Conceptualization of exterior powers of projective modules Let $A$ be a commutative noetherian ring, and $P$ a projective $A$ module with $rank(P)=n$. I know that $\wedge^nP \simeq L$ for some rank 1 projective $A$-module, $L$; but I'm not sure of how to think of this $L$. I understand that the top exterior power of a vector space with basis $\{e_1, e_2, \ldots, e_n\}$ is 1 dimensional with $e_1 \wedge e_2 \wedge \ldots \wedge e_n$ as its basis. But how does one expand this concept to a projective module where there is only a local basis? 
As a more specific question, let $A$ be the coordinate ring of the real 2-sphere, and $P$ the rank 2 projective module corresponding to the tangent space of $A$. $P$ is indecomposable, and there are no non-trivial rank 1 projective modules over $A$. So, $\wedge^2P \simeq A$, but I can't justify this any other way than by pigeonholing it. My main problem is that $P$ doesn't have a global basis, thus negating (as far as I can tell) the idea of taking the wedge product of basis elements.
 A: As requested, I am explaining the differential geometry point of view; I am assuming that you will be able to work out the commutative algebra on your own.

It is a well-known fact in differential geometry that a smooth manifold is orientable if and only if the top exterior power of the cotangent bundle has a global generator. The sphere is clearly orientable, so the vector bundle of differential $2$-forms must have a global generator – so we just need to find one.
First of all, let $M = \{ (x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1 \}$. Then, by abuse of notation, we have smooth functions $x, y, z : M \to \mathbb{R}$. The tangent bundle is the manifold
$$T M = \{ (x, y, z, u, v, w) \in \mathbb{R}^6 : x^2 + y^2 + z^2 = 1, x u + v y + z w = 0 \}$$
equipped with the evident projection down to $M$. The cotangent bundle is the fibrewise dual of this vector bundle, denoted by $T^* M$ or $\Omega^1_M$. The exterior derivative is a map of sheaves $\mathrm{d} : C^\infty_M \to \Omega^1_M$ such that $\langle \mathrm{d} f , X \rangle$ is the directional derivative of $f$ along the vector field $X$. In particular, we have global sections $\mathrm{d}x, \mathrm{d}y, \mathrm{d}z : M \to \Omega^1_M$, and these satisfy the equation
$$x \, \mathrm{d} x + y \, \mathrm{d} y + z \, \mathrm{d} z = 0$$
Since $M$ is an embedded submanifold of $\mathbb{R}^3$, it inherits a Riemannian metric; it is the global section $g : M \to T M \otimes T M$ given by
$$g = \mathrm{d}x \otimes \mathrm{d}x + \mathrm{d}y \otimes \mathrm{d}y + \mathrm{d}z \otimes \mathrm{d}z$$
By non-degeneracy, this induces an isomorphism of vector bundles $T M \cong T^* M$, so to show that $\Lambda^2 T M$ is a trivial vector bundle, it is enough to show that $\Lambda^2 T^* M$ is a trivial vector bundle. Now, we take local coordinates on $M$: let $(\theta, \phi)$ be parameters over the domain $(0, \pi) \times (-\pi, \pi)$ such that
$$\begin{align}
x & = \sin (\theta) \cos (\phi) &
\mathrm{d} x & = \cos (\theta) \cos (\phi) \, \mathrm{d} \theta - \sin (\theta) \sin (\phi) \, \mathrm{d} \phi \\
y & = \sin (\theta) \sin (\phi) &
\mathrm{d} y & = \cos (\theta) \sin (\phi) \, \mathrm{d} \theta + \sin (\theta) \cos (\phi) \, \mathrm{d} \phi \\
z & = \cos (\theta) &
\mathrm{d} z & = - \sin (\theta) \, \mathrm{d} \theta
\end{align}$$
One then computes that
$$\frac{1}{z} \, \mathrm{d} x \wedge \mathrm{d} y = \sin (\theta) \, \mathrm{d} \theta \wedge \mathrm{d} \phi$$
but the RHS is manifestly non-zero everywhere in the domain of parametrisation. Repeating this calculation for the various permutations of $(x, y, z)$ and various other parametrisations then shows that
$$\frac{1}{z} \, \mathrm{d} x \wedge \mathrm{d} y = \frac{1}{x} \, \mathrm{d} y \wedge \mathrm{d} z  = \frac{1}{y} \, \mathrm{d} z \wedge \mathrm{d} x$$
whereever these expressions make sense.
Of course, one might ask where I conjured up such a $2$-form. The answer has to do with the Riemannian metric. In local coordinates, we have
$$g = \mathrm{d} \theta \otimes \mathrm{d} \theta + (\sin \theta)^2 \, \mathrm{d} \phi \otimes \mathrm{d} \phi$$
and so we get a volume form
$$\omega = \sin (\theta) \, \mathrm{d} \theta \wedge \mathrm{d} \phi$$
By general theory, this is known to be a local expression for a single nowhere-vanishing global section, so all that remains is to find expressions for it in other charts.

Now, some words about algebraic geometry. If $A$ is a nice ring – say a finitely-generated integral domain over a field $k$ – one can define the module of Kähler differential forms as follows:
$$\Omega^1_{A / k} = \mathfrak{a} / \mathfrak{a}^2 \text{ where } \mathfrak{a} = \ker (f \otimes g \mapsto f g : A \otimes_k A \to A)$$
This is naturally an $A$-module, because $(A \otimes_k A) / \mathfrak{a} \cong A$. There is a canonical $k$-linear (not $A$-linear!) map $\mathrm{d} : A \to \Omega^1_{A / k}$ induced by the map $f \mapsto f \otimes 1 - 1 \otimes f : A \to A \otimes_k A$. One readily verifies that $\mathrm{d}$ satisfies the Leibniz rule, and it can be checked that $\Omega^1_{A/k}$ is generated as an $A$-module by $\{ \mathrm{d} f : f \in A \}$. So far so good. Then one can go on to define $\Omega^n_{A/k}$ to be the $n$-th exterior power (over $A$) of $\Omega^1_{A/k}$. It is a fact, at least when $k$ is algebraically closed and $A$ is a regular ring of dimension $n$, that $\Omega^n_{A/k}$ is a projective $A$-module of rank $1$.
Unfortunately, it is not at all clear from all this why $\Omega^2_{A / \mathbb{R}}$ should have a global generator of the form in the preceding discussion, when $A = \mathbb{R}[x, y, z] / (x^2 + y^2 + z^2 - 1)$. Yet it does, if we are prepared to regard $A$ as an affine scheme over $\operatorname{Spec} \mathbb{R}$: each of the three expressions is defined on a dense open subset of $\operatorname{Spec} A$, and they agree on the overlaps just as in differential geometry, so they patch together to give a global section that has non-zero stalks over all the points...
