Is $\mathbb{A}^3 - \{0\}$ a separated variety?

Question

To check if $X=\mathbb{A}^n -\{0\}$ is separated is equivalent to check that the diagonal is closed. In this case I think that: $X \times X= \{(x,y)| x,y \in X \}$ then $\Delta(X)=Z(y-x)- \{0\}\subset \mathbb{A}^n $ but i'm not sure if this is closed or not. I think that it is open can you confirm that?

Answer 1

There are several errors in your description of $\Delta(X)$. First, it lives inside $X\times X$, not $\Bbb A^n$. Secondly, you're using $x$ and $y$ for points of $X$ and then also as coordinates (when you subtract them), which doesn't really make sense. The second error is maybe not so bad, but the first is serious: it gives you the wrong answer. This variety is separated, as the diagonal copy of $X$ is in fact closed inside $X\times X$, exactly as the zero locus of $x_i-y_i$ where the $x_i$ and $y_i$ are coordinates on the first and second $X$, respectively.

Answer 2

The answer to the title question is yes: a locally closed subvariety of a separated variety is separated. Cf. these notes