Algebraic dimension of the projective space

Question

This is the exercise 2.1.3 from Huybrechts's "Complex Geometry - An Introduction".

Determine the algebraic dimension of the following manifolds: $\mathbb{P}^1$, $\mathbb{P}^n$, and the complex torus $\mathbb{C}/(\mathbb{Z}+i\mathbb{Z})$. For the latter, you might need to recall some basic facts on the Weierstrass $\wp$-function. How big is the function field of $\mathbb{C}$?

The algebraic dimension of $X$ is defined to be the transcendence degree of the function field $K(X)$, the field of meromorphic functions on $X$. And there is a propostion 2.1.9 (Siegel) states that the algebraic dimension of a compact connected manifold is not greater than the geometric (complex) dimension.

My attempt:

For $\mathbb{P}^1$ this is trivial: the algebraic dimension is $1$ since there are non-trivial meromorphic functions (e.g. the identity map is holomorphic) on the Riemann sphere, so the dimension is greater than $0$.

The function field of $\mathbb{C}$ should have an infinite algebraic dimension since the functions $f(z)=z^k, k=1, 2, \cdots$ are holomorphic.

Question: how about $\mathbb{P}^n$ (the projective space) and the complex torus? Note that the RR theorem are not studied yet so I'm seeking a solution without it.

Answer 1

Although $z^k$ is homolomorphic for $k = 1, 2, \ldots$, these functions are not algebraically independent. For instance $(z^j)^k - (z^k)^j = 0$, so $z^j$ and $z^k$ are algebraically dependent for any $j,k$. However, there are more exotic holomorphic (and meromorphic) functions on $\mathbb{C}$. Consider $e^z$, for example.

I think you've proven that $\mathbb{P}^1$ has dimension $\geq 1$, but why is it $=1$? (EDIT: Ah, now I see that you're probably arguing that the geometric dimension is $1$ and using the proposition you cited.) Try showing that the function field of $\mathbb{P}^1$ is just the usual field of rational functions $\mathbb{C}(z)$. This may also give you some inspiration for $\mathbb{P}^n$: try showing that its function field is isomorphic to $\mathbb{C}(z_1, \ldots, z_n)$, where $z_i = Z_i/Z_0$. (Here $Z_0, Z_1, \ldots, Z_n$ are the homogeneous coordinates on $\mathbb{P}^n$.)

Recall that the function field $\mathbb{C}(T)$ of a torus is generated by the Weierstrass $\wp$-function and its derivative. (This is proved, for instance, in Stein and Shakarchi's Complex Analysis, p. 271.) Since $\wp$ satisfies the differential equation $$ (\wp')^2 = \wp^3 + c_4 \wp + c_6 $$ $\wp$ and $\wp'$ are algebraically dependent. (This is because a complex torus is an elliptic curve; see here for more.) Thus $\mathbb{C}(T)$ has transcendence degree $1$.

Answer 2

For the projective space we can argue as follows:

We can define a rational function $f\in \Bbb{C}(\xi_1,....,\xi_n)$ on $U_0=\{z_0\ne 0\}\cong \Bbb{C}^n$, then we can extend it to a meromorphic function on $\Bbb{CP}^n$ since the transition function is rational, the composition of them is still rational. Therefore we have a map $$\Bbb{C}(\xi_1,...,\xi_n)\to \mathcal{M}(\Bbb{CP}^n)\\f\mapsto F$$ where $F$ is the extension of the rational function $f$ as we did above.

it's an injective map.

Therefore $ n = \text{trdeg} _{\Bbb{C}} \Bbb{C}(\xi_1,...,\xi_n) \le \text{trdeg} _{\Bbb{C}} \mathcal{M}(\Bbb{CP}^n)$

Finally you know the Siegel theorem, which gives bound in the other direction therefore:

$$\text{trdeg}_{\Bbb{C}}\mathcal{M}(\Bbb{CP}^n) = n$$