Symmetric power of a locally free sheaf in $\mathbb{P}^{n}$

Question

In Algebraic Geometry, by Hartshorne, we have:

Definition 1: Let $A$ be a ring and let $M$ be an $A$-module. Let $T^{n }(M)$ be the tensor $M \otimes \cdots \otimes M$ of $M$ with itself $n$ times for $n\geq 1$. Then $T(M) = \bigoplus_{n \geq 0} T^{n}(M)$ is a $A$-algebra, which we call the tensor algebra of $M$.

Definition 2: We define the symmetric algebra $S(M) = \bigoplus_{n \geq 0} S^{n}(M)$ of $M$ to be the quocient of $T(M)$ by the two-sided ideal generated by all expressions $x \otimes y - y \otimes x$, for all $x, y \in M$. Its component $S^{n}(M)$ in degree $n$ is called the $\text{nth}$ symmetric product of $M$.

Suppose that $\mathcal{F}$ is locally free sheaf of $\text{rank}$ $n$. then $S^{n}(\mathcal{F})$ is also locally free of $\text{rank}$, ${n+r-1 \choose n}$.

For example, according to definition 2 above, what would be $S^{1}(\mathcal{F})$ and $S^{2}(\mathcal{F})$ where $\mathcal{F} = \mathcal{O}_{X}(a+b)$ with $a, b \in \mathbb{Z}$ and $X= \mathbb{P}^{n}$.

I thank you in advance for your help.

Answer 1

Your formula is false: if $E$ is a vector bundle of rank $r$ then $S^kE$ as rank ${r+k-1}\choose{k}$ (see https://en.wikipedia.org/wiki/Symmetric_algebra)

First, for any $A$-module $M$ you have $S^1M=M$ as $S^1$ is a quotient of $T^1M=M$ by definition.

Now if $\mathcal{F}$ is a line bundle on your variety, you have a morphism $T^kF \to S^k\mathcal{F}$ given locally by the quotient map $q_U : T^k\mathcal{F}(U) \to S^k\mathcal{F} (U)$. But as $\mathcal{F}$ is of rank $1$, this latter map $q_U$ is an isomorphism: $T^k\mathcal{F}(U)$ is generated by some local section $x$ and $q_U$ is the quotient by the relation $x\otimes x - x \otimes x = 0$ which is a trivial relation.