The projective line is defined over $\mathbb Q$

Question

Notations:

A variety $X$ over a field $C$ is an integral $C$-scheme such that the structure morphism $p: X\longrightarrow \textrm{Spec } C$ is separated and of finite type.

We say that a variety $X$ is defined over a subfield $K$ of $C$ if does exist a variety $X_K$ over $K$ such that $X\cong X_K\times_{\textrm{Spec }K}\textrm{Spec }C$.

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The my question:

From now we suppose that $C$ is an algebraically closed field of characteristic $0$; so $\mathbb Q$ is the prime field of $C$. In many articles/books I read that

the projective line $\mathbb P^1_C$ is "obviously" defined over $\mathbb Q$

but I don't understand why this is true. I need a down-to-earth explaination of this statement.

Thanks in advance.

Answer 1

The projective line over $\mathbb Q$ is the variety $\mathbb P^1_\mathbb Q=\text {Proj}( \mathbb Q[T_1,T_2])$.
Since $$\mathbb P^1_C=\mathbb P^1_\mathbb Q\times _{\text {Spec} \mathbb Q} \text {Spec}\: C(=\text {Proj}( \mathbb Q[T_1,T_2]\otimes_\mathbb Q C)=\text {Proj}( \mathbb C[T_1,T_2]))$$ we see that according to your definition $\mathbb P^1_C$ is indeed defined over $\mathbb Q$.
Since being separated and being of finite type are properties stable after base change, and since $\mathbb P^1_\mathbb Q$ has these properties, so has $\mathbb P^1_C$ .
Notice that algebraic closedness of $C$ is completely irrelevant to this problem.

Answer 2

The down-to-earth intuition behind "defined over $F$" is that if you spell out the polynomials that you use to describe an algeraic object (a variety, an algebraic group, ...) can be chosen so that all coefficients are in $F$. It takes a bit of thought though to recognize this key fact behind the abstarct definition.