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It has been claimed without proof in several answers that an intersection of two finitely generated ideals in a coherent ring is finitely generated. Thus, the finitely generated ideals in a coherent ring form a lattice. However, can an infinite intersection of finitely generated ideals fail to be finitely generated? The typical examples of coherent rings which are not Noetherian (polynomial ring in infinitely many variables over $\mathbb{Z}$, entire functions on $\mathbb{C}$) seem to have the property that infinite intersections happen to be finitely generated, but for non-general reasons. Does anyone have a nice counterexample?

My motivation is that, given the existence of such a pathological intersection, the abelian category of finitely presented modules can fail to admit certain colimits. Without a reference for the proof of the statement about finite intersections, I don't know where to begin.

[commutative and non-commutative examples accepted]

Morgan Rogers
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    As for a reference for the claims in the links, you can go back to the pioneering work of S. U. Chase on coherence, Direct Products of Modules. Theorem 2.2 is what you want. – Badam Baplan Oct 28 '20 at 02:56

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I think this is a counterexample.

Let $F_2$ be the field of two elements, and consider the subring of $\prod_{i\in \mathbb N} F_2$ of "eventually constant sequences."

It is commutative and von Neumann regular, so the finitely generated ideals are all principle. It's coherent because it is semihereditary.

Now, you can take "unit vectors" $e_i$ in this ring, and $1-e_i$ generate a cyclic maximal ideal.

But $\bigcap_{i\in\mathbb N}(1-e_{2i})R$ is an ideal whose elements have to be zero on the "even indices". Therefore every element in it is eventually zero. But clearly it's not generated by a single element, because it contains elements that are nonzero on indices in a cofinal subset of $\mathbb N$.

rschwieb
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  • An eventually constant sequence which is non-zero only on the odd indices must be zero everywhere, so that intersection is the trivial ideal! – Morgan Rogers Oct 27 '20 at 16:55
  • @MorganRogers What? no. It just has to be zero on the tail. $e_1$ is in it, for example. The point is that it has things with support way out there, but no single generator can cover them all (since a generator would necessarily end in zeros at some point.) – rschwieb Oct 27 '20 at 17:20
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    Ah sorry you're right. I've been working with polynomial examples all day, so I flipped some of the bits in my head. Thanks for the explanation. – Morgan Rogers Oct 27 '20 at 17:48
  • @MorganRogers Thanks for the question. I'm not good with coherent rings, and I'm surprised this example occurred to me. – rschwieb Oct 27 '20 at 17:58
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    @MorganRogers See, in a commutative self-injective von Neumann regular ring, I think the lattice of principal ideals is complete, so it would intersect to a principle ideal. But this one is not self-injective. – rschwieb Oct 27 '20 at 18:01