Lemma 102.28.1. Let $R$ be a ring. Let $I \subset R$ be an ideal generated by a countable collection of idempotents. Then $I$ is projective as an $R$-module.

## 102.28 A projective module which is not locally free

We give two examples. One where the rank is between $0$ and $1$ and one where the rank is $\aleph _0$.

**Proof.**
Say $I = (e_1, e_2, e_3, \ldots )$ with $e_ n$ an idempotent of $R$. After inductively replacing $e_{n + 1}$ by $e_ n + (1 - e_ n)e_{n + 1}$ we may assume that $(e_1) \subset (e_2) \subset (e_3) \subset \ldots $ and hence $I = \bigcup _{n \geq 1} (e_ n) = \mathop{\mathrm{colim}}\nolimits _ n e_ nR$. In this case

Note that the transition maps $e_{n + 1}M \to e_ nM$ are given by multiplication by $e_ n$ and are surjective. Hence by Algebra, Lemma 10.85.4 the functor $\mathop{\mathrm{Hom}}\nolimits _ R(I, M)$ is exact, i.e., $I$ is a projective $R$-module. $\square$

Lemma 102.28.2. Let $R$ be a nonzero ring. Let $n \geq 1$. Let $M$ be an $R$-module generated by $< n$ elements. Then any $R$-module map $f : R^{\oplus n} \to M$ has a nonzero kernel.

**Proof.**
Choose a surjection $R^{\oplus n - 1} \to M$. We may lift the map $f$ to a map $f' : R^{\oplus n} \to R^{\oplus n - 1}$. It suffices to prove $f'$ has a nonzero kernel. The map $f' : R^{\oplus n} \to R^{\oplus n - 1}$ is given by a matrix $A = (a_{ij})$. If one of the $a_{ij}$ is not nilpotent, say $a = a_{ij}$ is not, then we can replace $A$ by the localization $A_ a$ and we may assume $a_{ij}$ is a unit. Since if we find a nonzero kernel after localization then there was a nonzero kernel to start with as localization is exact, see Algebra, Proposition 10.9.12. In this case we can do a base change on both $R^{\oplus n}$ and $R^{\oplus n - 1}$ and reduce to the case where

Hence in this case we win by induction on $n$. If not then each $a_{ij}$ is nilpotent. Set $I = (a_{ij}) \subset R$. Note that $I^{m + 1} = 0$ for some $m \geq 0$. Let $m$ be the largest integer such that $I^ m \not= 0$. Then we see that $(I^ m)^{\oplus n}$ is contained in the kernel of the map and we win. $\square$

Suppose that $P \subset Q$ is an inclusion of $R$-modules with $Q$ a finite $R$-module and $P$ locally free, see Algebra, Definition 10.77.1. Suppose that $Q$ can be generated by $N$ elements as an $R$-module. Then it follows from Lemma 102.28.2 that $P$ is finite locally free (with the free parts having rank at most $N$). And in this case $P$ is a finite $R$-module, see Algebra, Lemma 10.77.2.

Combining this with the above we see that a non-finitely-generated ideal which is generated by a countable collection of idempotents is projective but not locally free. An explicit example is $R = \prod _{n \in \mathbf{N}} \mathbf{F}_2$ and $I$ the ideal generated by the idempotents

where the sequence of $1$'s has length $n$.

Lemma 102.28.3. There exists a ring $R$ and an ideal $I$ such that $I$ is projective as an $R$-module but not locally free as an $R$-module.

**Proof.**
See above.
$\square$

Lemma 102.28.4. Let $K$ be a field. Let $C_ i$, $i = 1, \ldots , n$ be smooth, projective, geometrically irreducible curves over $K$. Let $P_ i \in C_ i(K)$ be a rational point and let $Q_ i \in C_ i$ be a point such that $[\kappa (Q_ i) : K] = 2$. Then $[P_1 \times \ldots \times P_ n]$ is nonzero in $A_0(U_1 \times _ K \ldots \times _ K U_ n)$ where $U_ i = C_ i \setminus \{ Q_ i\} $.

**Proof.**
There is a degree map $\deg : A_0(C_1 \times _ K \ldots \times _ K C_ n) \to \mathbf{Z}$ Because each $Q_ i$ has degree $2$ over $K$ we see that any zero cycle supported on the “boundary”

has degree divisible by $2$. $\square$

We can construct another example of a projective but not locally free module using the lemma above as follows. Let $C_ n$, $n = 1, 2, 3, \ldots $ be smooth, projective, geometrically irreducible curves over $\mathbf{Q}$ each with a pair of points $P_ n, Q_ n \in C_ n$ such that $\kappa (P_ n) = \mathbf{Q}$ and $\kappa (Q_ n)$ is a quadratic extension of $\mathbf{Q}$. Set $U_ n = C_ n \setminus \{ Q_ n\} $; this is an affine curve. Let $\mathcal{L}_ n$ be the inverse of the ideal sheaf of $P_ n$ on $U_ n$. Note that $c_1(\mathcal{L}_ n) = [P_ n]$ in the group of zero cycles $A_0(U_ n)$. Set $A_ n = \Gamma (U_ n, \mathcal{O}_{U_ n})$. Let $L_ n = \Gamma (U_ n, \mathcal{L}_ n)$ which is a locally free module of rank $1$ over $A_ n$. Set

so that $\mathop{\mathrm{Spec}}(B_ n) = U_1 \times \ldots \times U_ n$ all products over $\mathop{\mathrm{Spec}}(\mathbf{Q})$. For $i \leq n$ we set

which is a locally free $B_ n$-module of rank $1$. Note that this is also the global sections of $\text{pr}_ i^*\mathcal{L}_ n$. Set

Finally, set

This is a direct sum of finite locally free modules, hence projective. We claim that $M$ is not locally free. Namely, suppose that $f \in B_\infty $ is a nonzero function such that $M_ f$ is free over $(B_\infty )_ f$. Let $e_1, e_2, \ldots $ be a basis. Choose $n \geq 1$ such that $f \in B_ n$. Choose $m \geq n + 1$ such that $e_1, \ldots , e_{n + 1}$ are in

Because the elements $e_1, \ldots , e_{n + 1}$ are part of a basis after a faithfully flat base change we conclude that the chern classes

are zero in the chow group of

Since $f$ is the pullback of a function on $U_1 \times \ldots \times U_ n$ this implies in particular that

on the variety

over the field $K = \mathbf{Q}(C_1 \times \ldots \times C_ n)$. In other words the cycle

is zero in the chow group of zero cycles on $W$. This contradicts Lemma 102.28.4 above because the points $Q_ i$, $n + 1 \leq i \leq m$ induce corresponding points $Q_ i'$ on $(C_ n)_ K$ and as $K/\mathbf{Q}$ is geometrically irreducible we have $[\kappa (Q_ i') : K] = 2$.

Lemma 102.28.5. There exists a countable ring $R$ and a projective module $M$ which is a direct sum of countably many locally free rank $1$ modules such that $M$ is not locally free.

**Proof.**
See above.
$\square$

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