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Consider the ring homomorphism $\phi:\mathbb{C}[x,y]\rightarrow \mathbb{C}[z]$ sending $x\to z^2$ and $y\to z^3$. Show that $ker(\phi)=(x^3-y^2)$

I have shown the inclusion that $(x^3-y^2)\subseteq ker(\phi)$. But am stuck with the other side. I have tried using the divison algorithm but am unable to work it out.

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For every $f(x,y)$ we can write $f(x,y)= (x^3-y^2)g(x,y)+ x^2a(y)+xb(y)+c(y)$. If $f(x,y)\in\ker\phi$, we have $z^4a(z^3)+z^2b(z^3)+c(z^3)=0$. Note that powers of terms in $c(z^3)$ are equal $0$ modulo $3$, powers of terms in $z^2b(z^3)$ are equal $2$ modulo $3$, and powers of terms in $z^4a(z^3)$ are equal $1$ modulo $3$. This implies that there are no cancelations in $z^4a(z^3)+z^2b(z^3)+c(z^3)$, so coefficients of all polynomials $a$, $b$ and $c$ are $0$, i.e. $a=b=c=0$. Thus $f(x,y)=(x^3-y^2)g(x,y)$ and $f(x,y)\in \langle x^3-y^2\rangle$.

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