# Is the intersection of two f.g. projective submodules f.g.?

Let $R$ be a commutative unital ring and $M$ a finitely generated projective $R$-module. My question is: if $N_1$ and $N_2$ are f.g. projective submodules of $M$, is $N_1 \cap N_2$ f.g.? Is it projective?

(Surely the answer is no but I haven't been able to find a counterexample. Also, sorry for the lack of motivation, but my reason to ask is so convoluted I think including it would seriously lower the interest/size ratio of this question.)

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Of course, if “module” means “left module”, the question also makes sense for noncommutative rings. I'm more interested in the commutative case, but noncommutative comments are very welcome. –  PseudoNeo Aug 1 '12 at 21:58
I agree the answer is likely no in general (to both questions). I'd like explicit examples. –  Jack Schmidt Aug 1 '12 at 22:01

Thinking of a f.g. projective module as a vector bundle, it seems very likely the answer is no: consider the trivial bundle of rank 2 and two "twisted" subbundles of rank 1 whose intersection is 0-dimensional everywhere except over a closed subset with non-empty interior – so not a vector bundle, in particular.

Let's see where this thinking leads. Let $R$ be the ring of real-valued continuous functions on the real line $\mathbb{R}$. Let $M = R \oplus R$. Let $f : \mathbb{R} \to \mathbb{R}$ be the following continuous function: $$f (x) = \begin{cases} \exp (-(x+1)^{-2}) & x < 1 \\ 0 & -1 \le x \le 1 \\ \exp (-(x-1)^{-2}) & x > 1 \end{cases}$$

Consider the submodules below: \begin{align} N_1 & = \{ (s_1, s_2) \in M : f s_1 - s_2 = 0 \} & N_2 & = \{ (s_1, s_2) \in M : f s_1 + s_2 = 0 \} \end{align} $M$, $N_1$ and $N_2$ are all clearly f.g. free $R$-modules. Yet their intersection is not f.g. projective: $$N = N_1 \cap N_2 = \{ (s_1, s_2) \in M : f s_1 = 0, s_2 = 0 \}$$ Let $\mathfrak{m}$ be the maximal ideal of $R$ consisting of functions vanishing at $1$, and let $R_\mathfrak{m}$ be the localisation of $R$ at $\mathfrak{m}$. If $N$ were f.g. projective, then $N_\mathfrak{m}$ would be f.g. free over $R_\mathfrak{m}$. On the other hand, by construction, $N_\mathfrak{m}$ is annihilated by the germ of $f$, which is not zero, and $N_\mathfrak{m}$ is itself non-zero (take, for example, any bump function supported on $[-1, 1]$) – so $N_\mathfrak{m}$ cannot be free. Therefore $N$ is not f.g. projective.

Here is a picture of one of the monstrosities we constructed:

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Great answer! I was stupid not to think of this problem in terms of vector bundles. I will try to see if this gives other examples (for example, over Dedekind rings, the intersection will automatically be finitely generated but there is no reason for the rank of the intersection to be constant over $\mathrm{Spec}\mathop R$...). –  PseudoNeo Aug 2 '12 at 10:02
Hmmm... What I just said is rubbish, as a f.g. torsion-free module over a Dedekind domain is projective. I'd love to understand what forces this intersection to have a constant rank... –  PseudoNeo Aug 2 '12 at 10:18
I tried to construct an algebraic counterexample at first but I got quite confused because the set-theoretic intersection of the vector bundles does not necessarily correspond to the intersection of the corresponding modules of sections. (Try considering the bundles $\operatorname{Spec} k[x,y,z]/(x y - z)$ and $\operatorname{Spec} k[x,y,z]/(x y + z)$ over $\mathbb{A}^1_k$ and see what I mean!) –  Zhen Lin Aug 2 '12 at 10:46
yep, if the intersection of modules corresponded to the set-theoretic intersection, it would be easy to construct examples over $\mathbb A^1$, but because of the property of Dedekind domain I mentioned, it's not going to happen... –  PseudoNeo Aug 2 '12 at 10:56
Indeed, $N$ can't be projective at all, since local + projective implies free even without knowing f.g. –  Harry Altman Aug 2 '12 at 19:38