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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:$$Cl\left(\big\{\left((0,y,0),[0:1]\right)\mid y\in\mathbb A\big\}\;\cup\;\big\{\left((yt,y,t^2),\left[1:t\right]\right)\mid t>0\big\}\cup\big\{\left((-yt,y,t^2),\left[1:-t\right]\right)\mid t>0\big\}\right)$$

Rushabh Mehta
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1 Answers1

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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.

KReiser
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  • I thought that this blowup was still supposed to be singular at $0$, and required another blowup at the origin to fully resolve? I just computed the result again myself and am finding the Jacobian at the origin to be $0$. – Rushabh Mehta Jun 08 '20 at 00:59
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    @DonThousand I am fairly confident in my calculations in this post. Without seeing your calculations which show the Jacobian at the origin to be zero, I can't say anything to help you with that - one guess would be that perhaps you did not remove the exceptional divisor? And no further blowup should be necessary to resolve these singularities - you may need to invest in more blowups if you want the strict transform to meet the exceptional divisor transversally, which is occasionally worth caring about (though it is not clear to me that this is what you're after). – KReiser Jun 08 '20 at 01:20
  • I'll edit them in in a bit. But, I'll take a closer look and try to see if I've made a mistake anywhere. – Rushabh Mehta Jun 08 '20 at 12:53