Showing the following coordinate rings are not isomorphic How do I show that $\frac{k[x,y,z]}{\langle xz-y^2\rangle}$ is not isomorphic to $k[x,y,z]$? (I want to understand the coordinate ring of this surface) Is there a 'natural' evaluation map that realizes $\langle xz-y^2\rangle$ as its kernel? 
 A: The dimension of $k[x,y,z]/(xz-y^2)$ is $2$. In geometric terms, its spectrum is an algebraic surface. But the dimension of $k[x,y,z]$ is $3$.
This computation relies on the following general result (see Qing Liu, Algebraic Geometry and Arithmetic Curves, Prop. 2.5.23): If $A$ is a finitely generated $k$-algebra, where $k$ is a field and $A$ is a domain, and $\mathfrak{p}$ is a prime ideal of $A$, then $\dim(A/\mathfrak{p})+\operatorname{ht}(\mathfrak{p})=\dim(A)$. A corollary of this (2.5.26 loc. cit.) is that the dimension of $V(f) \subseteq X$ is $n-1$, where $X$ is an irreducible variety of dimension $n$ and $f$ is a regular function which is not nilpotent.

After your edit, the following might be of interest:
The homomorphism $k[x,y,z]/(xz-y^2) \to k[s,t]$, $x \mapsto s^2$, $y \mapsto st$, $z \mapsto t^2$ is injective, and the image identifies with $k[s^2,t^2,st]$.
For the sake of completeness (and the hope that someone will explain me how to improve this), here is the proof for injectivity: Since $y^2-xz \in k[x,z][y]$ is monic, we see that $k[x,y,z]/(y^2-xz)$ is free over $k[x,z]$ with basis $1,y$. Hence, it's free over $k$ with basis $x^{\mathbb{N}} z^{\mathbb{N}}$, $x^{\mathbb{N}} y z^{\mathbb{N}}$, which we can rewrite as $1,x^{\mathbb{N}^+} z^{\mathbb{N}^+}$, $x^{\mathbb{N}} y z^{\mathbb{N}}$. A typical element has the form $a + \sum_{i,j > 0} a_{ij} x^i z^j + \sum_{i,j > 0} b_{ij} x^i y z^j$. If it gets mapped to $0$, this means $a + \sum_{i,j > 0} a_{ij} s^{2i} t^{2j} + \sum_{i,j > 0} b_{ij} s^{2i+1} t^{2j+1}=0$. The three sums here don't share any monomial (consider the parities of the exponents). Thus they vanish individually. But also every monomial appears only once. Thus all coefficients vanish.
Actually this enables us to find a more ad hoc invariant (which coincides with the dimension, but one doesn't have to know  that) which distinguishes the two rings, namely the transzendence degree of the field of fractions over $k$. For $k[x,y,z]$ it is $3$. For $k[x,y,z]/(xz-y^2)$ the field $k(s^2,st,t^2)$ contains $a := s/t$ and $b := s^2$; one checks that $s,t$ are algebraic over $k(a,b)$, and the latter is purely transcendent. Thus $\{a,b\}$ is a transcendence basis and the transcendence degree is $2$.
A: Here's a way to avoid using Krull dimension. Given a $k$-algebra $A$ and a morphism $e_p : A \to k$ (which corresponds to evaluation at a point $p$ of $\text{Spec } A$), a derivation at $p$ is a linear map $D : A \to k$ such that
$$D(ab) = D(a) e_p(b) + e_p(a) D(b).$$
The space of all such $D$ is a vector space over $k$ which I'll denote by $T_p(A)$; geometrically it is the Zariski tangent space at $p$ ($D$ represents the directional derivative with respect to a tangent vector). More algebraically, a derivation at $p$ is a morphism 
$$\phi = e_p + \epsilon D : A \to k[\epsilon]/\epsilon^2$$
such that composition with the natural quotient map $k[\epsilon]/\epsilon^2 \to k$ gives $e_p$. 
This definition allows us to use the universal property of polynomial rings to conclude that a derivation at any point for $k[x, y, z]$ is freely determined by $D(x), D(y)$, and $D(z)$; consequently $\dim T_p(k[x, y, z]) = 3$ for all $p$. On the other hand, a derivation $D : k[x, y, z]/(xz - y^2)$ at any point $p = (x_0, y_0, z_0)$ with $y_0 \neq 0$ satisfies
$$D(y^2) = 2y_0 D(y) = D(xz) = z_0 D(x) + x_0 D(z)$$
hence $D(y)$ is determined by $D(x)$ and $D(z)$ (in characteristic not equal to $2$). Consequently $\dim T_p(k[x, y, z]/(xz - y^2)) = 2$. 
(Note that when $p = (0, 0, 0)$ we instead have $\dim T_p(k[x, y, z]/(xz - y^2)) = 3$. This is a singular point of the variety $V(xz - y^2)$, so $k[x, y, z]/(xz - y^2)$ also cannot be isomorphic to $k[x, y]$ even though they have the same dimension.) 
