Timeline for Does the Cantor-Schröder-Bernstein Theorem hold in the category opposite to the category of noetherian commutative rings?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Oct 10, 2019 at 16:07 | comment | added | YCor | @R.vanDobbendeBruyn it seems that $A$ and $B$ are not elementary equivalent: with a little effort one checks, if I'm correct, that $A$ satisfies ($\forall t$, if $t$ and $t-3$ are invertible then so is $t-1$), but $B$ doesn't. [I need $-3$ because in $A$, $u=x(x-1)$ satisfies $u,u-2$ invertible but not $u-1$.] | |
| Oct 10, 2019 at 14:45 | vote | accept | Pierre-Yves Gaillard | ||
| Oct 10, 2019 at 13:29 | comment | added | R. van Dobben de Bruyn | This example (very similar to the one linked at the end of the question) is also an alternative (Noetherian!) answer to this question. What fascinates me is that you don't use any structure inside the rings to conclude they are not isomorphic, but rather write down what an isomorphism (if any) might look like. (Although I guess in principle it might be possible to rephrase this as some complicated statement inside the rings...) | |
| Oct 10, 2019 at 12:50 | review | First posts | |||
| Oct 10, 2019 at 13:22 | |||||
| Oct 10, 2019 at 12:45 | history | answered | darx | CC BY-SA 4.0 |