Prove the Jacobian of a curve of genus g is a complex torus

Question

As stated in the title I am about to prove the Jacobian of a curve of genus g is a complex torus. Here is what I have done so far:

I know the first homology group of $X$ is $H_1(X,\mathbb{Z}) \cong \mathbb{Z}^{2g}$. Let $\alpha_1, \dots, \alpha_2g$ be a basis of it. Let $x_i \in \mathbb{R}, i=1, \dots, 2g$ such that as a $\mathbb{C}$-linear functional on $\Omega^1_{hol}(X) = \{ \omega \text{ holomorphic differential one form} \}$. $$ \sum_{i=1}^{2g} x_i \int_{\alpha_i} = 0 $$ So: $$ \sum_{i=1}^{2g} x_i \int_{\alpha_i} \omega = 0 \quad \forall \omega \in \Omega^1_{hol} $$ Taking the conjugate we have: $$ \sum_{i=1}^{2g} x_i \int_{\alpha_i} \overline{\omega} = 0 \quad \forall \omega \in \Omega^{1}_{hol} $$ And so by Hodge decomposition: $H^{1}(X, \mathbb{C}) = \Omega^{1}_{hol} \oplus \overline{\Omega^{1}_{hol}}$: $$ \sum_{i=1}^{2g} x_i \int_{\alpha_i} \omega = 0 \quad \forall \omega \in H^{1}(X, \mathbb{C}) $$ By De Rham's Theorem we then have: $\sum_{i=1}^{2g} x_i \alpha_i = 0$ as a chain in $H^{1}(X, \mathbb{C})$. I would now conclude using the linearity independece of these chains. Unfortunately I got it only on $\mathbb{Z}$. What I got convinced of, but I am not sure about, is that $2g$ vector in $\mathbb{Z}^{2g}$ linearly independent over $\mathbb{Z}$ are linearly independent over $\mathbb{C}$ in $\mathbb{C}^{2g}$. If it is true then I have finished because $\alpha_i$ are vectors in $\mathbb{Z}^{2g}$. Here's why I think this: Let $\Gamma$ be a subset of $\mathbb{Z}^ {2g}$ composed of $2g$ vector linearly independent over $\mathbb{Z}$. Let $V = \{ \text{intersection of all vector subspace in } \mathbb{C}^{2g} \text{ which contains } \Gamma\}$. Then $V = Span_{\mathbb{C}}(\Gamma)$. If dimV = k < 2g then let $\phi$ be the isomorphism between $V$ and $\mathbb{C}^k$. We get $\mathbb{Z}^{2g} \cong \Gamma = \Gamma \cap V \cong \phi(\Gamma \cap V) \cong \mathbb{Z}^k$ which is an absurd, right? If there is some mistake could someone give me some hint to work out a different solution?

Answer 1

Let me rephrase your actual statement

Suppose $\{v_i\}_{i=1,...n}$ is a linearly independent set of a $\mathbb Z$-module (abelian group) $M$. Then $\{v_i \otimes 1\}$ is linearly independent in $M \otimes_\mathbb Z \mathbb C$.

This is the same as your question, as $\mathbb Z^n \otimes_\mathbb Z \mathbb C \cong \mathbb C^n$.

Any set $\{v_i\}_{i=1,...,n}$ of an $R$-module $N$ lifts to a homomorphism, say $\alpha_{\{v_i\}}$, $R^n \rightarrow M$, by $(a_1,...,a_n) \mapsto \sum_{i=1}^n a_i v_i$ (that is simply saying that $R^n$ is the free $R$-module on an $n$-element set). Linear independence of the set $\{v_i\}_{i=1,...n}$ is equivalent to injectivity of $\alpha$. Hence we're asking: Given an injective $\mathbb Z$ homomorphism $\alpha_{\{v_i\}}$, is $\alpha_{\{v_i \otimes 1\}} \cong \alpha \otimes_{\mathbb Z} \mathbb C$ injective?

One of the definitions of flatness of a module $N$ is that $ - \otimes N$ maps injective maps to injective maps. Hence the answer to our question is yes if we know that $\mathbb C$ is flat over $\mathbb Z$. But that is the case, since $\mathbb Q$ is flat over $\mathbb Z$ (as every localization is flat) and $\mathbb C$ is flat over $\mathbb Q$ (as $\mathbb Q$ is already a field).