# Decomposition of $\pi\colon E\to\mathbb{P}^1_k$ as a direct sum of tensor powers of the tautological line bundle?

Suppose you have a vector bundle $\pi\colon E\to\mathbb{P}^1_k$, where $k$ is some field. Is it always possible to decompose the vector bundle into a direct sum of tensor powers of the tautological line bundle, at least up to isomorphism?

I know Grothendieck's theorem says that every vector bundle on $\mathbb{P}^1_F$ is a direct sum of line bundles. Is every line bundle a tensor power of the tautological line bundle in this case or something?

• All line bundles on $\mathbb P^1$ are of the form $\mathscr )(k)$ for $k \in \mathbb Z$. So all line bundles are tensor power of the tautological line bundle or its dual. – user99914 Apr 19 '15 at 6:31

Let $\mathcal{O}(-1)$ denote the tautological line bundle and $\mathcal{O}(1)$ denote its dual. For $a \in \mathbb{Z}$, define $\mathcal{O}(a)$ as $\mathcal{O}(1)^{\otimes\, a}$. To make sense of this definition when $a < 0$, note that we are working in the Picard group, so $\mathcal{O}(1)^{-1} = \mathcal{O}(1)^* = \mathcal{O}(-1)$.
The Birkhoff-Grothendieck theorem states that any algebraic vector bundle over $\mathbb{P}^1_k$ is the direct sum of line bundles of the form $\mathcal{O}(a)$. In particular, each line bundle in the decomposition is a power of the tautological bundle (if you include negative powers).