0
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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 $k[x,x^{-1}]$. $\endgroup$ – reuns Aug 12 at 21:43
4
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.