I'm looking for the dimension of $R=k[x,y,z]/(xy-z^2)$. In my correction it's written that $\{\bar x,\bar y\}$ is a transcendence basis of $R$, but I don't understand why (where $\bar x$ is the residue class of $x$). So my questions are the following :
1) Why $\{\bar x,\bar y\}$ are algebraically independent ?
2) Why $\{\bar x,\bar y,\bar z\}$ is not are algebraically independent ?
My attempts
1) I suppose by contradiction that there is a non-zero polynomial s.t. $f(\bar x,\bar y)=0$. But I can't get a contradiction.
2) No idea.
1)
Assume that $f(\bar x, \bar y) = 0_R = \bar 0$. Then $f(x,y) \in (xy-z^2)$, which means that $$f(x,y)=g(x,y,z) \cdot (xy-z^2)$$ for some $g \in k[x,y,z]$.
Consider this equality in $S[z]$, where $S = k[x,y]$. The degree (w.r.t. $z$) of $f(x,y) \in S[z]$ is $0$ (because $f \in S$), while the degree of $g(x,y,z) \cdot (xy-z^2)$ is $\mathrm{deg}(g) \cdot 2$. I let you think about why this fact implies that $f = 0 \in k[x,y,z]$.
2)
Hint: consider the non-zero polynomial $p(u,v,w) = w^2 - uv$. Does it give you an algebraic relation between $\overline x, \overline y$ and $\overline z$ ?