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I understand what it means (in classical terms) for a projective or affine variety to be "defined over a field $k$." A book that I am reading then talks about a quasi-projective variety being "defined over $k$," but doesn't give a definition of this term for quasi-projective varieties.

Since a quasi-projective variety is an open subset of a projective algebraic set, I can think of two reasonable definitions: write $V = C \cap X^c$, where $C$ and $X$ are projective algebraic sets. Then we could either require that both $C$ and $X$ are defined over $k$, or that only $C$ is defined over $k$.

It's not even obvious to me that either of those definitions is independent of the choice of $C$ and $X$, but these are my best guesses. What is the right definition and why?

Edit: What I mean by "classical terms" is that a projective variety is a subset of projective space over an algebraically closed field $\overline{k}$ defined by the vanishing of a collection of homogeneous polynomials. "Projective space" means the set of $n+1$-tuples of elements of $\overline{k}$ modulo linear equivalence, excepting the point whose coordinates are all zero. A projective variety $V$ is "defined over $k$" for some $k \subset \overline{k}$ if the ideal generated by the homogeneous polynomials that vanish at every point of $V$ has a set of generators whose coefficients all lie in $k$.

stillconfused
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    In the modern definition it would be an open subscheme $C \setminus X$ of a geometrically integral and closed subscheme $C$ of projective space over $k$, so I would say $X$ is supposed to be defined over $k$. But perhaps you can prove that it does not depend on the choice of $X$ or $C$. Anyway, you should give the precise definition of "classical" variety you are using – it is surprisingly subtle! – Zhen Lin Apr 05 '25 at 00:26
  • Thanks! I've added an edit, hopefully it clarifies the setting I am working in. – stillconfused Apr 06 '25 at 22:54

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I would say a quasiprojective variety defined over $k$ is the complement of an algebraic subset defined over $k$ of a projective variety defined over $k$.

First, observe that there can be pathologies if we do not require the complement to be defined over $k$: for example, $\mathbb{P}^1$ is defined over $k$, and $\mathbb{P}^1 \setminus \{ (x_0 : x_1) \}$ is isomorphic (over $k$) to an affine variety defined over $k$ if $x_0$ and $x_1$ are in $k$, but this fails if $x_0 / x_1$ and/or $x_1 / x_0$ is not in $k$.

Second, given algebraic sets $C$ and $X$ defined over $k$, there is a smallest algebraic set $C'$ defined over $k$ and containing $V = C \setminus X$, and $X' = C \cap X$ is an algebraic set defined over $k$, so we also have $V = C' \setminus X'$. Thus, provided $V$ is given as a set of points in projective space, there is a canonical algebraic set $C$ defined over $k$ containing $V$ that can be used to test whether $V$ is a quasiprojective variety defined over $k$.

Finally, I should note that the above also applies when the definition of "variety" includes irreducibility (which as far as I know is the more conventional choice), because $V$ is irreducible if and only if $C'$ is.

Zhen Lin
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