Why is the graph of polynomial closed in Zariski-Topology?

Question

The official wikipedia entry to Zariski-Topology states that:

"In the Zariski topology on the affine plane, this graph of a polynomial is closed."

I guess the argumentation would be the following:

For $f(x) \in \mathbb{R}[x]$ we have: $$Graphf = [(x,y) \in \mathbb{R}^2 \vert f(x)=y ] = [(x,y) \in \mathbb{R}^2 \vert f(x)-y=0 ] = [(x,y) \in \mathbb{R}^2 \vert \forall g(x,y) \in [f(x)-y]: f(x)=y] = V([f(x)-y]) \subset \mathbb{R}^2$$

But the graph of a polynomial are uncountably many points and closed sets in Zariski are only ever finite as they need to be roots of a polynomial. How is it possible to close that gap?

(Btw. how can I write { } on MSE? { and \textbraceleft don't work)

Answer 1

Zariski closed sets are not always finite !

As you said they are the roots of polynomials. For example the circle $S^1$ is the roots of the polynomial $$ X^2 + Y^2-1$$

Non constant polynomials in a single variable only have a finite set of roots but when working with more variables this is not always the case.

If you have a polynomial $f\in \mathbb R[x]$ then its graph is the set of points in $\mathbb R^2$ which verify $$ f(X) = Y$$ but this is exactly the set of roots of the $f(X)-Y$ (which is a polynomial in two variables).

Answer 2

The closed sets in the Zarisky topology on $\Bbb R$ are exactly the finite sets and the whole set. Which is to say, all zero sets of polynomials in $\Bbb R[x]$.

The Zarisky topology on $\Bbb R^2$ is very different. The closed sets are still all the zero sets of polynomials in $\Bbb R[x,y]$ (and their intersections, although in the special case of $\Bbb R$, that's unnecessary to specify). This includes every the graph of every polynomial in $\Bbb R[x]$, and a whole lot more.