Problem on the number of generators of some ideals in $k[x,y,z]$ [closed]

I have got stuck with two generator problems:

1. The ideal $(zx,xy,yz)$ can't be generated by $2$ elements.
2. The ideal $(xz-y^2,yz-x^3,z^2-xy)$ can't be generated by $2$ elements.

Here the ring is $k[x,y,z]$.

Thanks in advance for any help.

Solution 1:

Suppose that $I := (xy,yz,zx)$ can be generated by two elements, i.e. $I = (f_1, f_2)$. If $m := (x,y,z)$ is the homogeneous maximal ideal, then $\overline{f_1}, \overline{f_2}$ generate $I/mI$ as a $k[x,y,z]/m \cong k$-module, so $\dim_k I/mI \le 2$.

However, $\overline{xy}, \overline{yz}, \overline{zx}$ are $k$-linearly independent in $I/mI$: if $a \overline{xy} + b \overline{yz} + c \overline{zx} = \overline{0}$ with $a, b, c \in k$, then $axy + byz + czx \in mI = (x^2y,xy^2,xyz,y^2z,yz^2,zx^2,z^2x)$. But no element of $mI$ has a nonzero term of degree $2$, so $a, b, c = 0$, hence $\dim_k I/mI \ge 3$.

The same argument applies to $J := (xz-y^2, yz-x^3, z^2-xy)$, although $J$ is not homogeneous: if $a,b,c \in k$ and $a(xz-y^2) + b(yz-x^3) + c(z^2-xy) \in mJ$, then $a, b, c = 0$, since no element of $mJ$ has a nonzero $y^2$ (resp. $yz$, $z^2$) term.