Let $S=k[X,Y,Z]$, where $k$ is a field, and let $R=k[t^3,t^8,t^{10}]\in k[t]$. We consider the homomorphism $\varphi: S\to R, X\mapsto t^3,Y\mapsto t^8,Z\mapsto t^{10}$.
I want to prove that $\mathrm{Ker}(\varphi)=(X^2Z-Y^2,YX^4-Z^2,X^6-YZ)$.
Can someone help me? Thanks.
Given an element of the kernel using $z^2=x^4y$ we can assume that it has the form $$zf(x,y)+g(x,y)$$ And $$t^{10}f(t^3,t^8)+g(t^3,t^8)=0$$ Now consider a term in $f(x,y)$ say $x^ay^b$ and if you want take $10+3a+8b$ to be maximal. We can assume with out loss that there is only one such term. For if there were another, $x^cy^d$ with $$3a+8b=3c+8d$$ then $$a=c+8n$$ and $$b=d-3n$$ (or the same with the signs reversed). thus we have two terms $$x^cy^bx^{8n}, x^cy^by^{3n}$$ and their difference is
$$x^cy^b(x^{8n}- y^{3n})$$ and it is easy to show that $x^{8n}- y^{3n}$ is in the ideal.
Ok so the term $zx^ay^b$ in $f$ must cancel with with a term $x^py^q$ in $g$, which we can also assume unique by the last argument.
so we have the equation
$$10+3a+8b=3p+8q$$ which a simple congruence argument shows that $$p-a=8n+6$$ $$b-q=3n+1$$
Now note that
$$zx^ay^b-x^py^q=zx^ay^qy^{3n}y -x^ay^qx^{8n}x^6=x^ay^q(zy^{3n}y -x^{8n}x^6)$$
and $zy^{3n}y -x^{8n}x^6$ is in the ideal using $y^3-x^8$ and $x^6-yz$.