Affine twisted cubic realized as intersection of quadratic surfaces?

Question

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?

Answer 1

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.

Answer 2

Since projective twisted cubic is not intersection of two quadratics which contains an extra line, I would not expect this to be the case in affine as well.

This is not correct. When you go from projective to the affine space, you remove some points. Claims like "projective twisted cubic is not intersection of two quadratics" only hold in projective space, which is complete.

If this is not clear, consider even more basic statements like "in the projective plane, two distinct lines intersect in exactly one point." If you go to the affine plane, this isn't true.