shulman is forcing the language, he knows $R$ is ring but it's still true that in a sense, the constructive sense he refers to, it can be seen as a field. This paper by O’Connor might help.

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