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Suppose that $H^0(L^{\otimes n})|_ Y$ is non-zero for all $Y$ and $n$ sufficiently big, and vanishes somewhere on $Y$. Then it follows that the base set of $L$ is trivial: indeed, $L$ has a non-zero section on any complex subvariety, which includes the base set. This implies that the natural map $P_n:\; X\rightarrow {\mathbb P}(H^0(X, L^{\otimes n}))$ is holomorphic, for $n$ sufficiently big. Also from your assumption it follows that $P_n$ does not map any irreducible, positive-dimensional subvariety to a point (again, for $n$ sufficiently big). This implies that $P_n$ is a finite, proper map to a projective variety, hence $X$ is a ramified covering of a projective variety. A ramified covering of a projective variety is projective, which can be seen from vanishing of cohomology of powers of $L$ (a finite map is acyclic on coherent sheaves, hence the cohomology of $L^{\otimes n}$ on $X$ are the same as cohomology of ${\cal O}(1)$ on its image).

However, this works only when $H^0(L^{\otimes n} |_Y)\neq 0$; the implication $H^0(L^{\otimes n} |_Y)\neq 0$ $\Rightarrow$ $H^0(L^{\otimes n})|_Y\neq 0$ is not that easy.

If $Y$ does not lie in a zero divisor of $L^{\otimes n}$, we are done, otherwise we replace $Y$ with the zero divisor of $L^{\otimes n}$. Using induction on dimension, we may already assume that the restriction $L|_Y$ is ample. Consider the exact sequence $$ 0\rightarrow H^0(L^{\otimes n+k}) \rightarrow H^0(L^{\otimes n+k}) \stackrel r \rightarrow H^0(Y, L^{\otimes n+k} |_Y) \rightarrow H^1(L^{\otimes k}) \rightarrow H^1(L^{\otimes n+k}) \rightarrow 0. $$ The arrow $r$ of this sequence is the restriction map; if it vanishes, we have $\dim H^0(Y, L^{\otimes n+k}|_Y) \leq \dim H^1(L^{\otimes k})$ for all $n\gg 0$ and $k$. The last term of this exact sequence actually implies that $\dim H^1(L^{\otimes n+k})\leq \dim H^1(L^{\otimes k})$, for all $k$, and all $n\gg 0$, hence $H^1(L^{\otimes n+k})$ is bounded by a universal constant. This implies that $\dim H^0(Y, L^{\otimes n+k} |_Y)$ is also bounded, whenever $r=0$, which is impossible, because $L|_Y$ is ample.

Suppose that $H^0(L^{\otimes n})|_ Y$ is non-zero for all $Y$ and $n$ sufficiently big, and has a section which vanishes somewhere on $Y$. Then it follows that the base set of $L$ is trivial: indeed, $L$ has a non-zero section on any complex subvariety, which includes the base set. This implies that the natural map $P_n:\; X\rightarrow {\mathbb P}(H^0(X, L^{\otimes n}))$ is holomorphic, for $n$ sufficiently big. Also from this assumption it follows that $P_n$ does not map any irreducible, positive-dimensional subvariety to a point (again, for $n$ sufficiently big). This implies that $P_n$ is a finite, proper map to a projective variety, hence $X$ is a ramified covering of a projective variety. A ramified covering of a projective variety is projective, which can be seen from vanishing of cohomology of powers of $L$ (a finite map is acyclic on coherent sheaves, hence the cohomology of $L^{\otimes n}$ on $X$ are the same as cohomology of ${\cal O}(1)$ on its image).

However, this works only when $H^0(L^{\otimes n} |_Y)\neq 0$; the implication $H^0(L^{\otimes n} |_Y)\neq 0$ $\Rightarrow$ $H^0(L^{\otimes n})|_Y\neq 0$ is not that easy.

