Let $A:={\mathbb{C}}[x\_0,\ldots,x\_n]=\oplus\_{d=0}^{\infty} {\mathbb{C}}[x\_0,\ldots, x\_n]\_d$ be the graded complex algebra of polynomials in $n+1$ variables, graded by degree. 


Suppose $L$ is a line bundle over a projective manifold $X$ such that the ring of plurisections of $L$, i.e. $\oplus\_{d=0}^{\infty} H^0(X,L^d)$ is isomorphic to $A$ as graded complex algebras.


Does that imply that $X$ is necessarily complex projective space ${\mathbb{CP}}^n$ and $L$ is the tautological line bundle? 



Projective variety with no syzygies but not isomorphic to projective space
The answer is no. Take any projective manifold $X$ mapping $\pi\colon
X\rightarrow\mathbb P^n$ to such that $\pi\_\*\mathcal O\_X=\mathcal O\_{\mathbb
 P^n}$ and let $L$ be $\pi^\*\mathcal O(1)$. Then $H^0(X,L^{\otimes
 m})=H^0(\mathbb P^n,\mathcal O(m))$ Examples are $X=Y\times\mathbb P^n$ or a
blowing up of a closed smooth subvariety of $\mathbb P^n$.


I do not know if some power of $L$ is necessarily without base points (in which
case $L$ itself is). In any case we can blow up $X$ such that there is a map to
$\mathbb P^n$. The line bundle that gives that mapping may I guess have more sections.


**Addendum**: I was really rambling in the last paragraph. What I meant was that suppose we only know that $\bigoplus\_nH^0(X,L^{\otimes n})$ is a polynomial ring. Does that mean that $L$ is base-point free? It may very well be true but I don't see it at the moment.


