Surjectvity of multiplication map $H^0(X,L)^{\otimes t} \to H^0(X,L^{\otimes t})$ Given a proper, integral scheme $X$ over an arbitrary field and an invertible sheaf $L$ on $X$, does there exist any integer $k \gt 0$ such that the maps
$H^0(X,L^{\otimes k})^{\otimes t} \to H^0(X,L^{\otimes kt})$
become surjective for all integers $t \ge 0$? Or are there any counterexamples?
 A: No, such a $k\gt 0$ needn't exist.   
Take for $X$ a smooth projective curve of genus $g\gt 0$ and for $L$ take $\mathcal O(p)$, where $p$ is some rational point of $X$.
Then  $ \operatorname \:{dim } (H^0(X,L))=1$   
On the other hand, Riemann-Roch implies that  for $N=\operatorname {deg}(L^{\otimes N})\geq g+1$ we have $ \operatorname \:{dim }H^0(X,L^{\otimes N})\geq 2$.
Since of course $ \operatorname \:{dim } (H^0(X,L)^{\otimes N})=1$, the map   $H^0(X,L)^{\otimes N} \to H^0(X,L^{\otimes N})$ cannot be surjective and a fortiori the $k\geq 0$ you are asking about doesn't exist.  
Edit
The question was edited after I answered it. The original question was about surjectivity of $H^0(X,L)^{\otimes kt} \to H^0(X,L^{\otimes kt})$. So I stand by my answer with the caveat that the non-existence of $k$ refers to the $k $ of the original question, about surjectivity of $H^0(X,L)^{\otimes kt} \to H^0(X,L^{\otimes kt})$.  Thanks a lot to QiL'8 for drawing my attention to the modification of the question.
A: I think you meant the surjectivity of $H^0(X,L^k)^t\rightarrow H^0(X,L^{kt})$. In any case your statement is true for ample linebundles, see the Castelnuovo-Mumford regularity section of Lazarsfeld's Positivity.
