Here are nine examples of projective modules that are not free, some of which are finitely generated.

Direct Products

Consider the ring $R= \Z/2\times\Z/2$ and the submodule $\Z/2\times \{0\}$. It is by construction a direct summand of $R$ but certainly not free. And it’s finitely generated! Another example is the submodule $\Z/2\subset \Z/6$, though this is the same kind of thing because $\Z/6\cong\Z/2\times\Z/3$. This was the first example I ever saw of a nonfree projective module.

Infinite Direct Products

One can modify the above construction for infinite direct products of rings, too. For instance, $R = \prod_{i=1}^\infty \Z$ contains $\Z$ as a direct summand. Hence $\oplus_{i=1}^\infty\Z$ is a projective $R$ module, yet cannot be free since nonzero free modules are uncountable.

Ideals in Dedekind Domains

In a Dedekind domain $R$, take an ideal representing a nontrivial element in the class group. It will then be projective. As an example, the class number of $\Z[\sqrt{5}]$ is two, and the ideal $(2,1+\sqrt{5})$ represents the nontrivial element in the class group. It is not free since it is not principal, and it is finitely generated projective since it is invertible.

More generally, for any ring extension of commutative rings $R\subseteq S$, one may define invertible $R$-submodules of $S$ as it is done for Dedekind domains. Then any invertible $R$-submodule of $S$ will be finitely-generated and projective. For more details and a further example, see Lam’s ‘Lectures on Modules and Rings’, Sections 2B-2C.

Rings of Continuous Functions

This example is due to Irving Kaplansky: Let $R$ be the ring of real-valued continuous functions on the closed interval $[0,1]$. Since this space is compact and contractible, every vector bundle over it is trivial, and so finitely generated projective $R$-modules are free by Swan’s theorem. However, the ideal $I$ that consists of those functions vanishing on some open neighbourhood of $0$ is projective but not free and hence not finitely generated. The easy part is that $I$ is not free: for any $f\in I$ there exists a $g\in R$ such that $fg = 0$. This cannot be true for a nonzero free $R$-module.

Checking that $I$ is not projective is not difficult, but is better done with a few pictures and the short argument can be found in perhaps the best book on modules ever written: Lam’s book ‘Lectures on Modules and Rings’, Section 2.B, Example 2.12D.

Semisimple Rings

A ring $R$ is semisimple (by one possible definition) if every $R$-module is projective, or equivalently if every ideal is a direct summand.

The Wedderburn-Artin theorem states that semisimple rings are precisely finite direct products $M_{n_1}(D_1)\times\cdots\times M_{n_k}(D_k)$ of matrix rings over division rings $D_i$. Here is a subtle example central in the representation theory of finite groups: If $R$ is any semisimple ring and $G$ is a finite group such that $|G|$ is invertible in $R$ then the group ring $RG$ is also semisimple. For instance $\C[\Z/2]$ is semisimple: If $\sigma\in\Z/2$ is the nontrivial element, then it is isomorphic to the product $\C\times\C$ via the isomorphism $a + b\sigma\mapsto (a-b,a+b)$. For yet another example, $\C[S_3]\cong M_2(\C)\times M_1(\C)^2$.

Since every left ideal is projective but $R$ is Artinian, the proper left ideals cannot be free (and in fact the minimal left ideals form a complete set of representatives for isomorphism classes of simple left modules). So for instance, in a matrix ring over a field, the left ideal of column vectors for some column is a left ideal that is a projective module but not free. This is another finitely-generated example.

Upper Triangular Matrix Rings

One need not take a full matrix ring to get examples. In fact, any finite dimensional algebra $R$ over a field or division ring with nontrivial idempotent $e$ will give a finitely generated projective left module $Re$ or right module $eR$ that is not free. For instance, let $k$ be a field and let
$$
R = \left\{
\begin{pmatrix}
a & b\\
0 & c
\end{pmatrix} : a,b,c\in k\right\}.
$$
This is a $k$-algebra. If we consider the ideal right-ideal
$$
I = \left\{
\begin{pmatrix}
0 & 0\\
0 & c
\end{pmatrix} : c\in k\right\}.
$$
then $I$ is a right projective $R$-module. It can’t be free since as a vector space over $k$ it is one-dimensional, whereas $R$ is three-dimensional. However, it’s a direct summand, since the other factor is the right-ideal
$$
J = \left\{
\begin{pmatrix}
a & b\\
0 & 0
\end{pmatrix} : a,b\in k\right\}.
$$
This is also different than the semisimple example, since $R$ is not semisimple. Indeed, a easy but somewhat roundabout way to see this is that $R$ is not von Neumann regular. Recall that a ring $S$ is von Neumann regular if for every $a\in S$ there exists an $x\in S$ such that $axa = a$. It is not hard to see that a ring $S$ is von Neumann regular if and only if every module is flat. So if $R$ is not von Neummann regular, it cannot be semisimple, since semisimple rings are von Neumann regular.

In the upper triangular matrices ring, the calculation:
$$
\begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix}
\begin{pmatrix}
a & b\\
0 & c
\end{pmatrix}
\begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix}
=
\begin{pmatrix}
0 & 0\\
0 & 0
\end{pmatrix}
$$
shows that $R$ cannot be von Neumann regular, so there exists a nonflat $R$-module, hence a nonprojective $R$-module. Hence $R$ is not semisimple.

