# The Stacks Project

## Tag 05WG

### 101.28. 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 $\aleph_0$.

Lemma 101.28.1. 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.

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 10.85.4 the functor $\mathop{\mathrm{Hom}}\nolimits_R(I, M)$ is exact, i.e., $I$ is a projective $R$-module. $\square$

Lemma 101.28.2. 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 $$A = \left( \begin{matrix} 1 & 0 & 0 & \ldots \\ 0 & a_{22} & a_{23} & \ldots \\ 0 & a_{32} & \ldots \\ \ldots & \ldots \end{matrix} \right)$$ 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 101.28.2 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 $$e_n = (1, 1, \ldots, 1, 0, \ldots )$$ where the sequence of $1$'s has length $n$.

Lemma 101.28.3. 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 101.28.4. 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'' $$C_1 \times_K \ldots \times_K C_n \setminus U_1 \times_K \ldots \times_K U_n$$ 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 $A_0(U_n)$. 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 $$B_n = A_1 \otimes_{\mathbf{Q}} A_2 \otimes_{\mathbf{Q}} \ldots \otimes_{\mathbf{Q}} A_n$$ 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 $$L_{n, i} = A_1 \otimes_{\mathbf{Q}} \ldots \otimes_{\mathbf{Q}} M_i \otimes_{\mathbf{Q}} \ldots \otimes_{\mathbf{Q}} A_n$$ 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 101.28.4 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 101.28.5. 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. $\square$

The code snippet corresponding to this tag is a part of the file examples.tex and is located in lines 1956–2177 (see updates for more information).

\section{A projective module which is not locally free}
\label{section-projective-not-locally-free}

\noindent
We give two examples. One where the rank is between $0$ and $1$
and one where the rank is $\aleph_0$.

\begin{lemma}
\label{lemma-ideal-generated-by-idempotents-projective}
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.
\end{lemma}

\begin{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 $I = \bigcup_{n \geq 1} (e_n) = \colim_n e_nR$.
In this case
$$\Hom_R(I, M) = \Hom_R(\colim_n e_nR, M) = \lim_n \Hom_R(e_nR, M) = \lim_n e_nM$$
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 \ref{algebra-lemma-ML-exact-sequence}
the functor $\Hom_R(I, M)$ is exact, i.e., $I$ is a projective
$R$-module.
\end{proof}

\begin{lemma}
\label{lemma-map-cannot-be-injective}
\begin{slogan}
A map of finite free modules cannot be injective if the source has
rank bigger than the target.
\end{slogan}
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.
\end{lemma}

\begin{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 \ref{algebra-proposition-localization-exact}.
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
$$A = \left( \begin{matrix} 1 & 0 & 0 & \ldots \\ 0 & a_{22} & a_{23} & \ldots \\ 0 & a_{32} & \ldots \\ \ldots & \ldots \end{matrix} \right)$$
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.
\end{proof}

\noindent
Suppose that $P \subset Q$ is an inclusion of $R$-modules with $Q$ a
finite $R$-module and $P$ locally free, see
Algebra, Definition \ref{algebra-definition-locally-free}.
Suppose that $Q$ can be generated by $N$ elements as an $R$-module.
Then it follows from
Lemma \ref{lemma-map-cannot-be-injective}
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 \ref{algebra-lemma-finite-projective}.

\medskip\noindent
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 $n$.

\begin{lemma}
\label{lemma-ideal-projective-not-locally-free}
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.
\end{lemma}

\begin{proof}
See above.
\end{proof}

\begin{lemma}
\label{lemma-chow-group-product}
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 $U_i = C_i \setminus \{Q_i\}$.
\end{lemma}

\begin{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''
$$C_1 \times_K \ldots \times_K C_n \setminus U_1 \times_K \ldots \times_K U_n$$
has degree divisible by $2$.
\end{proof}

\noindent
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
$$B_n = A_1 \otimes_{\mathbf{Q}} A_2 \otimes_{\mathbf{Q}} \ldots \otimes_{\mathbf{Q}} A_n$$
so that $\Spec(B_n) = U_1 \times \ldots \times U_n$ all products
over $\Spec(\mathbf{Q})$. For $i \leq n$ we set
$$L_{n, i} = A_1 \otimes_{\mathbf{Q}} \ldots \otimes_{\mathbf{Q}} M_i \otimes_{\mathbf{Q}} \ldots \otimes_{\mathbf{Q}} A_n$$
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
$$B_\infty = \colim_n B_n \quad\text{and}\quad L_{\infty, i} = \colim_n L_{n, i}$$
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 \ref{lemma-chow-group-product}
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 $[\kappa(Q_i') : K] = 2$.

\begin{lemma}
\label{lemma-projective-not-locally-free}
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.
\end{lemma}

\begin{proof}
See above.
\end{proof}

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