Zariski topology over $\mathbb R$

Question

What is a "Zariski topology on $\mathbb R$"? I don't think I quite understand the definition of a "Zariski topology". Thank you.

Can you explain what exactly you don't understand in the definition of the Zariski topology in the case of $\mathbb R$? — Asaf Karagila, Oct 10 '11 at 08:33

score 5 · Answer 1 · answered Oct 10 '11 at 08:32

5

Taking $\mathbb{R}$ as the affine line over reals, the Zariski topology of $\mathbb{R}$ consits of all its subsets whose complement is either finite or all of $\mathbb{R}$ (i.e. the finite complement topology.)

answered Oct 10 '11 at 08:32

Dinesh

1,737

2

Indeed, as follows from the simple observation that the set of zeroes of a polynomial is a finite set, and all finite sets can occur in this way. – Henno Brandsma Oct 10 '11 at 19:09
@Henno Brandsma yes – Dinesh Oct 11 '11 at 03:01

score 0 · Answer 2 · answered Jun 11 '23 at 16:52

If $(\Bbb{R},\tau)$ is the reals with the Zariski topology then we say $U \in \tau$ if and only if $\Bbb{R} \setminus U = \{x_1,…,x_k\}$ I.E., the complement with respect to the whole space, of the open sets, is finite. It is worth noting this space is compact but not Hausdorff. However Hausdorff in the finite case as singletons are open. This is also known as the cofinite topology.

Zariski topology over $\mathbb R$

2 Answers2