Vector Bundles on Spheres

Swan proved in ‘Vector Bundles and Projective Modules’ that on a compact Hausdorff space, there is a bijection between the set of vector bundles and finitely generated projective modules. (The analogous statement for affine schemes was proven by Serre). Applying this to spheres and a bit of clever algebra, Swan showed that the following module $P$ is not free:

Let $n$ be a nonnegative integer with $n\not\in \{0,1,3,7\}$ (do you recognise these numbers?!). Let $R = \R[x_0,\dots,x_n]/(x_0^2 + \cdots + x_n^2 – 1)$, where $\R$ denotes the real numbers. Let $F$ be the free $R$-module on generators $s_0,\cdots,s_n$ and define the $R$-module homomorphism on generators by $g(s_i) = \sum_{j=0}^n x_ix_js_j$. Set $P = {\rm ker}(g)$.

We can show that $P$ is projective as follows: first we calculate
$$
\begin{align*}
g(g(s_i)) &= \sum_j x_ix_j\sum_k x_jx_ks_k\\
&= x_i\sum_j\sum_k x_j^2x_ks_k\\
&= \sum_kx_ix_ks_k\\
&= g(s_i).
\end{align*}
$$
Hence $g:F\to F$ is an idempotent $R$-module homomorphism. If we define $f:F\to\ker g$ via $f(x) = x – g(x)$, then we see that $f(x) = x$ for $x\in\ker g$ so $f$ is surjective, and the inclusion $\ker g\to F$ is a section showing that $\ker g$ is projective.

Swan’s paper is highly recommended and very readable, and it’s a good source if you want to construct even more crazy projective modules over affine algebras.

Relation Modules for Infinite Group Rings

Groups provide a rich source of examples: we’ve already seen that when $G$ is a finite group of order $n$ over a semisimple ring $R$ in which $n$ is invertible, then $RG$ is semisimple. Of course, this isn’t anything more than what we already had via Wedderburn-Artin, but what about infinite groups? They give some of the most interesting examples because infinite groups are much more subtle and tricky!

The following example is due to Berridge and Dunwoody in their paper ‘Non-Free Projective Modules for Torsion-Free Groups’: take the trefoil group $T = \langle a,b ~|~ a^2 = b^3\rangle$, and define for each $i=0,1,2,\dots$ a homomorphism
$$
\begin{align*}
\mu_i:\Z T\oplus\Z T&\longrightarrow \Z T,\\
(t_1,t_2)&\longmapsto (1 + a + \cdots + a^{2i})t_1 + (b^{-1}a + b^{-2}a – 1)t_2.
\end{align*}
$$
Let $N_i = \ker\mu_i$. The authors prove in their paper that each of these $N_i$ is a finitely generated projective module of rank 1. Moreover, they prove that if $N_i$ and $N_j$ are isomorphic, then the set of prime divisors of $i(i+1)$ is the same as the set of prime divisors of $j(j+1)$. Hence we see that the $N_i$ give infinitely many rank $1$ projectives that are pairwise nonisomorphic.

Projective But Not Locally Free

It’s true that if $R$ is any ring and $e$ is an idempotent then the left $Re$ is a projective left $R$-module. Here is a theorem and proof I learned from the elephantine Stacks Project, with modified notation so that it works for a noncommutative ring as well:

Theorem. If $I$ is a left ideal generated by countably many idempotents, then it is projective.
Proof. Let $I$ be the left ideal of $R$ generated by $e_1,e_2,e_3,\dots$. We can replace $e_2$ by $x = e_1 + (1-e_1)e_2$. Indeed, since $e_1\in I$ we see that $x – e_1 + e_2e_1 = e_2$. Similarly, inductively replace $e_n$ by $e_n + (1-e_n)e_{n+1}$. Then, we get that $Re_1\subseteq Re_2\subseteq Re_3\subseteq\cdots$. Via these inclusions, we get that $I = \varinjlim Re_i$.

Now, let $0\to A\to B\to C\to 0$ be an exact sequence of left $R$-modules. For each $i$, we can apply the functor ${\rm Hom}_R(Re_i,-)$ to get an exact sequence
$$
0\to {\rm Hom}_R(Re_i,A)\to{\rm Hom}_R(Re_i,B)\to{\rm Hom}_R(Re_i,C)\to 0
$$
which is exact, and hence forms a short exact sequence in appropriate category of diagrams of abelian groups. Moreover each ${\rm Hom}_R(Re_i,A)$ (or with $B$ or $C$) is a directed system, and to get ${\rm Hom}_R(I,-)$ applied to the sequence $0\to A\to B\to C\to 0$ we have to take the inverse limit. Moreover, it is well-known that if all the transition maps ${\rm Hom}_R(Re_{i+1},A)\to{\rm Hom}_R(Re_i,A)$ are surjective (more generally, if the system is Mittag-Leffler), then the sequence will remain exact upon applying the inverse limit.

Hence, it suffices to show ${\rm Hom}_R(Re_{i+1},A)\to{\rm Hom}_R(Re_i,A)$ is surjective. For this, it suffices to observe that the inclusion $Re_i\to Re_{i+1}$ is a split injection of $R$-modules via $Re_{i+1}\to Re_i$ given by $x\mapsto xe_i$. This is indeed a splitting because $e_i^2 = e_i$ by definition.

For example, take $R=\prod_{i=1}^\infty F_i$ to be an infinite product of fields $F_i$ and $e_i$ to be the idempotent defined by $e_i(j) = 1$ for $j \leq i$ and $e_i(j) = 0$ otherwise. This is an infinite collection of idempotents, and the ideal that they generate is projective. One can check out the Stacks Project, Tag 05WG, for a short argument on why this module is not even locally free.

More Examples

If readers have more examples that are significantly different than the ones above, please write them in the comments!

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