Does the embedding of the Grassmannian in the projectivization of the exterior algebra require the axiom of choice?

Question

According to the wikipedia page for the Grassmannian, the embedding in the projectivization of the exterior algebra of $V$ requires the choice of, for every r-dimensional subspace, a basis for that space. The point in $P(\Lambda (V))$ is associated with the equivalence class of the wedge product of the basis vectors (or at least that's how I've been thinking about it). The choice of basis vectors requires a function from each subspace to a set of basis vectors, right? And the existence of such a function is only permitted by the axiom of choice, right? Or is there an alternate construction where this choice is unnecessary?