# Why is $k-\{0\}$ not a closed in $k$ in the zariski topology?($k$ algebraically closed)

Why is $$k-\{0\}$$ not a closed in $$k$$ in the zariski topology?($$k$$ algebraically closed)

I am trying to prove that if $$h:X\to Y$$ is a morphism of varieties, then $$h(X)$$ is not necessarily a subvariety of $$Y$$, for this I am taking $$X=V(xy-1)$$ and $$Y=k$$ ($$k$$ algebraically closed) to thus have a morphism $$h:V(xy-1)\to k$$, given by $$h(x,1/x)=x$$. In this case $$h(X)$$ would be $$k-\{0\}$$, but I don't know how to prove that this is not a subvariety of $$k$$, could someone please help me? Thank you!

• So the affine variety corresponding to $k-\{0\}$ is $V(xy-1) \subset \Bbb{A}^2$, you need to add variables/dimensions to remove closed subsets of an irreducible algebraic set. Also to each subset $U$ you should think to the corresponding ring of rational functions regular on $U$, here $k[x,x^{-1}]$. – reuns Aug 12 at 21:43

There is no polynomial over $$k$$ which has roots at all nonzero points but not at $$0$$ (as $$k$$ is infinite).
The only polynomial in one variable that has infinitely many roots is the zero polynomial, but $$V(0) = k$$ (algebraically closed fields cannot be finite).