Let $X=\operatorname{Spec}\mathbb{K}[x,y,z]/\langle x^2-yz\rangle$ an affine scheme. It is singular because only at the rational point $0$ corresponding to the ideal $\langle x,y,z\rangle$, the jacobian matrix has rank zero, so $\operatorname{dim} T_{X,0}=3$ and $\operatorname{dim} T_{X,x}=2$ in all the other points.
I know that the dimension of $\mathbb{K}[x,y,z]/\langle x^2-yz\rangle$ is $2$, but I can't prove it. However, my question is:
I want to prove that $X$ is normal. In order to do this, I have built a ring homomorphism $\phi: \mathbb{K}[x,y,z] \mapsto \mathbb{K}[Y,Z]$ sending $x \mapsto YZ$, $y \mapsto Y^2$ and $z \mapsto Z^2$. Then $\langle x^2-yz\rangle \subset \operatorname{Ker} \phi$. Is the homomorphism correct? How can I proof the reverse inclusion?
Let $\phi: K[X,Y,Z]\to K[s,t]$ given by $\phi(X)=s^2$, $\phi(Y)=st$, and $\phi(Z)=t^2$. Let's prove that $\ker\phi=(Y^2-XZ)$.
Consider $f\in\ker\phi$, and divide this at $Y^2-XZ$ in $K[X,Z][Y]$. We get $f=(Y^2-XZ)g+r$ with $\deg_Yr\le1$. Then $r$ writes as $aY+b$ with $a,b\in K[X,Z]$. From $f(s^2,st,t^2)=0$ we get $a(s^2,t^2)st+b(s^2,t^2)=0$ and this entails $a=b=0$ (why?).
Now you can conclude that $\dim K[X,Y,Z]/(Y^2-XZ)=2$ since it is isomorphic to $K[s^2,st,t^2]$, and $K[s^2,st,t^2]\subset K[s,t]$ is an integral extension. (Of course we can also provide some other arguments for the ring is $2$-dimensional.)