# coherent ring whose nilradical is not finitely generated

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?

• 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"). – nfdc23 Jan 29 '17 at 13:49
• @nfdc23 I tried to understand your example but Can you provide some other example? – Anoop singh Jan 29 '17 at 15:32
• Similar example: $k[[T^{1/n} : n \in {\mathbb N}]]/(T)$ – David Lampert Jan 29 '17 at 16:49
• 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. – nfdc23 Jan 29 '17 at 16:50

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).