TITLE: How to show that a polynomial maps an algebraic set to an algebraic set?
QUESTION [5 upvotes]: Sorry if this is an ignorant question. I am studying algebraic geometry. This isn't an exercise problem. It is an assumption I can use to prove something else. I think it must be obvious, but I don't know how to prove it. 
My attempt:
Let $V=(\{F_i\}_{i\in I})$ be an algebraic set. Let $f$ be a map. Then I couldn't get to anywhere. 
I tried an example: $V=\{(x-1)(x-2), (x-1)(x-2)(x-3)\}=\{1,2\}$. Let $f(x)=x^2$. Then $f(V)=\{1,4\}$. They are zero loci of $\{(x-1)(x-4), (x-1)(x-4)(x-3)\}$. But that would change the $F_i$'s randomly. And for higher dimensions, you don't have a finite set to deal with.
Please help.

REPLY [4 votes]: Let $\phi\colon k^n \to k^m$ be given by polynomials $\phi_1,\ldots,\phi_m\in k[x_1,\ldots,x_n]$. You want to show that for any algebraic set $V\subseteq k^n$, the set $\phi(V)$ is again algebraic?
If so, then I have to disappoint you. This only holds under additional assumptions. A counterexample would be the hyperbola $V=Z(1-xy)\subseteq k^2$ (assume $k$ algebraically closed to avoid problems) and the projection map $\phi:k^2\to k$ given by $(x,y)\mapsto x$. You can see that the image of $V$ is $\phi(V)=k\setminus\{0\}$, which is not an algebraic set because it is not finite.