0

Is complement of an algebraic set also algebraic in $\mathbb{A}^n(k)$, where $k$ is arbitrary field?

My Solution: Let $V$ be an algebraic set i.e. $V = V(F)$ for some $F =\{f_{\alpha}: \alpha \in \Lambda \}$. Assume that $U := \mathbb{A}^n(k) -V$ is also an algebraic set. Then, $U$ and $V$ are both Zariski open and closed set in $\mathbb{A}^n(k)$. Thus, $\mathbb{A}^n(k)$ will be disconnected.

Doubt: I am not sure how to conclude after this. Is it true that $\mathbb{A}^n(k)$ is connected in Zariski topology? If yes, why?

LearningMath
  • 570
  • 1
    If by $\Bbb A^n(k)$ you mean $k^n$ with the Zariski topology, then there's a bifurcation depending on whether $k$ is infinite or not. If $k$ is finite, the space is discrete; if not, it is connected (and even irreducible, see here). Is that what you mean by $\Bbb A^n(k)$? – KReiser Mar 17 '21 at 00:28

2 Answers2

1

This is false. For example, consider the case $n=1$. If $X\subsetneq\mathbb{A^1}(k)$ is algebraic then $X=V(I)$ for some ideal $0\ne I\subseteq k[x]$. Since $k[x]$ is a Noetherian ring we know $I$ is finitely generated, and so $X=V(f_1,...,f_r)$ for some nonzero polynomials $f_1,...,f_r\in k[x]$. Every polynomial in one variable has only finitely many roots, and so $X$ is finite.

Conversely, if $X\subseteq\mathbb{A^1}(k)$ is a one point set $X=\{x_0\}$ then obviously $X=V(x-x_0)$, and so $X$ is algebraic. It follows that every finite set is algebraic, because a finite union of closed sets is closed.

So if $n=1$ then the Zariski topology is simply the cofinite topology on $k$. If $k$ is an infinite field then the complement of a finite set is not finite, and so not algebraic.

Mark
  • 39,605
  • Can we prove it for general case? – LearningMath Mar 17 '21 at 00:42
  • Yes. The general case is written in the linked question. It is a simple generalization of what I did to a general $n$. If the field is infinite then the Zariski space is irreducible. – Mark Mar 17 '21 at 00:59
1

Adding on @KReiser comment, it is sufficient to prove that if $\mathbb{A}^n(k)$ is irreducible, then $\mathbb{A}^n(k)$ is connected. Use the result that $V$ is irreducible $\iff I(V)$ is prime to get the result.

MUH
  • 1,377