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Describe the kernel of the ring homomorphism $\phi\colon \mathbb{C}[x, y, z]\to \mathbb{C}[t]$ defined by $\phi(x) = t$, $\phi(y) = t^2$, $\phi(z) = t^3$.

Some hints?

I know $\ker\phi = \{ a \in\mathbb{C}[x, y, z] \mid \phi(a) = 0\}$ is closed under addition and multiplication. Also, it is an ideal of $\mathbb{C}[x, y, z]$.

Using egreg hint, I came up with

$\ker \phi$ ={$ (x^2 -y)r_1 + (xy-z)r_2 + (x^3-z)r_3 | r_1,r_2, r_3 \in\mathbb{C}[x, y, z] $} ? I am not sure if this is right.

sarah
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  • Hint: $\phi(x^2)=\phi(y)$ and $\phi(x^3)=\phi(z)$. – egreg Nov 30 '13 at 23:23
  • Related http://math.stackexchange.com/questions/568528/the-twisted-cubic-is-an-affine-variety and http://math.stackexchange.com/questions/140302/a-problem-about-the-twisted-cubic and... –  Dec 08 '13 at 21:05

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Using twice the division algorithm we get $$ f(x,y,z)=(x^3-z)g(x,y,z) + (x^2-y)h(x,y) + r(x). $$ If $f\in\ker\phi$ then $\phi(f)=0$, that is, $f(t,t^2,t^3)=0$ and thus $r(t)=0$. Since $r=0$ we get $f\in(x^3-z,x^2-y)$, therefore $\ker\phi\subseteq(x^3-z,x^2-y)$. (The other inclusion is trivial.)