Are there any homogeneous prime ideals of $k[x,y,z]/(xz,yz,z^2)$ not containing $z$?

Question

Let $R=k[x,y,z]/(xz,yz,z^2)$. I would like to find the homogeneous prime ideals of $R$ not containing $z$. In other words, I would like to find the points of $D_+(z) \subset \operatorname{Proj} R$.

Any homogeneous prime ideal of $R$ corresponds to a homogeneous prime ideal $P \subset k[x,y,z]$ such that $(xz, yz, z^2) \subset P$. Since $(xz, yz, z^2)=(x,y,z)\cdot (z)$, then $P$ must contain $(x,y,z)$ or must contain $(z)$. In any case, $z \in P$. So, there should be no homogeneous prime ideals of $R$ not containing $z$.

However, we also have $$D_+(z) \cong \operatorname{Spec}(R_z)_0 \cong \operatorname{Spec} (k[z]_z)_0 \cong \operatorname{Spec} k.$$ So, $D_+(z)$ is a singleton.

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Am I confused about something? Are there homogeneous prime ideals of $R$ not containing $z$? If so, what is it?

Answer 1

If $P\subset k[x,y,z]$ is a prime ideal containing $(xz,yz,z^2)$ then in particular $z\in P$, so indeed every homogeneous prime ideal of $R$ contains $z$. Your second line of reasoning fails because $$R_z=\big(k[x,y,z]/(xz,yz,z^2)\big)_z=k[x,y,z]_z/(xz,yz,z^2)=0,$$ because $z^2$ is a unit in $k[x,y,z]_z$.

Answer 2

$$R_z=(k[x,y,z]/(xz,yz,z^2))_z \cong k[x,y,z]_z/(xz,yz,z^2) = k[x,y,z]/(x,y,1) \cong 0.$$

So, $(R_z)_0 =0$ hence $D_+(z) = \varnothing$.