A question about prime ideal I want to show that $(xz-y^2,yz-x^3,z^2-x^2y)$ in $Q[x,y,z]$ is prime.
I'm considering the map $\phi:$ $x\rightarrow t^3,y \rightarrow t^4, z \rightarrow t^5$. If we show that ker$(\phi)=(xz-y^2,yz-x^3,z^2-x^2y)$ then $Q[x,y,z]/(xz-y^2,yz-x^3,z^2-x^2y)$ is isomorphic to a subset of $Q[t]$ which is a domain then we conclude that $(xz-y^2,yz-x^3,z^2-x^2y)$ is a prime ideal.
It is obvious that $(xz-y^2,yz-x^3,z^2-x^2y)\subset$ ker$(\phi)$. 
I don't know how to show that ker$(\phi) \subset (xz-y^2,yz-x^3,z^2-x^2y)$
 A: Let's try a (partial) answer.
For a monomial we define $\deg_\varphi x^ay^bz^c:=3a+4b+5c$. With this we lift the canonical graduation of $\mathbb{Q}[t]$ to $\mathbb{Q}[x,y,z]$, such that $\mathbb{Q}[x,y,z]=\bigoplus_k R_k$ is a graded ring. In particular for $f=\sum f_k; f_k\in R_k$ we get 
$$f\in\ker(\varphi)\iff f_k\in \ker(\varphi)\ \forall k.$$
With respect to $\deg_\varphi$ the ideal $\ker(\varphi)$ and $(xz-y^2,yz-x^3,z^2-x^2y)$ are graded ideals. For $0\leq k\leq 10$ we can check 
$$\ker(\varphi)\cap R_k = (xz-y^2,yz-x^3,z^2-x^2y)\cap R_k\qquad (*)$$
because there are only a few monomials with $\deg_\varphi\leq 10$.
It remains to show that $\ker(\varphi)$ doesn't contain another homogeneous element. Let $f_k\in \ker(\varphi)\cap R_k;k>0$. Because of $k>10$ there is no monomial $a\cdot z$. Therefore we can use the terms $xz-y^2,yz-x^3,z^2-x^2y$ to eliminate all monomial that contain $z$. Using $x^4-y^3=y(xz-y^2)+x(yz-x^3)$ we can reduce the exponents of $y$ and get finally a $g_k=a_0x^{l_0} +a_1x^{l_1}y+a_2x^{l_2}y^2$ such that $f_k-g_k\in(xz-y^2,yz-x^3,z^2-x^2y)$. Because the three summands of $g_k$ have different $\varphi$-degrees. Two of them are zero and therefore $$g_k=0\Rightarrow f_k\in(xz-y^2,yz-x^3,z^2-x^2y).$$
(I'm not sure if I got it right. Feel free to comment!)
