Resolution of the $E_8$ singularity with a weighted blowup

Question

I am reading Miles Reid's notes on weighted projective spaces, and I'm a little confused about a particular paragraph (notes here, page 8):

A famous case is the $E_8$ singularity $X: (x^2+y^3+z^5=0)$, which is naturally weighted homogeneous with weights 15,10,6. The $\mathbb{G}_m$ quotient morphism $X \to \mathbb{P}^1$ defined by the ratio $x^2:y^3:z^5$ has stabiliser of order 2, 3, and 5. The weighted blowup $Y \to X$ (the graph of the quotient morphism $X \to \mathbb{P}^1$) is a surface having cyclic quotient singularities of order 2,3,5 at the 3 points, giving rise to the Dynkin diagram of $E_8$.

I'd like to see this very explicitly. I agree that $\mathbb{P}(15,10,6) \cong \mathbb{P}^2$, and I can see that the equation $x^2 + y^3 + z^5$ becomes $u+v+w$ in $\mathbb{P}^2$, with coordinates $(u,v,w)$, so I agree that $X \to \mathbb{P}^1$. However, I am having trouble writing the equations for the graph and observing the singularities Reid describes.

Here is what I can do: The map $\mathbb{A}^3 \setminus \{0\} \to \mathbb{P}(15,10,6)$ is given by $(x,y,z) \mapsto [x:y:z]$, and the isomorphism $\mathbb{P}(15,10,6) \to \mathbb{P}^2$ is $[x:y:z] \mapsto [x^2:y^3:z^5]$. The graph of this map is $$ \Gamma = \{(x,y,z) \times [u:v:w] \,|\, uy^3=vx^2, uz^5=wx^2,wy^3=vz^5\} $$ Restricting to $w=-u-v$ I get the equations $(uy^3=vx^2, uz^5=(-u-v)x^2,(-u-v)y^3=vz^5)$. Taking partial derivatives, this appears to be singular everywhere. What have I done wrong?

Answer 1

Edit: Your computation is correct. The weighted blowup $Y \to X$ as defined in Example 3.7 of Reid's notes (i.e. the graph of the quotient morphism $X \to \mathbb{P^1}$) is singular at all the points of the exceptional line. However, the statement about $Y$ having 3 cyclic singularities becomes true when you replace $Y$ by its normalization $Y'$.

As you write, $\Gamma$ is defined in $X \times \mathbb{P}^2$ by the equations $$uy^3 = vx^2,\ uz^5 = wx^2,\ wy^3 = vz^5,\ w+u+v = 0$$ Now $w+u+v = 0$ defines a hypersurface in $X \times \mathbb{P}^2$ isomorphic to $ X \times \mathbb{P}^1$, with coordinates $((x,y,z), [u:v])$. Now $\Gamma$ is defined in this $ X \times \mathbb{P}^1$ by your equations $$uy^3 = vx^2,\ uz^5 = -(u+v)x^2,\ -(u+v)y^3 = vz^5$$ It follows that on $ X \times \mathbb{A}^1$ where $v \neq 0$, $\Gamma$ is defined by $$u'y^3 = x^2$$ where $u' := u/v$. Let $U := \Gamma \cap \{v \neq 0\}$. As you noted, $U$ is singular at all the points on $u'$-axis. In particular, $U$ is not normal. Indeed, it is straightforward to note that $x/y, z^2/y$ and $xz/y^2$ are integral over the coordinate ring of $U$. It turns out that you need to only adjoin these elements to get the normalization $U'$ of $U$; i.e. $U'$ is the closure in $U \times \mathbb{A}^3$ of the graph of the map $U\setminus\{y=0\} \to \mathbb{A}^3$ given by $(u',x,y,z) \mapsto (x/y, z^2/y, xz/y^2)$. $U'$ has $2$ singular points: $(0,\ldots,0) $ and $(-1,0,\ldots, 0) $ (where the first coordinate corresponds to $u'$) - corresponding respectively to the stabilizers of $x=0$ and $z=0$ on $X$. The other singular point (corresponding to the stabilizer of $y=0$ on $X$) on $Y'$ is on (the normalization of) the chart $\Gamma \cap \{u \neq 0\}$.