Say I have the affine variety $X=\{y^2=\sqrt2x\}\subset\mathbb{C}^{2}$. Clearly $X$ is "definable" over the field $K=\mathbb{Q}(\sqrt2)$, and not over $\mathbb{Q}$. My question is, how do I define this in algebraic geometry form?
At first glance it seems something like "there exists a morphism $X\to spec(K) $ which is locally of finite type", but really this definiton doesn't seem that good, or even very comprehensible.
My question is:
Given a general variety (scheme) $X$, what does it mean in algebraic geometric terms that $X$ is "definable" over $K$?
Given a scheme $X\to \operatorname{Spec}L$ over a field $L$, it is definable over $K\subset L$ if there is another scheme $Y\to \operatorname{Spec}K$ such that the base change $L\times_K Y$ is isomorphic to $X$.
For example, say $X$ is affine, of the form $$X=\operatorname{Spec} \frac{\mathbb C[x_1,\ldots ,x_n]}{(f_1,\ldots ,f_m)}.$$ If all the coefficients of the $f_i$'s belong to some smaller field $K$, then $X$ is defined over $K$, as we can take $$Y=\operatorname{Spec} \frac{ K[x_1,\ldots ,x_n]}{(f_1,\ldots ,f_m)}.$$ Notice that $Y$ has no reason to be unique. For example, the circle $\{x^2+y^2 = 1\}\subset \mathbb C^2$ is the base change of both $\operatorname{Spec} \frac{ \mathbb Q[x, y]}{(x^2+y^2-1)}$ and $\operatorname{Spec} \frac{ \mathbb Q[x, y]}{(x^2+y^2+1)}$, which are not isomorphic, since the former has $\mathbb Q$ points and the latter doesn't.