Grauert's criteria for ample line bundles In their book "Compact complex surfaces", W.P. Barth, K. Hulek, C.A.M. Peters and A. Van de Ven refer to the following theorem:
Let $X$ be a compact complex space and $L$ a holomorphic line bundle on $X$. Then $L$ is ample if and only if the folowing holds: given any irreducible analytic subset $Y$ of strictly positive dimension on $X$, there exist an $n=n(Y)$ such that $L^{\otimes n} |_Y$ has a section which has at least one zero, but does not vanish identically.
They don't give a proof in their book.  Instead they refer to Grauert's original paper Über Modifikationen und exzeptionelle analytische Mengen which was published in Math. Ann. 1962.
I do not know German. So my question is: Can I find the proof of this theorem somewhere else? Or instead, some comments on the idea of proof will also be very helpful.
Finally, this paper of Grauert is among one of the papers I want to read with greatest enthusiasm. Is there a translation? Or can I find some books or papers which give an explanation of the results of this paper?
 A: You can find a proof in Kleiman's famous paper Toward a Numerical Theory of Ampleness, Theorem 1 page 317. 
A: 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.
