13

Note that without any assumption on the singularities the answer to your question is yes. For instance, consider the hypersurface $Y = \{x_0^2+x_1^3+x_2^4\}\subset\mathbb{P}^3$. Then $Sing(Y) = [0:0:0:1]$. The projective tangent cone of $Y$ in the singular point is not reduced. Therefore, the singularity is not ordinary. Let $\pi:X\rightarrow\mathbb{P}^3$ be ...

12

I believe that $(Y,B)$ is always klt for some boundary $B$. In fact by Theorem 5.2 of https://projecteuclid.org/download/pdf_1/euclid.jdg/1090347529, after passing to a truncation of the pluricanonical rings, there is a klt pair of log general type $(Y',B')$ with isomorphic pluricanonical ring (more precisely $R(K_X)^{(a)}\cong R(K_{X'}+B')^{(b)}$ for ...

11

I assume you want $Y \to X$ to be proper. The answer is a definite no, in general. For example, take a polynomial $f: \mathbb A^2 \to \mathbb A^1$; such a $Y$ would have to be finite over $\mathbb A^2$, and birational, so $Y = \mathbb A^2$. There are lots of counterexamples in higher dimension too: for example, it follows from the purity theorem that you ...

11

As pointed out by Alex in his comment, this is in general not true.
For instance, consider the case $N=d=n=m=2$. Then $|L|$ is the linear system of plane curves of degree $2$ passing through $p_1$ and $p_2$ with multiplicity $2$. Of course there is only one such a curve, namely the line through $p_1$ and $p_2$ counted with multiplicity $2$. In particular, $|...

answered Mar 16 '14 at 0:50

Francesco Polizzi

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10

That particular singularity only can have a crepant resolution if it has a small resolution.
Terminal singularities
If a variety has terminal singularities (like yours does), the then every further blow-up also has discrepancies $> 0$. (In other words, whether or not the variety has terminal singularities can be read off from a single log resolution). ...

10

The weighted projective plane $\mathbb{P}(1,1,n)$ can be viewed as $\mathbb{P}(1,1,n)=\mathbb{C}^3\setminus \{0\}/(x,y,z)\sim (\lambda x,\lambda y,\lambda^n z)$. For $n=1$ we obtain the standard projective plane. For $n>1$, the point $(0,0,1)$ is the unique singular point, and the blow-up of this point is a Hirzebruch surface $\mathbb{F}_n$, with ...

10

The answers are the following.
(1) It is well known that the singularity at the vertex of the cone over the Veronese surface is isomorphic to a quotient singularity of type $\frac{1}{2}(1, \, 1, \,1)$, that is the isolated singularity given by the quotient of $\mathbb{C}^3$ by the action of $\mathbb{Z}/2 \mathbb{Z}$ of the form $(x, \,y, \,z) \mapsto (-x, \,...

answered Dec 30 '14 at 15:38

Francesco Polizzi

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10

Consider $A=k[[x^3,x^2y,y^3]]\subset k[[x^3, x^2y, xy^2, y^3]]=B$. $B$ is the integral closure of $A$, $A$ is a hypersurface, but $B$ is not Gorenstein.

9

Put $d:=\deg(X)$. From the exact sequence
$$0\rightarrow \mathcal{O}_X(-d)\rightarrow \Omega ^1_{\mathbb{P}^n|X}\rightarrow \Omega ^1_X\rightarrow 0$$you get an exact sequence $\ 0\rightarrow T_X\rightarrow T_{\mathbb{P}^n|X}\rightarrow \mathcal{O}_X(d)\rightarrow \mathcal{E}xt^1(\Omega ^1_X,\mathcal{O}_X)\rightarrow 0$.
From this you find easily $H^i(X,...

ag.algebraic-geometry deformation-theory projective-geometry singularity-theory resolution-of-singularities

9

It is easy to see that the canonical class of $X'$ is trivial (combine the formula for the canonical class of the blowup of $A^3$ and the adjunction formula). Hence $X'$ is crepant. But a crepant resolution of a surface is unique, so $X = X'$.
By the way, if you look at the exceptional divisor of $X'$, you will note that this is a union of two curves (in ...

