2
$\begingroup$

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.

  • 0
    so? you have any ideas?2011-08-03

1 Answers 1

4

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.

  • 0
    Interesting. Thanks for the information.2011-08-04