Closure of image by polynomial of irreducible algebraic variety is also irreducible algebraic variety How to show that closure of image by polynomial of irreducible algebraic variety is also irreducible algebraic variety.
$A$ -irreducible algebraic variety, $F$ - polynomial map ($F:A \to B$), proof that $\bar B$ - is irreducible algebraic variety.
Closure of some set is always algebraic variety as definition of closure. But how to show, that is irreducible ???
 A: There are two questions :


*

*The image of an irreducible subset is irreducible 

*The closure of an irreducible subset is irreducible


Recall that a set $A \subset X$ irreducible if for any closed (for induced topology) subset $F_1, F_2 \subset A$, $A = F_1 \cup F_2 \Rightarrow A = F_1$ or $A = F_2$. 


*

*First question : let $f : X \to Y$ continuous and $A \subset X$ irreducible. If $B := f(A)$ is not irreducible, there is two closed subset $F_1, F_2$ with $F_1 \neq B, F_2 \neq B$. Now the equality $A = f^{-1}(F_1) \cup f^{-1}F_2$ contradicts that $A$ was irreducible, which proves the first claim. 

*Second question : let $A \subset X$ be an irreducible subset. Let $\overline A$ be the
closure of $A$, we will show that $\overline A$ is irreducible.
Again, assume that $F_1 \cup F_2 = \overline A$ are two proper closed
subsets. Then, $F_1' := F_1 \cap A$ and $F_2' := F_2 \cap A$ are by
definition proper closed subset of $A$, this shows that $A$ was not
irreducible which is again a contradiction.
Finally, notice that the image of an algebraic variety by a polynomial map does not need to be a algebraic variety. The simplest example is the map $\Bbb C^2 \to \Bbb C^2, (z,w) \mapsto (z, zw)$.
