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Updated: The early problem was not correct. Thanks to Georges Elencwajg.

Does the following statement hold?

Let $N$ be a compact connected $n$-dimensional complex manifold. Let $X$ be the total space of the holomorphic tangent bundle of $N$ and let $R$ be the ring of holomorphic functions on $X.$ Since $X$ is connected, $R$ is a domain and denote $K$ as its quotient field. If $\text{tr. deg}_{\mathbb{C}} K \geq n,$ then $N$ is algebraic.

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so? you have any ideas? – ulead86 Aug 3 '11 at 9:53
up vote 4 down vote accepted

I think this is not true.

Take for $N$ any non-algebraic $n$-dimensional compact connected complex manifold with trivial tangent bundle : a non-algebraic torus for example.
The tangent bundle $T(N)$ being isomorphic to $N\times \mathbb C^n$ , lifting holomorphic functions from $\mathbb C^n$ to $T(N)=N\times\mathbb C^n $ via the second projection shows that $\mathcal O(\mathbb C^n) \subset \mathcal O(T(N))=R$.
Of course $\mathbb C[T_1,....,T_n] \subset O(\mathbb C^n)$ and consequently $\mathbb C(T_1,....,T_n) \subset Frac(O(\mathbb C^n))\subset Frac(O(T(N)))=K$, which proves that $trdeg_{\mathbb C} K \geq n$, even though $N$ is not algebraic.

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Very nice! By the way, is it even true that a compact complex manifold with enough meromorphic functions (i.e. transcendence degree equal to the dimension) is algebraic? I am aware that these come from algebraic spaces, but not necessarily from schemes. – Akhil Mathew Aug 4 '11 at 4:34
Dear Akhil: the compact complex manifolds with enough meromorphic functions you mention are called Moishezon manifolds . They are algebraic projective in dimension 2, by an old theorem of Chow and Kodaira . However in dimension $\geq 3$ they may be non-algebraic. But in that case you can blow them up to projective algebraic manifolds: this is the main result in Moishezon's paper where he introduced these manifolds (under another name!) – Georges Elencwajg Aug 4 '11 at 9:01
Interesting. Thanks for the information. – Akhil Mathew Aug 4 '11 at 13:10

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