# Is there an obvious way for showing singularities are quotient?

I'm stuck on a technicality concerning singularities.

Basically, I have to show that the singularities of a $\mathbf{certain}$ normal projective variety over $\mathbf{C}$ are rational. (I won't bother you with the exact set-up. For, it's possible that I'm even wrong.)

My idea was to use a theorem of Viehweg and show that all singularities are quotient. But I don't have a clue of how to do the latter. What are the standard techniques used in such a proof? That is, say you want to show that all singularities are quotient. What would be your first idea to apply? Do problems like this become easier in low dimensions or does it really not matter?

Note. Forgive me if the question is ill-posed/vague.

Added later. In view of Karl's remark, I decided to give the set-up.

Let $X$ be a smooth projective variety over $\mathbf{C}$ (of any dimension). Let $\pi:Y\longrightarrow X$ be a finite morphism, where $Y$ is a normal projective variety over $\mathbf{C}$. We also are given a flat morphism $h:X\longrightarrow C$, where $C$ is a smooth projective curve. Of course, it can happen that $Y$ has singularities that are not rational. (Example?) But I would like to show that the singularities of $Y$ are rational in the situation I will describe now.

Let $V\longrightarrow U$ be a connected finite etale covering of $U=X-D$, where $D$ is a simple normal crossings divisor on $X$. Define $Y$ to be the normalization of $X$ in the function field of $V$.

So the set-up is quite general, i.e., I don't have any equations. Maybe one could try to apply arguments based on fundamental groups to show that the singularities of $Y$ are quotient? I think I can show that the singularities of $Y$ occur in the inverse image under $\pi$ of $D^{sing}$, where $D^{sing}$ is the singular locus of $D$.

And Karl, what are the techniques in characteristic $p>0$ that you mention below?

• I don't have a good answer to your question, but depending on the setting, showing that a certain variety has rational singularities might be pretty easy. For example, do you have equations for the variety (there might be some easy checks you can do using char p > 0 techniques then)? Does the variety map to or from some other variety that you understand better in some natural way? Do other special conditions apply (is it a hypersurface, or even Gorenstein, is it a surface, etc?) – Karl Schwede Apr 30 '10 at 17:14
• I see. Well allow me to elaborate a bit on the set-up. Since it's hard to read it in the comment section, I will edit my question. – Ariyan Javanpeykar Apr 30 '10 at 17:52
• In terms of the characteristic $p > 0$ stuff, either look up some papers of Mehta-Ramanathan-Ramanan, Frobenius splitting techniques can be used to prove rational singularities. You may also want to check out the more modern book of Brion & Kumar on the same subject. Likewise, you should see papers of Karen Smith, Nobuo Hara, and Mehta-Srinivas relating "$F$-rational singularities" with rational singularities. – Karl Schwede Apr 30 '10 at 19:34
• Sorry for the confusion! That was a typo. So I define $Y$ to be the normalization of $X$ in the function field of $V$. In the 2-dimensional case, one can show that all singularities of $Y$ are cyclic quotient. Since there are examples for dim Y >2 where this is no longer true, I suspect the singularities to be at least quotient. – Ariyan Javanpeykar Apr 30 '10 at 19:51
• I will take a look at all the articles you mentioned. Thank you very much! – Ariyan Javanpeykar Apr 30 '10 at 19:51

Say $X=\mathbb A^n$ and $D_1,\dots, D_n$, the components of $D$, are the coordinate hyperplanes $x_i=0$, for simplicity. You can assume that WLOG, because your $(X,D)$ is isomorphic to this one in étale topology.

$\pi_1(X\setminus D) = \mathbb Z^n$. So the cover $V\to U$ corresponds to a finite quotient of $\mathbb Z^n$, which is a finite abelian group $G$. So the ring of regular functions on $V$ is generated by the roots of monomials in $x$. You can write these as $x^m$ for some $m\in \mathbb Q^n$. The lattice $H$ generated by $m_i$ contains $\mathbb Z^n$, and the quotient $H/\mathbb Z^n$ is the dual abelian group $G^{\vee}$.

So what is $Y$ now? It is the normalization of the ring $k[x_1,\dots,x_n]$ in the bigger field $k(x^{m_i})$. So it is a toric problem now. The normalization is generated by monomials in the lattice $H$ which lie in the cone $(\mathbb R_{\ge0})^n$.

So $Y$ is toric and simplicial, and every such singularity is an abelian quotient singularity.

The condition $\pi_1(X\setminus D) = \mathbb Z^n$ fails in char $p$ (indeed, the fundamental group in that case is huge), so this argument and the statement both fail in char $p$.

• Doesn't one need the assumption of Q-factoriality to conclude that a "toric singularity is an abelian quotient singulaity"? (of course, this condition is satisfied above.) – naf May 1 '10 at 8:52
• ok, sure. edited. – VA. May 1 '10 at 12:46

I'm not sure if I understand what you are doing correctly, but let me take a guess.

Say you have $f : Y \to X$ which is finite and etale outside of a simple normal crossings divisor $D$ on $X$.

Consider the pair $(X, (1-\varepsilon)D)$ where $\varepsilon > 0$ is a very small positive number. This pair is "Kawamata-log-terminal". See for example the book by Koll\'ar and Mori.

It follows that $(Y, f^*(1-\varepsilon)D - Ram_f)$ is also Kawamata log terminal (see Proposition 5.20 in Kollar and Mori's book), here $Ram_f$ is the ramification divisor. Because $\varepsilon$ is very very small, $f^*(1-\varepsilon)D - Ram_f$ is effective (we are also using the fact that we are working in characteristic zero, so no wild ramification). It follows that $Y$ has rational singularities essentially by a theorem of Elkik (explicitly, see Theorem 5.22 in Kollar-Mori).

If you'd like, I can give a slicker proof of klt (Kawamata log terminal) => rational too.

This is essentially Abhyankar's lemma.

What VA says is correct. However, one can simply remark that the subgroups $(d\mathbb Z)^n$ for $d>0$ form a cofinal system of subgroups of finite index, and the normalization $Y_d$ of $X$ in the corresponding covering of $X \smallsetminus D$ is smooth. Since the fundamental group is abelian, the covering $Y \to X$ is Galois. If $G$ is the Galois group, the kernel $K$ of $\pi_1(X\smallsetminus D) \to G$ contains some $(d\mathbb Z)^n$; and then $Y$ is a quotient of $Y_d$ by the action of $K/(d\mathbb Z)^n$.