8

$\hat{X}$ can be as bad as you want. For example, take your favorite non-Gorenstein variety $\hat{X}$ in $\mathbb{A}^N$. By Noether Lemma there is a finite morphism $\hat{X} \to \mathbb{A}^n =: Y$. Take $X$ to be a resolution of singularities of $\hat{X}$. Then $X \to Y$ is a quasifinite morphism between smooth varieties.

8

The only restriction I see is that $\hat{X}$ must be normal (because $X$ is): if $\phi$ is a rational function on (some affine open subscheme of) $\hat{X}$ which is integral over $\mathscr{O}_\hat{X}$, then $\phi\circ\pi$ is integral over $\mathscr{O}_{X}$, hence in $\mathscr{O}_{X}$. In other words, $\phi$ lies in $\pi_*\mathscr{O}_{X}=\mathscr{O}_\hat{X}$.

answered Mar 26 '15 at 14:59

Laurent Moret-Bailly

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8

Edit. This follows from the Elkik-Fujita Vanishing Theorem. There is a more general vanishing theorem due to Elkik and Fujita. One version of this theorem (where I read the theorem) is Theorem 1.3.1 of the following article.
MR0946243 (89e:14015)
Kawamata, Yujiro; Matsuda, Katsumi; Matsuki, Kenji
Introduction to the minimal model problem. Algebraic ...

7

This is false in general. Take $f:\mathbb A^2\to \mathbb A^1$, $(x,y)\mapsto xy$. Any attempt to resolve the singular fiber will bring in a new component in the fiber, so it remains singular.

7

1
The statement is still false with the most recently added conditions.
The problem is that $-E$ is not necessarily ($\pi$-)nef. I will discuss below what one can do to fix the statement so it will hold. First, here is an example:
Let $W$ be a two dimensional rational double point that's either a $D_n$ or an $E_n$ singularity. The point is to have an ...

7

Here is an answer for the affine case. Assume $f:A\to B$ is a homomorphism of Noetherian rings. Then $B$ is a filtered colimit of smooth (finitely generated) $A$-algebras iff $f$ is regular (flat with geometrically regular fibers). This is due to Popescu and Spivakovsky; see for instance Teissier's Bourbaki talk
http://www.math.jussieu.fr/~teissier/documents/...

answered Mar 6 '13 at 8:15

Laurent Moret-Bailly

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7

First let us describe the resolution explicitly. Let $A = k[x,y,z,w]/(xy-(z-w^k)(z+w^k))$. Note that the ideal $I = (x,z-w^k)$ has a resolution of the form
$$
A\oplus A \to A\oplus A \to I \to 0,
$$
where the map is given by the matrix $\left(\begin{smallmatrix} x & z+w^k \\ z-w^k & y \end{smallmatrix}\right)$. It follows that the blowup $X$ of the ...

7

For the Hermitian symmetric $G/P$, Nicolas Perrin explicitly classified all the Schubert varieties admitting a small resolution. (More explicitly, he classifies all the minimal models and quotes a theorem that says a small resolution is a smooth minimal model.) If I remember correctly, except for some very small rank exceptions, type A, and the obvious ...

7

There are six small crepant resolutions of this singularity, it being the (pullback via the Weyl group of the) versal deformation space of the $\mathbb{Z}_3$ surface singularity. The six small resolutions are in one-to-one correspondence with the Weyl group, namely $S_3$, and they correspond to orderings of $\{x,y,z\}$.
To see how this works, if you blow-...

7

It is a good exercise to make the computation in the simplest possible case: $X=\mathbb{C}^2$, $G=\mathbb{Z}/2$ acting by the antipodal involution, so that $Y$ is a quadratic cone, $\tilde{Y}$ obtained by blowing up the vertex of the cone. Locally $\tilde{X} :=X\times _Y\tilde{Y} $ is given in $\mathbb{C}^3$ by the equations $x(y-tx)=y(y-tx)=0$. That is, $\...

