Blowing up the whitney umbrella over the z-axis My professor gave us the example of the Whitney Umbrella as an example of a non-trivial resolution of singularities. 
I'm aware that to resolve the singularities of the Whitney Umbrella, I need to first, blow up the Whitney Umbrella over the $Z$-axis.
As per my understanding, since the ideal generating the $z$-axis in $\mathbb A[x,y,z]$ is $(x,y)$, the blowup of $W =\mathbb V(x^2-zy^2)$ is a subset of $\mathbb A^3\times\mathbb P^1$, and is defined as ($Cl$ is the closure) $$Cl\{((x,y,z),[x:y])\mid (x,y)\neq(0,0),\;x^2=zy^2\}$$ But, I'm just not making any progress in figuring out what the blow up actually is. 
It's easy to see that in the blow-up, $y\neq0$ and $z\geq0$. Letting $t=\sqrt z$, we can partition the blowup as follows:$$Cl\left(\big\{\left((0,y,0),[0:1]\right)\mid y\in\mathbb A\big\}\;\cup\;\big\{\left((yt,y,t^2),\left[1:t\right]\right)\mid t>0\big\}\cup\big\{\left((-yt,y,t^2),\left[1:-t\right]\right)\mid t>0\big\}\right)$$
 A: The blowup of $\Bbb A^3$ along the $z$-axis is the subset of $\Bbb A^3\times \Bbb P^1$ cut out by $xt=yu$, where $x,y,z$ are coordinates on $\Bbb A^3$ and $t,u$ are homogeneous coordinates on $\Bbb P^1$. This is covered by two affine patches, each isomorphic to $\Bbb A^3$: $t=1$, which has coordinate algebra $k[x,y,z,u]/(x-yu)\cong k[y,z,u]$, and $u=1$, which has coordinate algebra $k[x,y,z,t]/(xt-y)\cong k[x,z,t]$.
Call our blowup map $\pi$. The preimage $\pi^{-1}(W)$ of the Whitney umbrella in $Bl_{(x,y)}\Bbb A^3$ is called the total transform, and it decomposes as a union of the exceptional divisor $V(x,y)$ lying over the locus we blew up, and the strict transform $\widetilde{W}$. Since the affine patches in the above paragraph cover $Bl_{(x,y)}\Bbb A^3$, we can compute these on each affine patch and glue together.
On the patch where $u=1$, the preimage of the Whitney umbrella is $V(xt-y,x^2-zy^2)$. We can rewrite this as $V(x^2-zt^2x^2)$, and the equation defining this factors as $x^2(1-zt^2)$: the locus where $x=0$ corresponds to the exceptional divisor, and the zero set of $1-zt^2$ is the strict transform. We can check that the strict transform is smooth: the Jacobian is $(0,-t^2,-2zt)$, and $t,z\neq 0$.
On the patch where $t=1$, the preimage of the Whitney umbrella is $V(x-yu,x^2-zy^2)$. We can rewrite this as $V(y^2u^2-zy^2)$, and the equation defining this factors as $y^2(u^2-z)$: the locus where $y=0$ corresponds to the exceptional divisor, and the zero set of $u^2-z$ is the strict transform. We can check this is smooth: the Jacobian is $(0,-1,2u)$.
Since these two patches are smooth, the strict transform is smooth and therefore it is a resolution of singularities. All that's left is to figure out how the two patches glue together.
Now we observe something: there are no points with $t=0$ on our strict transform. So our strict transform is located entirely inside the coordinate patch where $t=1$, where it's given as $V(u^2-z)\subset \Bbb A^3$ with coordinates $y,z,u$. This is just a copy of $\Bbb A^2$ with coordinates $y,u$, and this gives the map $\Bbb A^2\to W$ as $x=yu$, $y=y$, $z=u^2$ where the LHS of these equations are the coordinates on $W$ inherited from the embedding in $\Bbb A^3$ and the RHS are the coordinates on the strict transform.
