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I was wondering what would be an example of two affine varieties $A, B$ over the field of complex numbers $\mathbb{C}$ such that $A\setminus B$ is not an affine variety? I was initially thinking about $A = V(x^2 + y^2 - 1)$ and $B = V((x - 1)(y - 1))$, so that $A\setminus B = A\cap B^c = \{(z, w) \in \mathbb{C}^2\mid z^2 + w^2 = 1\land (z - 1)(w - 1) \neq 0\}$ with the reasoning that $(z, w) = (1, 0)$ and $(z, w) = (0, 1)$ would still be vanishing points for the circle. But then I realized that we are probably adding some polynomial to $x^2 + y^2 - 1$ such that it does not vanish at $(1, 0), (0, 1)$.

Epsilon Away
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    What ways of showing an open set of an affine variety is not affine do you know? The classic example in this situation is $\mathbb{C}^2\setminus{(0,0)}$. – Sergey Guminov Apr 18 '22 at 17:15
  • @SergeyGuminov What ways? None, I'm a complete beginner to this subject. So is $\mathbb{C}^2$ a variety? – Epsilon Away Apr 18 '22 at 17:41
  • $\mathbb{C}^2$ is a variety, although I probably should have written $\mathbb{A}_\mathbb{C}^2$ instead. It is just the complex plane with the Zariski topology. One of the most important facts is that an affine variety is uniquely identified by its ring of functions. Once you know that, you can show that removing a point from the plane does not change the ring of functions. This has been discussed in multiple questions on this site. – Sergey Guminov Apr 18 '22 at 17:51

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