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In many sources it was written that *$R$ is ring and *$R$ can't be a field because it's not possible that $d^2=0$ in field.

But in some sources it was written that *$R$ is a field.

How can *$R$ be a field in Synthetic Differential Geometry or Smooth Infinitesimal Analysis?

Thanks.

Mike_bb
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    Which sources are you referring to? – Mark Saving Dec 16 '23 at 06:43
  • @MarkSaving https://www.math.ru.nl/~landsman/scriptieTim.pdf (page 16) - ring. https://pages.physics.ua.edu/staff/fabi/InvitationSDG.pdf (page 3) - "Nevertheless most of the literature refers to R simply as a ring while a few authors still refer to R as a field" – Mike_bb Dec 16 '23 at 06:52
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    Why do you use $^\ast !R$ do denote the line of SDG? Both sources you linked use simply $R$. – Mikhail Katz Dec 21 '23 at 13:22
  • @MikhailKatz It's my mistake. *$R$ was in Giordano's source which I read. – Mike_bb Dec 21 '23 at 13:40
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    @Mike_bb, Giordano is not doing Synthetic Differential Geometry. Furthermore, Giordano did not use ${}^\ast!R$ but rather ${}^\bullet! R$. – Mikhail Katz Dec 21 '23 at 13:43
  • @MikhailKatz Ok. I find now in Giordano's source that IDG differ from SDG. It's true. Thx. – Mike_bb Dec 21 '23 at 13:48

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Since SDG uses intuitionistic logic, it is important to specify how exactly you define a field. If you want the property that if $x\not=0$ then $x$ is invertible, then this still holds. Nilsquare infinitesimals do not provide a counterexample to this, since they are not provably nonzero.

Mikhail Katz
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