If $Y$ does not lie in a zero divisor of $L^{\otimes n}$, we are done, otherwise we replace $Y$ with the zero divisor of $L^{\otimes n}$. Using induction on dimension, we may already assume that the restriction $L|_Y$ is ample. Consider the exact sequence $$ 0\rightarrow H^0(L^{\otimes k}) \rightarrow H^0(L^{\otimes n+k}) \stackrel r \rightarrow H^0(Y, L^{\otimes n+k} |_Y) \rightarrow H^1(L^{\otimes k}) \rightarrow H^1(L^{\otimes n+k}) \rightarrow 0. $$ The arrow $r$ of this sequence is the restriction map; we need to prove that $r$ does not vanish. If it vanishes, we have $\dim H^0(Y, L^{\otimes n+k}|_Y) \leq \dim H^1(L^{\otimes k})$ for all $n\gg 0$ and $k$. The last term of this exact sequence actually implies that $\dim H^1(L^{\otimes n+k})\leq \dim H^1(L^{\otimes k})$, for all $k$, and all $n\gg 0$, hence $H^1(L^{\otimes n+k})$ is bounded by a universal constant. This implies that $\dim H^0(Y, L^{\otimes n+k} |_Y)$ is also bounded, whenever $r=0$, which is impossible, because $L|_Y$ is ample.

The proof is actually pretty easy. Suppose that $H^0(L^{\otimes n})|_ Y$ is non-zero for all $Y$ and $n$ sufficiently big, and vanishes somewhere on $Y$. Then it follows that the base set of $L$ is trivial: indeed, $L$ has a non-zero section on any complex subvariety, which includes the base set. This implies that the natural map $P_n:\; X\rightarrow {\mathbb P}(H^0(X, L^{\otimes n}))$ is holomorphic, for $n$ sufficiently big. Also from your assumption it follows that $P_n$ does not map any irreducible, positive-dimensional subvariety to a point (again, for $n$ sufficiently big). This implies that $P_n$ is a finite, proper map to a projective variety, hence $X$ is a ramified covering of a projective variety. A ramified covering of a projective variety is projective, which can be seen from vanishing of cohomology of powers of $L$ (a finite map is acyclic on coherent sheaves, hence the cohomology of $L^{\otimes n}$ on $X$ are the same as cohomology of ${\cal O}(1)$ on its image).

However, this works only when $H^0(L^{\otimes n} |_Y)\neq 0$; the implication $H^0(L^{\otimes n} |_Y)\neq 0$ $\Rightarrow$ $H^0(L^{\otimes n})|_Y\neq 0$ is not that easy.

Suppose that $H^0(L^{\otimes n})|_ Y$ is non-zero for all $Y$ and $n$ sufficiently big, and vanishes somewhere on $Y$. Then it follows that the base set of $L$ is trivial: indeed, $L$ has a non-zero section on any complex subvariety, which includes the base set. This implies that the natural map $P_n:\; X\rightarrow {\mathbb P}(H^0(X, L^{\otimes n}))$ is holomorphic, for $n$ sufficiently big. Also from your assumption it follows that $P_n$ does not map any irreducible, positive-dimensional subvariety to a point (again, for $n$ sufficiently big). This implies that $P_n$ is a finite, proper map to a projective variety, hence $X$ is a ramified covering of a projective variety. A ramified covering of a projective variety is projective, which can be seen from vanishing of cohomology of powers of $L$ (a finite map is acyclic on coherent sheaves, hence the cohomology of $L^{\otimes n}$ on $X$ are the same as cohomology of ${\cal O}(1)$ on its image).

However, this works only when $H^0(L^{\otimes n} |_Y)\neq 0$; the implication $H^0(L^{\otimes n} |_Y)\neq 0$ $\Rightarrow$ $H^0(L^{\otimes n})|_Y\neq 0$ is not that easy.

