# Algebraic analog of a geometric result

There is a famous topological result:

Let $$X$$ be a smooth manifold of dimension $$n$$, $$E$$ be a vector bundle of rank $$k > n$$, then $$E$$ contains a trivial line bundle.

So, I guess that (enlightened by Hartshorne's Exercise 2.8.2):

Let $$A$$ be a ring (some more conditions are needed, say Noetherian, universally catenary,...) of Krull dimension $$n$$ and let $$P$$ be a projective module of rank $$k > n$$. Then there exists a split injection $$A \hookrightarrow P$$

I wonder if there's a result similar to this one.

• You want $A$ to be a direct summand of $M$. Since a projective module is torsion free, any nonzero element gives rise to an injection $A\hookrightarrow M$. – abx May 14 at 16:11
• Serre proved that any vector bundle on an affine scheme X of rank larger than the dimension of X has a trivial bundle as a summand; this is Theoreme 1 in numdam.org/article/SD_1957-1958__11_2_A9_0.pdf. – skd May 14 at 16:24
• Serre proved his result for arbitrary varieties over an infinite filed. If $E$ is a globally generated vector bundle of rank greater than the dimension, a general section is nowhere vanishing. Of course, this does not give a splitting in general, except for affine varieties. – Mohan May 14 at 16:36
• Here's a concrete example: let $E/\mathbb C$ be an elliptic curve. A vector bundle of rank $2$ is equivalent to a representation $\pi_1(E) \to GL_2(\mathbb C)$ up to conjugacy, that is, two commuting matrices up to conjugacy. These have a common eigenvector, but the vector bundle does not split unless the commuting matrices are simultaneously diagonalizable. – Joshua Mundinger May 14 at 16:43
• @JoshuaMundinger: "A vector bundle of rank 2 is equivalent to a representation" ― this doesn't seem quite right. On $\mathbf P^1$ there are interesting vector bundles such as $\mathcal O \oplus \mathcal O(1)$, but $\pi_1(\mathbf P^1)$ is trivial... – R. van Dobben de Bruyn May 14 at 17:00