Geometric intuition behind $V(M)=\operatorname{Spec}(\operatorname{Sym}(M))$? In the equivalence between geometric vector bundles and locally free sheaves we assign to a locally free sheaf $M$ the bundle $V(M)=\operatorname{Spec}(\operatorname{Sym}(M))$.
I don't doubt the proofs of the anti-equivalence but I would really like to understand $V(M)$ geometrically. What's the symmetric algebra doing there? How should one think of this bundle geometrically? How could one have "guessed" this equivalence?
 A: For simplicty, I start from the case of the trivial vector bundle over a scheme.
Let $\mathbb{V}$ be a vector space over a field $\mathbb{K}$ of finite dimension; defined
\begin{equation*}
\mathbb{K}[\mathbb{V}^{\lor}]=\bigoplus_{n\in\mathbb{N}}S^n(\mathbb{V}^{\lor})\equiv\operatorname{Sym}(\mathbb{V})
\end{equation*}
where $\mathbb{V}^{\lor}$ is the dual space and $S^n(\cdot)$ is the $n$-th symmetric power of $\cdot$; we get
\begin{equation*}
\mathscr{V}=\operatorname{Spec}\mathbb{K}[\mathbb{V}^{\lor}],
\end{equation*}
because
\begin{equation*}
\forall A\in Ob(\mathbf{Alg}_{\mathbb{K}}),\,\mathscr{V}(\operatorname{Spec} A)=\hom_{\mathbf{Sch}}(\operatorname{Spec}A,\mathscr{V})\simeq\hom_{\mathbf{Alg}_{\mathbb{K}}}(\mathbb{K}[\mathbb{V}
^{\lor}],A)\simeq\\
\simeq\hom_{\mathbf{Mod}_{\mathbb{K}}}(\mathbb{V}^{\lor},A)\simeq\mathbb{V}\otimes_{\mathbb{K}}A;
\end{equation*}
where the categories are obvious.
This construction is generalizable to free modules $M$ over a (commutative with unit) ring $R$, getting
\begin{equation*}
\mathscr{M}=\operatorname{Spec}R[M^{\lor}];
\end{equation*}
from all this, because a locally free sheaf $\mathcal{F}$ over a scheme $X$ is locally isomorphic to a free sheaf $\widetilde{M}=\widetilde{R^{\oplus n}}$, you understand because we need of the affine schemes $\mathscr{M}$!
Is it all clear?
