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As Varieties:

  1. Isn't $\Bbb A^1-\{0\}\cong V(xy-1)$ for $xy-1\in \Bbb{C}[x,y]$?
  2. But $\Bbb{A}^1-\{0\}$ isn't closed surely?

As (affine) Schemes:

  1. A generic point is any point of the spectrum, where this point isn't topologically closed?
  2. $(xy-1)\in \text{Spec}(\Bbb{C}[x,y])$, is generic. Any point $(x-a,y-b)$ such that $f(a,b)=0$, is in $V((xy-1))$ (and those points are all closed)?
  3. The 'variety' $V(xy-1)$(this $V$ is the variety version) is homeomorphic to the closed points of some scheme. So whichever scheme corresponds to this variety has to have precisely these closed points. These closed points are all in $V((xy-1))$ (scheme version), so we want to throw away everything else I guess.
  4. We want to consider $\text{Spec}(K[x,y])$ but throw away $D((xy-1))$? What scheme is this?
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Varieties 1: Yes, there is an isomorphism between $k[x]_x$ and $k[x,y]/(xy-1)$ which gives an isomorphism of varieties.

Varieties 2: It depends where you embed things - talking about whether a topological space is closed or not is meaningless unless you give an embedding of that space into another topological space. $V(xy-1)\subset \Bbb A^2$ is closed and not open, but $\Bbb A^1\setminus \{0\}\subset \Bbb A^1$ is open and not closed.

Schemes 1: No, a generic point $x\in X$ should have $\overline{\{x\}}=X$. Nonclosed points are all generic points of some closed subscheme of $X$, though - just take the closure of that nonclosed point.

Schemes 2: Per the above paragraph, the point $(xy-1)\in \operatorname{Spec} \Bbb C[x,y]$ is not a generic point, but it is the generic point of $V((xy-1))$, and any maximal ideal $(x-a,y-b)$ belongs to $\overline{(xy-1)}$ if and only if $ab=1$.

Scheme 3: I don't agree with your definition of variety, but if you want to do things this way, yes, sure, take the closed points of $V((xy-1))\subset \operatorname{Spec} \Bbb C[x,y]$.

Scheme 4: It's $V((xy-1))$. $D(I)$ is the open set of things not in $V(I)$.

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