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Let $A$ be a commutative ring with $1$.We say that $A$ is coherent if and only if every finitely generated ideal of $A$ is finitely presented.

Does there exist a coherent ring such that nil-radical of $A$ is NOT finitely generated?

In other words, by the definition of coherent rings, nil-radical of $A$ is not finitely presented!

Actually, my question is, Does there exist a coherent structure sheaf $ \mathcal O_X$ such that nilradical sheaf $ \mathcal N_X$ is not coherent?

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    $\begingroup$ 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"). $\endgroup$
    – nfdc23
    Commented Jan 29, 2017 at 13:49
  • $\begingroup$ @nfdc23 I tried to understand your example but Can you provide some other example? $\endgroup$
    – Anoop singh
    Commented Jan 29, 2017 at 15:32
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    $\begingroup$ Similar example: $k[[T^{1/n} : n \in {\mathbb N}]]/(T)$ $\endgroup$
    – David Lampert
    Commented Jan 29, 2017 at 16:49
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    $\begingroup$ 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. $\endgroup$
    – nfdc23
    Commented Jan 29, 2017 at 16:50
  • $\begingroup$ @DavidLampert What's a good way to see that is coherent? $\endgroup$
    – rschwieb
    Commented Nov 22 at 13:55

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Consider the ring $$ R = (\mathbb{Z}/2)[y_0,y_1,y_2,\dotsc]/ (y_0^3+y_0y_1,\;y_1^3+y_1y_2,\;y_2^3+y_2y_3,\dotsc). $$ This is also generated by the elements $x_n=\sum_{i\leq n}y_{n-i}^{2^i}$. It can be shown that the nilradical is generated by the elements $x_nx_m$ with $n\neq m$, and that it is not finitely generated. One can also give $R$ a grading with $|y_i|=2^i$. It can be shown that $R$ is coherent in the graded sense: any homogeneous ideal with a finite set of homogeneous generators has a finite homogeneous presentation. I am not sure whether it is coherent as an ungraded ring. All of this is in the paper "Large self-injective rings and the generating hypothesis", by Leigh Shepperson and myself (Theorem 1.6).

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  • $\begingroup$ @Neil Strickland But your definition for coherent graded ring in the Def1.2 in your paper is same as I have given for any commutative ring with 1. So I guess this must be true for ungraded one? $\endgroup$
    – Anoop singh
    Commented Jan 29, 2017 at 15:29
  • $\begingroup$ @Anoop: Check out lines 4-5 of the introduction. $\endgroup$
    – Fred Rohrer
    Commented Jan 29, 2017 at 19:59

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