TITLE: Blowup of $xy-z^2$
QUESTION [8 upvotes]: I try to compute the blowup of $f=xy-z^2 \subset \mathbb{A}^3$ at the origin, but got something I could not explain:
Let $A=\mathbb{C}[x,y,z]/(f)$, and  $m=(x,y,z)$ be the ideal corresponding to the origin. Then the blowup is 
$${\rm Proj}(A\oplus (m/f)t\oplus \cdots \oplus(m^k + (f)/(f))t^k\oplus\cdots).$$
Set graded ring $G:=A\oplus (m/f)t\oplus \cdots \oplus(m^k + (f)/(f))t^k\oplus\cdots$ (I insert variable $t$ in order to keep track of grading),  $$\Gamma:=A[u,v,w]$$ with $u,v, w$ variables, then one has a surjective graded ring morphism:
$$\Gamma \to G$$
sending $u,v,w$ to $xt,yt,zt$ respectively.
The kernel should be ideal $I:=(uy-vx, uz-wx,vz-wy)$, and hence we have an isomorhism $\Gamma/I \cong G$, and the blowup is ${\rm Proj}(\Gamma/I)$. We can take an affine piece of this projective variety, say $w \neq 0$. Then it can be written as
$${\rm Spec}((A)[u/w,v/w]/J)$$
where $J=(uy/w-vx/w, uz/w-x, vz/w-y)$. In order to simplify the notation, we can assume $w=1$, and set $u/w=u$ etc. Under this notation, we can rewrite above by 
$${\rm Spec}((\mathbb{C}[x,y,z]/(xy-z^2))[u,v]/J')$$ where $J'=(uy-vx,uz-x,vz-y)$. And hence can be simplified by
$${\rm Spec}(\mathbb{C}[uz,vz,z]/(uvz^2-z^2))[u,v],$$ and furthermore
$${\rm Spec} (\mathbb{C}[u,v,z]/(uvz^2-z^2)).$$
One can show this affine piece maps to ${\rm Spec}\mathbb{C}[x,y,z]/(xy-z^2)$ by maps $x,y,z$ to $uz,vz,z$ respectively.
Question: 
(1)Does $${\rm Spec} (\mathbb{C}[u,v,z]/(uvz^2-z^2))$$ corresponds to the blowup of $f$ restricted to an affine piece?
(2)The above variety is an union of $z^2=0$ and $uv-1=0$, what does each piece mean? $z^2=0$ does not seem like an exceptional divisor because it is dimension $2$, and  the exceptional divisor should have dimension $1$.

REPLY [5 votes]: The kernel of $\Gamma\to G$ is bigger than $I$: note that as $A$ is an integral domain, and 
$$uvz^2=(uz)(vz)=wxwy=w^2z^2,$$ 
we have 
$uv-w^2=0$ in $G$. But $uv-w^2\notin  I$.