Kernel of the homomorphism $\mathbb C[x,y,z] → \mathbb C[t]$ defined by $x→t,y→ t^{2},z→ t^{3}$.

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I think we have $z-x^3$, $y-x^2$, and $z^2-y^3$ as elements of the kernel of the homomorphism $\mathbb C[x,y,z] → \mathbb C[t]$ defined by $x→t,y→ t^{2},z→ t^{3}$. But why the kernel is not generated by all the 3 elements, and only by $z-x^3$, $y-x^2$? I think maybe it is because of $z^2-y^3$ is in $\left<z-x^3, y-x^2\right>$, but I am not able to write $z^2-y^3$ as a linear combination of the $z-x^3$, $y-x^2$.

So is that true? How can I conclude that $z^2-y^3$ is not a generator? I think I do not need a claim-and-justify method to determine the kernel. Could someone let me just see the fact?

Thanks!

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Here's the idea: in the quotient, $z^2 = x^3 z = xyz = x^4 y = y^3$ (only using the relations from $z = x^3, y = x^2$). It might be easier to try to figure out the relations its induced by, rather than the combination of ideal elements it can be written as (at least at first). To go from this to writing $z^2 - y^3$ as a sum of ideal elements: \begin{align*} z^2 - x^3z &= z(z - x^3)\\ x^3 z - xyz &= -xz(y - x^2)\\ xyz - x^4 y &= xy(z - x^3)\\ x^4 y - y^3 &= -y(y^2 - x^4)\\ &= -y (y + x^2)(y - x^2)\\ \implies z^2 - y^3 &= (z^2 - x^3 z) + (x^3 z - xyz) + (xyz - x^4 y) + (x^4 y - y^3)\\ &= z(z - x^3) - xz(y - x^2) + xy(z - x^3) - (y^2 + x^2y)(y - x^2)\\ &= (xy + z)(z - x^3) - (y^2 + x^2y +xz)(y - x^2). \end{align*}