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
9 events
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Oct 10, 2019 at 14:45 | history | edited | Pierre-Yves Gaillard | CC BY-SA 4.0 |
typo
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Oct 10, 2019 at 14:45 | vote | accept | Pierre-Yves Gaillard | ||
Oct 10, 2019 at 14:03 | comment | added | YCor | @Pierre-YvesGaillard oh thanks. Indeed Eric's example is immediately adaptable to answer the question. | |
Oct 10, 2019 at 12:45 | answer | added | darx | timeline score: 18 | |
Oct 10, 2019 at 10:42 | comment | added | Pierre-Yves Gaillard | @YCor - I hope you saw Angelo's comment. (At the end of the question I mentioned Eric Wofsey' example of a non-surjective epimorphic endomorphism of a noetherian ring.) | |
Oct 10, 2019 at 7:51 | comment | added | Angelo | The embedding $\mathbb{Z} \times \mathbb{Z} \subseteq \mathbb{Z} \times \mathbb{Q}$ is such an example. | |
Oct 10, 2019 at 7:17 | comment | added | YCor | Do you have an example of a non-surjective epimorphism $A\to B$ of noetherian rings with $A,B$ with the Krull dimension of $A$ $\le$ that of $B$? (Say, with finite Krull dimension— otherwise the Krull dimension should be defined as an ordinal, namely as $\mathrm{Kdim}(A)=\sup(\mathrm{Kdim}(A/P)+1)$ for $A\neq 0$, where $P$ ranges over non-minimal prime ideals and $\sup\emptyset=0$.) | |
Oct 8, 2019 at 12:51 | history | edited | Pierre-Yves Gaillard | CC BY-SA 4.0 |
typos caught by Steven Landsburg
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Oct 8, 2019 at 12:32 | history | asked | Pierre-Yves Gaillard | CC BY-SA 4.0 |