已知簇有理参数化必不可约，问询拟仿射/射影簇的相关结论

Awesome question! Let’s break this down clearly, starting with quasi-affine varieties, then moving to quasi-projective ones, and wrapping up with general takeaways.

First, recall that a quasi-affine variety is a non-empty open subset of an affine variety (in the Zariski topology). If such a variety $X$ admits a rational parameterization (meaning there’s a dominant rational map from some affine space $\mathbb{A}^n$ to $X$, where "dominant" means the map’s image is dense in $X$), here’s what we can conclude:

• $X$ must be irreducible. The reasoning is straightforward:
  1. The source space $\mathbb{A}^n$ is irreducible.
  2. The Zariski closure of the image of a dominant rational map from an irreducible variety is itself irreducible.
  3. Since $X$ is an open subset of its own Zariski closure (which is an affine variety), and non-empty open subsets of irreducible varieties are always irreducible, $X$ inherits irreducibility.
• Your example of the $n$-dimensional sphere minus a point fits perfectly: the full sphere $S^n$ is an irreducible real algebraic variety, removing a single point (a Zariski-closed set) leaves a quasi-affine variety that’s still irreducible, and stereographic projection gives a dominant rational map (in fact, a birational equivalence) to $\mathbb{R}^n$.

Quasi-projective varieties are non-empty open subsets of projective varieties, and the logic here mirrors the quasi-affine case:

• Any rationally parameterizable quasi-projective variety is irreducible. The key points are:
  1. Rational parameterizations typically come from dominant maps from $\mathbb{P}^n$ (projective space) or $\mathbb{A}^n$, both of which are irreducible.
  2. The closure of the map’s image is irreducible, and $X$ is an open subset of this closure (or a larger irreducible projective variety), so $X$ itself can’t be split into two disjoint non-empty closed subsets.
• A simple example: take $\mathbb{P}^2$ (the projective plane) minus a single line. This is a quasi-projective variety, and it’s directly parameterizable by $\mathbb{A}^2$ (via the standard affine open embedding), which confirms its irreducibility.

To distill this into general rules:

1. All rationally parameterizable quasi-affine or quasi-projective varieties are irreducible. This is a direct consequence of three facts: the source spaces for rational parameterizations are irreducible, dominant rational maps preserve irreducibility of the closure, and open subsets of irreducible varieties stay irreducible.
2. The reverse doesn’t hold: not every irreducible quasi-affine/quasi-projective variety can be rationally parameterized. For example, higher-dimensional abelian varieties are irreducible quasi-projective varieties, but they are not rational (no dominant rational map from $\mathbb{A}^n$ exists), so they can’t be parameterized this way.
3. Note: We’re assuming the standard definition of "rational parameterization" as a dominant rational map (image is dense). If you only have a non-dominant rational map to $X$, the irreducibility conclusion doesn’t apply—but this isn’t usually what’s meant by a "parameterization" in algebraic geometry.

内容的提问来源于stack exchange，提问作者Javi