Why is $\mathbb{A}^{1} \backslash 0$ a quasi projective variety in “An Invitation to Algebraic Geometry.”

algebraic-geometry

I like the textbook, yet whenever I first read this I couldn't think which way to interpret locally closed. It's actually a closed set contained in an open subset which is a subspace of $\mathbb{P}^{n}$ by the induced topology though. So I was just reviewing it and the problem came up again but I forgot about how exactly $\mathbb{P}^{n}$ is supposed to be a subspace with this random closed set I was thinking of in it when it is the whole space, I remembered it's the complement of the variety $V = \{ \emptyset \}$ right? The only thing is I still have to ensure taking open or closed subsets in the quasi – projective variety is closed.

To take a closed subset or open subset of a quasi projective variety; first I've apparently got to consider both the open subspace that contains it and the closed set, then second define a closed subset of a quasi projective variety to be a closed subset of the particular variety I was first examining or instead any closed subset of the open subspace? Likewise should an open subset of a quasi – projective variety be any open subset in the subspace or just open subsets of the subspace intersecting with the variety. It's a bit ambiguous to me because of that $\mathbb{P}^{n}$ is an open subset of the topological space $\mathbb{P}^{n}$ which includes $\emptyset$ as a closed set. I think my first intuition is still useful because I was only trying to consider the first thing I read as a variety: the vanishing set of a collection of polynomials.

4.1

In either case when I look at the example:

4.1 example

I can't figure out why $\mathbb{A}^{1} \backslash \{ 0 \}$ is a closed set. I must have mistook it as a perforated type of line at first, or now that I think about it again that could be it. There is no subset marker compared to the second variety however I am being more doubtful now. Will someone tell me why it is closed?

Best Answer

$\mathbb{A}^{1} \backslash 0$ is the closed set of the open subspace $\mathbb{A}^{1} \backslash 0 \subset \mathbb{A}^{1}$ because it is the vanishing set of the polynomial $0$ in the complementary set of the vanishing set of polynomials with no constant term.

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