I'm trying to catch up on Algebraic Geometry for a potential Masters project. I'm just now trying to get my head around determining the projective closure, and wanted to check my reasoning.

So in the exercise underway at the moment I've been given a set of zeroes $Z(yt-x^2, xy-zt, y^2-xz)$ in $\mathbb{P}^3$ - the question being to show that this is, in fact, the Zariski closure of the zeroes $U$ of the affine twisted cubic in $\mathbb{A}^3$.

I can show that if you identify $\mathbb{A}^3$ with the affine chart for $\mathbb{P}^3$ at $t \neq 0$ then you get the right zeroes, so $U \subset Z$. I'm trying to follow the argument for why $Z$ is $\bar{U}$.

So where I've got to is that it's to do with the points in $Z$ that aren't necessarily in $U$ - which will be any 'points at infinity' that aren't in the affine chart (i.e. not $(0:0:0:1)$) but happen to be in $Z$, right? Of which there is only one - $(0:0:1:0)$ (all the others don't give zeroes).

Here I reach a logical block that I can't think around. So the projective closure of $U$ is $U \cup \{\infty\}$ where ${\infty}$ is that particular point. Do i just rely on the fact that a single point is closed in $\mathbb{P}^3$ and $U$ is closed in $\mathbb{A}^3$ and so as a union of closed sets $Z$ is closed? That feels wrong - like I shouldn't be invoking closures in two different topological spaces. I have some notes that point in this direction but they're a bit terse and ambiguous.


Z is closed in the Zariski topology $\mathbb{P}^3$ because it's the vanishing set of a homogeneous polynomial. Alternatively, your argument works because If $T$ is a topological space, and $U_\alpha$ is an open cover of $T$. Then for some set $C$, $C$ is closed iff $C\cap U_\alpha$ is closed for all $U_\alpha$. So in one chart, you get the twisted cubic, which is closed in that chart. In your second chart, you get a point, which is closed in that chart. In the third chart, you get $\emptyset$.


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