7

I assume that you mean $x_1=x_2=0$ in your description of the divisors. I believe that you only really get two different resolutions, but yes, they are both projective varieties. Hodge numbers of birational complex Calabi-Yau varieties are always the same.
You can think of this as a determinantal variety for the matrix $A$ of the form
$$
\Big(\begin{array}{...

7

Here is a proof of $d = \frac{1}{2}$. Then you can argue as Francesco did in point two of his answer.
Consider the action:
$$
\begin{array}{ccc}
\mu_{2}\times\mathbb{A}^{3} & \longrightarrow & \mathbb{A}^{3}\\
(\epsilon,x_{1}, x_{2}, x_3) & \longmapsto & (\epsilon x_{1},\epsilon x_{2}, \epsilon x_3)
\end{array}
$$
The ring of invariants is ...

7

I think the answer is no. In your notation, take $n=5$, $p=1$ and $q=2$. If you consider the blowup at the origin $X\to \mathbb C^2$, then $\mathbb Z_5$ acts on $X$ with two isolated fixed points: at one of the points the action has $p=q=1$ and at the other the action has $p=2$, $q=4$, which is the same as $p=1$, $q=2$. So the blow up introduces an isolated ...

6

Actually, there is a rather straightforward way to see that this singularity is factorial:
Let $X=\mathrm{Spec}\ k[x_1,x_2,x_3,x_4,x_5]/(x_1x_2+x_3x_4+x_5^2)$ and let $D=(x_2=0)\subseteq X$ be a principal divisor. Notice that then $D= \mathrm{Spec}\ k[x_1,x_3,x_4,x_5]/(x_3x_4+x_5^2)$, so it is a cone over a quadric cone. In particular, it is irreducible ...

6

Note that $-1 = i^2$. Thus your action is $(x,y)\mapsto (i^2x,iy)$. You can interpret this in the following way. Let $\epsilon\in \mu_4$ and consider tha action given by $(x,y)\mapsto (\epsilon^2 x,\epsilon y)$. The quotient $X = \mathbb{C}^2/\mu_4$ is given by $X = (w^2-uv = 0)\subset\mathbb{A}^3$. This is because as you observed the invariant polynomials ...

6

Let me start being a little nitpicking with the formulation of the question. The fact that $X$ is $\mathbb Q$-factorial does not in itself imply that such $a$ and $b$ exists. One also needs the fact that the Picard number of $X$ is $1$. This is indeed true, but perhaps should be mentioned. In fact, the Picard group of $X$ is $\mathbb Z$ which was silently ...

6

For what it's worth, one can say the following sort of thing.
Since $Y$ is log terminal so is $(\hat{X}, -\mathrm{Ram})$. This doesn't mean much since in the pair, the boundary has a negative coefficients (ie, the singularities of $\hat{X}$ can be arbitrarily bad). But it does say things like:
if $\hat{X}$ has really bad singularities at some points, ...

6

I am turning Giulia's comment into a CW answer. At over 400 pages, I think that "Travaux de Gabber", published in Astérisque, ought to count as a "comprehensive treatment of [some aspects of ] étale cohomology" using the method of alterations. The style is very SGA, and indeed there is a nonempty intersection among the set of authors.

6

By the Hochster-Roberts Theorem it is Cohen-Macaulay
The canonical sheaf of $\mathbb A^n$ is $G$-invariant (because the elements of $G$ have det=1) and hence it descends to the canonical sheaf of the quotient, which is then a line bundle.

6

Choose five points $P_i$ in general linear position (all choices are
equivalent under ${\rm PGL}_4$, so you might as well put four of them
at the coordinate vectors and the fifth at $(1:1:1:1)$); then
at each $P_i$ the condition of a triple point imposes $1+3+6 = 10$
linear conditions on the space of quintics, which has dimension
${8 \choose 5} = 56$, so you ...

answered Sep 3 '17 at 3:42

Noam D. Elkies

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