If $Y$ does not lie in a zero divisor of $L^{\otimes n}$, we are done, otherwise we replace $Y$ with the zero divisor of $L^{\otimes n}$. Using induction on dimension, we may already assume that the restriction $L|_Y$ is ample. Consider the exact sequence $$ 0\rightarrow H^0(L^{\otimes n+k}) \rightarrow H^0(L^{\otimes n+k}) \stackrel r \rightarrow H^0(Y, L^{\otimes n+k} |_Y) \rightarrow H^1(L^{\otimes k}) \rightarrow H^1(L^{\otimes n+k}) \rightarrow 0. $$ The arrow $r$ of this sequence is the restriction map; if it vanishes, we have $\dim H^0(Y, L^{\otimes n+k}|_Y) \leq \dim H^1(L^{\otimes k})$ for all $n\gg 0$ and $k$. The last term of this exact sequence actually implies that $\dim H^1(L^{\otimes n+k})\leq \dim H^1(L^{\otimes k})$, for all $k$, and all $n\gg 0$, hence $H^1(L^{\otimes n+k})$ is bounded by a universal constant. This implies that $\dim H^0(Y, L^{\otimes n+k} |_Y)$ is also bounded, whenever $r=0$, which is impossible, because $L|_Y$ is ample.

The proof is actually pretty easy. Suppose that $H^0(L^{\otimes n})|_ Y$ is non-zero for all $Y$ and $n$ sufficiently big, and vanishes somewhere on $Y$. Then it follows that the base set of $L$ is trivial: indeed, $L$ has a non-zero section on any complex subvariety, which includes the base set. This implies that the natural map $P_n:\; X\rightarrow {\mathbb P}(H^0(X, L^{\otimes n}))$ is holomorphic, for $n$ sufficiently big. Also from your assumption it follows that $P_n$ does not map any irreducible, positive-dimensional subvariety to a point (again, for $n$ sufficiently big). This implies that $P_n$ is a finite, proper map to a projective variety, hence $X$ is a ramified covering of a projective variety. A ramified covering of a projective variety is projective, which can be seen from vanishing of cohomology of powers of $L$ (a finite map is acyclic on coherent sheaves, hence the cohomology of $L^{\otimes n}$ on $X$ are the same as cohomology of ${\cal O}(1)$ on its image).

It remains only to show that $H^0(L^{\otimes n} |_Y)\neq 0$ implies $H^0(L^{\otimes kn})|_Y\neq 0$, for some $k>0$ depending on the codimension of $Y$.

If $Y$ does not lie in the zero divisor of $L^{\otimes n}$, we are done, because $H^0(L^{\otimes n})|_Y\neq 0$. Otherwise we use induction in codimension of $Y$, reducing the statement to the situation when $Y$ is a component of the zero divisor of a section of $L^{\otimes n}$. Then we may replace $Y$ with the zero divisor of a section of $L^{\otimes n}$, and obtain an exact sequence $$ 0\rightarrow H^0(L^{\otimes n}) \rightarrow H^0(L^{\otimes 2n}) \stackrel r \rightarrow H^0(Y, L^{\otimes n} |_Y)\rightarrow 0 $$ Since the rightmost arrow $r$ is just the restriction map, this implies that $H^0(L^{\otimes 2n})|_Y$ is non-zero.

The proof is actually pretty easy. Suppose that $H^0(L^{\otimes n})|_ Y$ is non-zero for all $Y$ and $n$ sufficiently big, and vanishes somewhere on $Y$. Then it follows that the base set of $L$ is trivial: indeed, $L$ has a non-zero section on any complex subvariety, which includes the base set. This implies that the natural map $P_n:\; X\rightarrow {\mathbb P}(H^0(X, L^{\otimes n}))$ is holomorphic, for $n$ sufficiently big. Also from your assumption it follows that $P_n$ does not map any irreducible, positive-dimensional subvariety to a point (again, for $n$ sufficiently big). This implies that $P_n$ is a finite, proper map to a projective variety, hence $X$ is a ramified covering of a projective variety. A ramified covering of a projective variety is projective, which can be seen from vanishing of cohomology of powers of $L$ (a finite map is acyclic on coherent sheaves, hence the cohomology of $L^{\otimes n}$ on $X$ are the same as cohomology of ${\cal O}(1)$ on its image).

However, this works only when $H^0(L^{\otimes n} |_Y)\neq 0$; the implication $H^0(L^{\otimes n} |_Y)\neq 0$ $\Rightarrow$ $H^0(L^{\otimes n})|_Y\neq 0$ is not that easy.