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]
Best Answer
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 $\mathbb N$.