Continuous maps for the Zariski topology

I am trying the show the following:

(10) Show that polynomials $F\in k[x_1,\dotsc,x_n]$ are continuous maps $f\colon\mathbb{A}^n\to\mathbb{A}^1$ (for the Zariski topology on $\mathbb{A}^n$ and on $\mathbb{A}^1$).

$\textbf{My Attempt:}$

In this case, to show that this map is continuous we need to show that the preimage of a closed set is closed. We know $f$ is a function that maps the zeroes of a certain affine set to the zeroes of another affine set. I'm not sure how to proceed further. Would there be a simple topological argument hidden in here?

• What are the closed subsets of $A^1$? What are their preimages under $f$? – Mariano Suárez-Álvarez Mar 21 '18 at 0:14
• @MarianoSuárez-Álvarez the closed subsets of $A^1$ are a finite number of points, and their pre-images are the surfaces corresponding to the zero sets of $A^n$? – user119264 Mar 21 '18 at 0:17
• Is the preimage of point of A1 closed in An? – Mariano Suárez-Álvarez Mar 21 '18 at 0:40

let $Z$ be a closed set. Then $f(v) \in Z$ if and only if $f(z)$ belongs to the zero locus of some collection of polynomials $h_1,\dots h_n$. But then $0=h_1\circ f=\dots =h_n\circ f$. But these are all polynomials, so $f^{-1}(Z)$ must be closed.
I've written it this way, since it generalizes, but really the point is that we can argue for a single point, since if the preimage $f^{-1}(a)$ is closed, then the preimage of a finite set of points is a finite union of closed sets, which is closed.
$f(v)=a$ if and only if $f(v)-a=0$, which is a specialization of $h_1 \circ f$ when $h_1(x)=x-a$.
• can you explain why you used the composition of $h_n$ with $f$? – user119264 Mar 21 '18 at 0:47
• hmmm... well the way I see it, you belong to a closed set $Z$ if and only if you are in the zero locus of some polynomials. These are $h_i$. – Andres Mejia Mar 21 '18 at 0:49