# In algebraic geometry, why do we use $\mathbb C$ instead of the algebraic closure of $\mathbb Q$?

In algebraic geometry, why do we use $\mathbb C$ instead of the algebraic closure of $\mathbb Q$? What properties of algebraic varieties use the topological completeness of our field? I'd be interested in hearing either general perspectives or specific results that might fail for $\bar{\mathbb Q}$.

• It might be the same reason we use $\mathbb R$ than $\bar{\mathbb Q}\cap\mathbb R$. Excluding transcendental numbers from the field might not be worth. Dec 5 '13 at 19:58
• I imagine at least part of the reason is to draw connections with complex manifolds and Kahler geometry. Dec 5 '13 at 20:20
• Just to amplify Jesse Madnick's comment: Hodge theory.
– user64687
Dec 6 '13 at 9:59
• I'm note sure that in general ${\mathbb C}$ is used. I've seen many texts starting with "In this paper, $k$ denotes an algebraically closed field of characteristic 0." Dec 6 '13 at 18:56

In contrast to the above answer (which is a good general guideline), the fact that $\bar{\mathbb Q}$ is countable can sometimes cause trouble. For example, the Noether-Lefschetz theorem says that for a very general hypersurface $X$ of degree $d \geq 4$ in $\mathbb P^3$, the Picard group is generated by $\mathcal O_X(1)$. Here "very general" means that this works for all hypersurfaces off of some countable union of proper subvarieties of the parameter space of degree $d$ hypersurfaces.
Over $\mathbb C$ (or any uncountable field), this is great -- a countable union has measure $0$, so certainly there exists a hypersurface $X$ with the claimed property. Over $\bar{\mathbb Q}$, who knows? A priori our countable union might exclude every single point of the parameter space, so we can't conclude that there exists even a single such surface.
• This is an excellent point. Similarly, it is highly nontrivial to find a $K3$ surface over $\mathbf Q$ with (geometric) Picard number 1.
One of the reasons is that for almost all practical purposes, there is no difference. According to the Lefschetz Principle (which is a theorem), all algebraically closed fields of characteristic $0$ are born equal. But $\mathbf C$ is more equal than others, because it has a very rich topology which is quite useful.