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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?

user26857
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Anoop singh
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2 Answers2

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After the user originally posted here, they crossposted to mathoverflow, where a few examples were given.

There's a simple one hidden in the comments which does not have a lot of justification.

Take a field $k$ and indeterminate $T$, and form $R=k[\{T^{1/n}\mid n\in\mathbb N^{> 0}\}]/(T)$.

  • it has a unique, nil, non-fintely-generated maximal ideal $(\{T^{1/n}\mid n\in\mathbb N^{> 0}\})$
  • since the units are things with nonzero constant term, every element is a unit times $T^q$ for some nonnegative rational $q < 1$. This means that the principal ideals (and hence all ideals) are linearly ordered
  • the last point causes the f.g. ideals to be all principal.
  • finally, observe that for a nonunit $x$, $ann(x)=ann(uT^q)=(T^{1-q})$ for some unit $u$ and rational $q$ with $0<q<1$, so the point-annihilators are principal.

From this last point we can see that $0\to ann(x)\to R\to (x)\to 0$ with $ann(x)$ principal witnesses that every f.g. ideal is finitely presented (because they're all of the form $(x)$.)

rschwieb
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Let $k$ be a field and set $A:= k[x_i]/(x_i^2)$, which is the filtered colimit of the subrings $A_n := k[x_1, \ldots, x_n]/(x_1^2, \ldots, x_n^2)$.

It is patent that the nilradical of $A$ is generated by the set of $x_i$ and cannot be finitely generated.

To see coherence of $A$, one can proceed directly by checking that annihilators and intersections of finitely generated ideals can be calculated in suitable $A_n$.

Alternatively, one may note that every transition map $A_i \subseteq A_j$ in our colimit makes $A_j$ into a free $A_i$-module with basis $x_{i+1}, \ldots, x_{j}$.

Since each $A_n$ is Noetherian, and a fortiori coherent, we see that $A$ is the filtered colimit of a system of coherent rings with flat transition maps. Such rings are coherent. See Proposition 20 in Soublin's Anneaux et Modules Cohérents.

Although in the particular case of $A$ it is easy to check its coherence directly, the proposition of Soublin makes it easy to construct other examples. For example, if a ring $B$ is coherent, then so is any superring $B \subseteq C$ which is finitely presented as a module. Thus we can always add as many sufficiently generic roots of monic polynomials as we like to a coherent ring without disturbing coherence. On the other hand, adding roots of monic polynomials will typically cause the nilradical to blow up, providing myriad examples like the one OP seeks.

Badam Baplan
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