This is a purely affine variety. Define $F_1=y-x^2, F_2=z-xy, F_3=xz-y^2$.
It is clear that $(x,x^2,x^3)$ is a solution to $F_1=F_2=F_3=0$. Since projective twisted cubic is not intersection of two quadractics which contains an extra line, I would not expect this to be the case in affine as well. However, $F_1\cap F_2$ seems to be exactly twisted cubic curve and I could not see any extra information on the other line. The other two intersections(i.e. $F_2\cap F_3$ and $F_1\cap F_3$ do yield two extra lines.
What have I done wrong here?
Best Answer
1) It is perfectly correct that the affine curve $$C=\{(u:u^2:u^3)\vert u\in k\}\subset \mathbb A^3_k=\mathbb A^3_{x,y,z}=\mathbb A^3\:$$ is the complete intersection of the affine smooth quadrics $Q_1=V(F_1), Q_2=V(F_2)\subset \mathbb A^3$.
2) The curve $C$ has as closure in $\mathbb P^3=\mathbb P^3_{[t:x:y:z]}$ the smooth complete curve $\overline C=C\cup \{[0:0:0:1]\}$.
3) The quadrics $Q_1, Q_2$ have as closures in $\mathbb P^3$ the quadrics $\overline {Q_1}=V(\overline {F_1}), \overline {Q_2}=V(\overline {F_2})\subset \mathbb P^3$ where $\overline {F_1}=ty-x^2, \overline {F_2}=tz-xy$.
4) However it is not true that $\overline C=\overline {Q_1}\cap \overline {Q_2}$ : actually $\overline {Q_1}\cap \overline {Q_2}=\overline C\cup L$ where $L\subset \mathbb P^3$ is the line given by the equations $ t=x=0$ (thus, a line included in the plane at infinity $t=0$).
5) Worse still, the curve $\overline C$ is not the complete intersection of any two surfaces in $\mathbb P^3$.
We have to take the intersection of at least three hypersurfaces to get that curve: for example $\overline C=\overline {Q_1}\cap \overline {Q_2}\cap \overline {Q_3}$ where $\overline {Q_3}=V(F_3)=V(xz-y^2)$.
6) Explanation of the paradox
To get rid of the parasitic component $L$ of $\overline {Q_1}\cap \overline {Q_2}$ and obtain just $\overline C$ we had to call $\overline Q_3$ to our rescue.
However restricting the situation to $\mathbb A^3$ gets rid of $L$ for free, since $L\subset \mathbb P^3\setminus \mathbb A^3$.
And that is why $C$ is already the complete intersection $C=Q_1\cap Q_2\subset \mathbb A^3$ of just two quadrics, whereas $\overline C$ cannot be obtained as the complete intersection of two surfaces.