Let $R$ be a ring.Why $A＝R[x,y,z・・・] /（x^2,y^2,z^2・・・）$ is $0$ dimmensional? I think if $R$ is algebraically closed, then there are bijection between $A$'s maximal ideal and $V（x^2,y^2,z^2・・・）$, so $A$ is local ring, and $R$ is PID, so $R$ has just one prime ideal. But if $R$ is not algebraically closed, I think this discussion does not work.
I guess $（x,y,z・・・）$ is the only prime ideal, but I cannot prove this. I would be appreciated if you could teach me how to show $A$ has just one prime ideal, thank you in advance ,my teachers.
$A$ needs not be $0$-dimensional. In fact, notice that $\sqrt{(x^2,y^2,\cdots)}=(\operatorname{nilrad R},x,y,\cdots)$ and $R[x,y,\cdots]/(\operatorname{nilrad} R,x,y,\cdots)\cong R/\operatorname{nilrad}R$. Therefore we have that $$\dim A=\dim (A/\operatorname{nilrad}A)=\dim (R/\operatorname{nilrad} R)=\dim R$$