4
$\begingroup$

If we have an irreducible subvariety of complex affine space $A_{\mathbb{C}}$, is the image under a regular (i.e., given by polynomials) map also irreducible? is it an irreducible subvariety of the target space?

$\endgroup$
2
  • 1
    $\begingroup$ Hint: This is true for any continuous map. It's similar as to the proof with continuity. $\endgroup$ Mar 29 '14 at 11:12
  • $\begingroup$ It is irreducible, but not necessarily a subvariety. It is thought "constructible" (finite union of locally closed subsets). $\endgroup$
    – Cantlog
    Mar 29 '14 at 20:49
4
$\begingroup$

Just to have this answered:

Yes, the image of an irreducible subvariety of $\mathbb{A}^n$ is irreducible. In fact, the image of any irreducible space under a continuous map is irreducible. This is just an exercise in point-set theory. It is proved in a similar way to the fact that the image of a connected set is connected.

The image of a variety need not be a subvariety though (by this I assume you mean a closed algebraic subset), as Cantlog points out in the comments. The most common example is the morphism $\mathbb{A}^2\to\mathbb{A}^2$ given by $(x,y)\mapsto (x,xy)$. The image of $\mathbb{A}^2$ is not closed. It is "constructible" though, as Cantlog points out. It turns out that the image of any variety under a morphism is constructible (this falls under the more general Chevalley's Theorem).

$\endgroup$
2
  • 1
    $\begingroup$ The image of $\mathbb A^2$ in this example is not open (not even locally closed) either. $\endgroup$
    – Cantlog
    Mar 30 '14 at 13:43
  • $\begingroup$ @Cantlog The perfect counterexample :) $\endgroup$ Mar 30 '14 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.