Timeline for coherent ring whose nilradical is not finitely generated
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 3 at 18:46 | comment | added | rschwieb | If what i've written down is correct, I think maybe the annihilator of $T^q$ is principal, and hence the f.g. ideals (which are necessarily principal in a uniserial ring) all are finitely presented in $R\to R\to (T^q)$ | |
| Dec 3 at 18:39 | comment | added | rschwieb | @nfdc23 I can tell that $k[T^{1/n} : n \in {\mathbb N}]/(T)$ is unserial with infinitely generated nilpotent radical... but what would you suggest as rationale that it's coherent? | |
| Nov 22 at 13:55 | comment | added | rschwieb | @DavidLampert What's a good way to see that is coherent? | |
| Jan 29, 2017 at 16:50 | comment | added | nfdc23 | I prefer not to say more about it, leaving further details as an exercise, since one only needs elementary considerations with ramification in discrete valuation rings (as I indicated). Lampert's comment conveys it as well (in his example, one can replace [[ with [ and replace ]] with ] since he works mod $T$ in the end). This example has the virtue that such rings arise very naturally in practice (i.e., it is not just a weird pathology whose only purpose for being is as a counterexample to things), in the context of Raynaud's approach to non-archimedean geometry via formal schemes. | |
| Jan 29, 2017 at 16:49 | comment | added | David Lampert | Similar example: $k[[T^{1/n} : n \in {\mathbb N}]]/(T)$ | |
| Jan 29, 2017 at 15:32 | comment | added | Anoop singh | @nfdc23 I tried to understand your example but Can you provide some other example? | |
| Jan 29, 2017 at 13:56 | answer | added | Neil Strickland | timeline score: 2 | |
| Jan 29, 2017 at 13:49 | comment | added | nfdc23 | Yes. Let $R$ be a complete discrete valuation ring with uniformizer $\pi$, $K$ its fraction field, and $\overline{K}/K$ an algebraic closure. The integral closure $\overline{R}$ of $R$ in $\overline{K}$ is a valuation ring, so all finitely generated ideals in $\overline{R}$ are even principal. Thus, $\overline{R}$ is a coherent ring, so its quotient $A =\overline{R}/(\pi)$ is a coherent ring. But all non-units in $A$ are nilpotent and the maximal ideal is not finitely generated (since any non-unit comes from a non-unit in some finite extension of $K$, so has controlled "ramification"). | |
| Jan 29, 2017 at 12:58 | history | asked | Anoop singh | CC BY-SA 3.0 |