Properties of Quasi-Coherent Modules

Question

Let $X=\mathrm{Spec}\,A$ be an affine scheme and $M$ an $A$-module. Show that the following two conditions are equivalent:

(a) $\tilde{M}$ is a locally free $\mathcal{O}_{X}$-module of finite type.

(b) $M$ is a finitely generated projective $A$-module.

How would one prove this equivalence? Any help would be valuable!

Also, it seems that one could find projective $A$-modules $M$ such that $\tilde{M}$ is not locally free, if we remove the condition of finite generation on $M$. What would be a good example of this? I thought something along the lines of the following: $\mathfrak{m}=k^{(J)}$ is a projective ideal where $k$ is a field and $J$ is an infinite indexing set. Set $A=k^{(J)}$, $X=\mathrm{Spec}\,A$. Then it seems we could find some $\mathfrak{\tilde{m}}$ that is not a locally free $\mathcal{O}_{X}$-module.

Answer 1

This beautiful result is standard and can be found in the usual books (EGA I, Bourbaki, etc.), but let me sketch it so that you can fill in the details as an exercise.

$(a) \Rightarrow (b)$ Choose $f_1,\dotsc,f_n \in A$ generating the unit ideal such that $M_{f_i}$ is free over $A_{f_i}$ of finite rank, say of rank $\leq d$. Deduce that $M \otimes_A (\prod_i A_{f_i})$ is a direct summand of a free module of rank $d$ over $\prod_i A_{f_i}$, hence finitely generated and projective. Since $\prod_i A_{f_i}$ is faithfully flat over $A$, it follows that $M$ is finitely generated projective over $A$. For this, prove the general fact that "finitely generated" and "finitely generated projective" descend along faithfully flat ring extensions.

$(b) \Rightarrow (a)$ Let $M$ be finitely generated projective, choose an epimorphism $F \to M$ with $F$ finitely generated free.

• Case 1: $A$ is a field. Done by linear algebra.
• Case 2: $A$ is local with residue field $k$. Choose a direct summand $F' \subseteq F$ such that $F' \otimes k \to M \otimes k$ is an isomorphism (every generating set of a finite-dimensional vector space contains a basis). Let $0 \to K' \to F' \to M \to 0$ be exact. Since $M$ is projective, the sequence splits, hence $0 \to K' \otimes k \to F' \otimes k \to M \otimes k \to 0$ stays exact. It follows $K' \otimes k = 0$ and then $K'=0$ by Nakayama (OK since $K'$ is finitely generated). Hence $M \cong F'$ is free.
• Case 3: $A$ is arbitrary. If $\mathfrak{p} \in X$, then $M_\mathfrak{p}$ is free of, say, rank $n$ over $A_{\mathfrak{p}}$ by Case 2. Hence, $\tilde{M}$ and $\mathcal{O}_X^n$ are both $\mathcal{O}_X$-modules of finite presentation which are isomorphic at $\mathfrak{p}$. It follows that they are isomorphic around a neighborhood of $\mathfrak{p}$. For this, prove the general fact for $\mathcal{O}_X$-modules $F,G$ that $\underline{\hom}(F,G)_x \to \hom(F_x,G_x)$ is an isomorphism when $F$ is of finite presentation. Deduce that $\underline{\mathrm{isom}}(F,G)_x \cong \mathrm{isom}(F_x,G_x)$ if $F,G$ are both of finite presentation.

In Case 2, the same proof works under the (a priori) weaker assumption that $M$ is flat and of finite presentation. In Case 3 we only needed local isomorphisms and finite presentation. And there is no need to restrict to affine schemes. Hence, if $X$ is an arbitrary scheme (or even algebraic stack!), and $M$ is a quasi-coherent module on $X$, the following are equivalent:

• $M$ is locally free of finite rank.
• For every morphism $\mathrm{Spec}(A) \to X$ the $A$-module $M|_A$ is finitely generated and projective.
• $M$ is flat and of finite presentation.
• $M$ is of finite presentation and $M_x$ is free of finite rank over $\mathcal{O}_{X,x}$ for all $x \in X$ [if $X$ is a scheme].

All finiteness assumptions are essential here.