I guess there is only one from the rational numbers to the complex numbers, since they form the prime-field.
However I have no idea, how to approach the question of how many homomorphisms from the algebraic closure of the rational numbers to the complex numbers there are, that keep the rational numbers fixed.
If someone has an idea on how to approach this question, I'd be really curious. Thank you and have a good day!
They are infinitely many. A Zorn lemma argument shows that for any algebraic extension $L/\mathbb{Q}$, you can extend any $\mathbb{Q}$-morphism $L\to\mathbb{C}$ to a $\mathbb{Q}$-morphism $L\to\mathbb{C}$
Now for any integer $n$, pick an extension of degree $n$, say $L=\mathbb{Q}(\sqrt[n]{2})/\mathbb{Q}$. You have $n$ morphisms $L\to\mathbb{C}$, extending to morphisms $\mathbb{Q}$-morphism $\mathbb{Q}$-morphism $L\to\mathbb{C}$ , necessarily distinct, since their restrictions to $L$ are.
Hence, for all $n$, $\vert Hom_\mathbb{Q}(\overline{\mathbb{Q}},\mathbb{C})\vert\geq n$, QED.
