Is the image of a curve under the projection morphism of a fiber product again a curve?

Question

If $X$ is a variety defined over the rationals and $f:X_\mathbb{C}\to X$ is the projection map of the fibered product, then is $f(C)$ is a curve in $X$? (Where $C$ is a curve in $X_\mathbb{C}$.)

Answer 1

No. Let $X=\Bbb A^2_{\Bbb Q}$ and let $C=V(x-\pi)\subset X_{\Bbb C}$. The preimage of the ideal $(x-\pi)\subset \Bbb C[x,y]$ in $\Bbb Q[x,y]$ is the zero ideal since $\pi$ is transcendental over $\Bbb Q$. Thus $f(C)$ contains a point with residue field of transcendence degree $2$ over $\Bbb Q$, so no matter what subscheme structure you put on it, it cannot be a curve over $\Bbb Q$.

Let me point out that what you're doing is a bit strange - in general, the question "what subscheme structure do I put on $f(Z)$ for a subscheme $Z\subset X$ and a map $f:X\to Y$" doesn't really have a good answer and may not even be answerable at all.