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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 $\bar{\mathbb Q}$.

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  • $\begingroup$ 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. $\endgroup$ – Carlos Eugenio Thompson Pinzón Dec 5 '13 at 19:58
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    $\begingroup$ I imagine at least part of the reason is to draw connections with complex manifolds and Kahler geometry. $\endgroup$ – Jesse Madnick Dec 5 '13 at 20:20
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    $\begingroup$ Just to amplify Jesse Madnick's comment: Hodge theory. $\endgroup$ – user64687 Dec 6 '13 at 9:59
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    $\begingroup$ 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." $\endgroup$ – Magdiragdag Dec 6 '13 at 18:56
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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.

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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.

Maybe for the Noether-Lefschetz theorem the answer is known (I have no idea), but "very general" conditions crop up fairly often, and certainly there are cases where it's not known whether there's an algebraic point or not.

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    $\begingroup$ This is an excellent point. Similarly, it is highly nontrivial to find a $K3$ surface over $\mathbf Q$ with (geometric) Picard number 1. $\endgroup$ – user64687 Feb 18 '14 at 15:52

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