Lemma 110.34.1. 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.
110.34 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
\mathop{\mathrm{Hom}}\nolimits _ R(I, M) = \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{colim}}\nolimits _ n e_ nR, M) = \mathop{\mathrm{lim}}\nolimits _ n \mathop{\mathrm{Hom}}\nolimits _ R(e_ nR, M) = \mathop{\mathrm{lim}}\nolimits _ n e_ nM
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 10.86.4 the functor \mathop{\mathrm{Hom}}\nolimits _ R(I, M) is exact, i.e., I is a projective R-module. \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 10.78.1. Suppose that Q can be generated by N elements as an R-module. Then it follows from Algebra, Lemma 10.15.7 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 10.78.2.
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
e_ n = (1, 1, \ldots , 1, 0, \ldots )
where the sequence of 1's has length n.
Lemma 110.34.2. 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 110.34.3. 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 \mathop{\mathrm{CH}}\nolimits _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 : \mathop{\mathrm{CH}}\nolimits _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”
C_1 \times _ K \ldots \times _ K C_ n \setminus U_1 \times _ K \ldots \times _ K U_ n
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 \mathop{\mathrm{CH}}\nolimits _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
B_ n = A_1 \otimes _{\mathbf{Q}} A_2 \otimes _{\mathbf{Q}} \ldots \otimes _{\mathbf{Q}} A_ n
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
L_{n, i} = A_1 \otimes _{\mathbf{Q}} \ldots \otimes _{\mathbf{Q}} L_ i \otimes _{\mathbf{Q}} \ldots \otimes _{\mathbf{Q}} A_ n
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
B_\infty = \mathop{\mathrm{colim}}\nolimits _ n B_ n \quad \text{and}\quad L_{\infty , i} = \mathop{\mathrm{colim}}\nolimits _ n L_{n, i}
Finally, set
M = \bigoplus \nolimits _{i \geq 1} L_{\infty , i}.
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
\bigoplus \nolimits _{1 \leq i \leq m} L_{m, i}.
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
c_ i(\text{pr}_1^*\mathcal{L}_1 \oplus \ldots \oplus \text{pr}_ m^*\mathcal{L}_ m), \quad i = m, m - 1, \ldots , m - n
are zero in the chow group of
D(f) \subset U_1 \times \ldots \times U_ m
Since f is the pullback of a function on U_1 \times \ldots \times U_ n this implies in particular that
c_{m - n}(\mathcal{O}_ W^{\oplus n} \oplus \text{pr}_1^*\mathcal{L}_{n + 1} \oplus \ldots \oplus \text{pr}_{m - n}^*\mathcal{L}_ m) = 0.
on the variety
W = (C_{n + 1} \times \ldots \times C_ m)_ K
over the field K = \mathbf{Q}(C_1 \times \ldots \times C_ n). In other words the cycle
[(P_{n + 1} \times \ldots \times P_ m)_ K]
is zero in the chow group of zero cycles on W. This contradicts Lemma 110.34.3 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 110.34.4. 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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