Counterexamples about Infinitely Generated Modules over Commutative Rings

This post is about examples of some nice things that could fail for modules that are not finitely generated (or not finitely presented when the ring is not Noetherian).

1. Nakayama’s Lemma (Krull’s Intersection Theorem)

2. \mathrm{Supp}M =V(\mathrm{ann M}).

3. \mathrm{Hom}_A(M,N)\otimes B= \mathrm{Hom}_B(M\otimes_A B, N\otimes_A B) for B a flat A-module

4. M\otimes _R \hat{R} =\hat{M}

5. Finitely Generated Nil Ideal is Nilpotent

6. Finitely presented flat module is projective

Counterexamples that Tensor Product of a Non-finitely Generated R-module M with the Completion of R is not the Completion of M

Let A be a commutative noetherian ring, m an ideal of A and M a finitely generated A-module, then we have an isomorphism M\otimes_A \hat{A} \to\hat{M} where the completion is with respect to the m-adic topology [III. 3.4 Theorem 3, Commutative Algebra, Bourbaki].

Note that when A is not noetherian, M being finitely presented would suffice.

Here are two counterexamples showing that the map above need not be injective or surjective when M is not finitely generated:

Checking the ring k[x,y,u,v]/(xy-uv) is a normal domain

Let A= k[x,y,u,v]/(xy=uv), is this ring normal (integral closed in its field of fraction)?

Edit (see comments below): Because p is of height 1, then there exists one among x,y,u,v, say x such that f(x)\notin p for any f(x)\in k[x]. Otherwise, if there are polynomials in one variable for each of the variables x,y,u,v, contained in p, then p is of height >1. Thus f(x) becomes invertible in A_p for any f(x)\in k[x].