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

Question

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!

Answer 1

There is no polynomial over $k$ which has roots at all nonzero points but not at $0$ (as $k$ is infinite).

Answer 2

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).