Projective variety with no syzygies but not isomorphic to projective space 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?  
 A: Here is an example for Torsten's Addendum: a line bundle with the required properties which is not semi-ample (no multiple is globally generated):
Take an $\pi:X\to \mathbb P^n$ as in Torsten's example, that is, with $\pi_*\mathcal O_X\simeq \mathcal O_{\mathbb P^n}$. Also assume that there is an exceptional divisor $E$, so let's just say it is a birational morphism contracting the divisor $E$ to a point. Now take $\mathcal L=\pi^*\mathcal O_{\mathbb P^n}(1)$ and consider the short exact sequence:
$$
0\to \mathcal L^{\otimes n} \to (\mathcal L(E))^{\otimes n} \to \mathcal O_{nE}(nE) \to 0
$$
The sheaf on the right has no global sections, so the other two have the same $H^0$. Therefore 
$$
H^0(X,(\mathcal L(E))^{\otimes n}) = H^0(\mathbb P^n, \mathcal O_{\mathbb P^n}(n))
$$
Finally $(\mathcal L(E))^{\otimes n}$ cannot be globally generated, because 
$H^0(X,\mathcal L^{\otimes n})=H^0(X,(\mathcal L(E))^{\otimes n})$, so
 ${\rm supp}\,E$ is always in the base locus (or in other words, if it had a section whose zero section did not contain ${\rm supp}\,E$, then its restriction to $nE$ would give a nonzero section of $\mathcal O_{nE}(nE)$).

And here is what you can assert based on your condition:
Naive statement: Essentially the only reason your desired claim can fail is via Torsten's example.
More precise statement: Let $k$ be a field, $S:=k[x_0,…,x_n]$ and suppose $\mathcal L$ is a line bundle on a projective variety $X$ (over $k$) such that $R(X,{\mathcal L})=\oplus_{d=0}^\infty H^0(X,\mathcal L^d)\simeq S$ as graded $k$-algebras. Then there exists a diagram 
$$
X \overset\sigma\longleftarrow \widetilde X \overset\phi\longrightarrow \mathbb P^n 
$$
such that $\sigma$ is a projective birational morphism and $\phi$ is surjective. Furthermore, $\sigma^*\mathcal L\supseteq \phi^*\mathcal O_{\mathbb P^n}(1)$. Finally if $n=\dim X$, then $\phi$ is also birational and hence $X$ is a rational variety. 
Proof: 
By the assumption ${\rm Proj}\, R(X,\mathcal L)\simeq \mathbb P^n$ and there exists a dominant rational map $\psi: X\dashrightarrow \mathbb P^n$. Let 
$$
X \overset\sigma\longleftarrow \widetilde X \overset\phi\longrightarrow \mathbb P^n 
$$
be the resolution of indeterminacies of $\psi$ via blow-ups on $X$. Then $\sigma$ is a projective birational morphism and hence $\widetilde X$ is also projective and $\sigma^*\mathcal L\supseteq \phi^*\mathcal O_{\mathbb P^n}(1)$. It follows that the induced morphism $\phi$ has to be surjective (since $\widetilde X$ is projective and the morphism is defined everywhere). Since this morphism is given by global sections of  sufficiently high powers of a line bundle (i.e., $\phi^*\mathcal O_{\mathbb P^n}(m)$ for some $m>0$),  it follows that it has connected fibers and $\phi_*\mathcal O_{\widetilde X}\simeq \mathcal O_{\mathbb P^n}$ and hence if $n=\dim X$, then $\phi$ is birational. 
A: 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.
