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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.

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    $\begingroup$ 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 $A\hookrightarrow M$. $\endgroup$
    – abx
    Commented May 14, 2020 at 16:11
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    $\begingroup$ 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. $\endgroup$
    – skd
    Commented May 14, 2020 at 16:24
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    $\begingroup$ 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. $\endgroup$
    – Mohan
    Commented May 14, 2020 at 16:36
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    $\begingroup$ 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. $\endgroup$
    – Joshua Mundinger
    Commented May 14, 2020 at 16:43
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    $\begingroup$ @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... $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 14, 2020 at 17:00

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