basic question on varieties (algebraic geometry)

Question

I study basic algebraic geometry and I saw this exercise: V is the complement of the twisted cubic in $$ A_c^3. $$ i.e. $$ V = A_c^3 - \{(t^3, t^4, t^5) \mid t\in c\}. $$ 1. How can I proove that V is not an affine variety (and is a quasi-affine)?
2. How should I present V as a union of affine neighborhoods?

Thanks!

It depends on what you know. Do you know that any regular function on an open set $V$ of an irreducible affine variety $X$ can be written as a fraction, $f/g$ where $f,g$ are regular functions on $X$? — Mohan, Sep 08 '15 at 17:13
@GeorgesElencwajg, Of course, my mistake. I meant rational functions. Thanks. — Mohan, Sep 08 '15 at 21:05
Don't worry , dear @Mohan: we all sometimes write (or say) one thing and mean another... — Georges Elencwajg, Sep 08 '15 at 21:11

score 1 · Answer 1 · edited Apr 13 '17 at 12:21

If $V$ were affine, so so would be its intersection with the (closed!) plane $z=0$, because a closed subset of an affine variety is affine .
But that intersection is the punctured affine plane $\mathbb A^2_{x,y,0}\setminus\{(0,0)\}$, which is well known not to be affine.
The open subset $V$ is the union of the two open subsets $U_1: xz-y^2\neq0$ and $U_2:x^3-yz\neq 0$ of $\mathbb A^3$.
This is equivalent to proving that given $a,b,c\neq 0$ with $ac-b^2=a^3-bc=0$, we can write $a=t^3,b=t^4, c=t^5$ for $t=\frac ba$ .
The open subsets $U_i$ are affine varieties because the complement of a hypersurface (like $xz-y^2=0$) in $\mathbb A^3$ is affine.
Thus we have written $U=U_1\cup U_2$ as the union of two affine open subsets of $\mathbb A^3$.

basic question on varieties (algebraic geometry)

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