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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?

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    $\begingroup$ Hint: This is true for any continuous map. It's similar as to the proof with continuity. $\endgroup$ Mar 29, 2014 at 11:12
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    $\begingroup$ It is irreducible, but not necessarily a subvariety. It is thought "constructible" (finite union of locally closed subsets). $\endgroup$
    – Cantlog
    Mar 29, 2014 at 20:49

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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).

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    $\begingroup$ The image of $\mathbb A^2$ in this example is not open (not even locally closed) either. $\endgroup$
    – Cantlog
    Mar 30, 2014 at 13:43
  • $\begingroup$ @Cantlog The perfect counterexample :) $\endgroup$ Mar 30, 2014 at 15:00

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