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edited Mar 23 2011 at 5:56 |
An easy way to produce a $\mathbb Q$-Cartier but not Cartier canonical divisor is by a quotient. For instance for the quotient
$$X=\mathbb A^3/(x,y,z)\sim (-x,-y,-z)$$
$2K_X$ is Cartier, but $K_X$ is not.
I leave it for you to prove that $2K_X$ is Cartier. Here is how to see that $K_X$ is not:
Clearly $X={\rm Spec}k[x^2,y^2,z^2,xy,yz,xz]$ in other words, $X$ is the affine cone over the Veronese surface $\mathbb P^2\simeq V\subset \mathbb P^5$. Blowing up the cone point gives a resolution of singularities $\pi: Y\to X$ with exceptional divisor $E\simeq V$. In fact $E^2\sim -2L$ where $L$ is the class of a line. This follows by considering the blow up as a blow up of the ambient $\mathbb A^6$ (the cone over $\mathbb P^5$) and noticing that $\deg V=2$ in $\mathbb P^5$ so the square of the exceptional divisor of the blow up of $\mathbb A^6$, which is $-1$-times the hyperplane in $\mathbb P^5$ restricts to $-2L$ on $Y$. Now write $K_Y\sim_{\mathbb Q} \pi^*K_X + aE$ and use the adjunction formula ($Y$ is smooth!) to get
$$
(a+1)E^2= \deg K_{\mathbb a+1)E^2\sim K_E=K_{\mathbb P^2} =-3.
\sim -3L.
$$
Solving for $a$ shows that $a=\dfrac 12$ which shows that $K_X$ cannot be Cartier.
Interesting to note that the same construction does not give a desired example in dimension $2$: The quotient $\mathbb A^2/(x,y)\sim(-x,-y)$ is a cone over a conic which is a surface in $\mathbb P^3$. In particular it is Gorenstein and hence $K_X$ is Cartier.
As for the adjunction formula, it definitely works as long as $K_X+D$ is Cartier and it works up to torsion if it is $\mathbb Q$-Cartier. If it is not $\mathbb Q$-Cartier, it is not clear what the adjunction formula should mean, but even then one can have a sort of adjunction formula involving $\mathscr Ext$'s but this is almost Grotherndieck Duality then.
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edited Feb 16 2011 at 16:58 |
An easy way to produce a $\mathbb Q$-Cartier but not Cartier canonical divisor is by a quotient. For instance for the quotient
$$X=\mathbb A^3/(x,y,z)\sim (-x,-y,-z)$$
$2K_X$ is Cartier, but $K_X$ is not.
I leave it for you to prove that $2K_X$ is Cartier. Here is how to see that $K_X$ is not:
Clearly $X={\rm Spec}k[x^2,y^2,z^2,xy,yz,xz]$ in other words, $X$ is the affine cone over the Veronese surface $\mathbb P^2\simeq V\subset \mathbb P^5$. Blowing up the cone point gives a resolution of singularities $\pi: Y\to X$ with exceptional divisor $E\simeq V$. In fact $E^2\sim -2L$ where $L$ is the class of a line. This follows by considering the blow up as a blow up of the ambient $\mathbb A^6$ (the cone over $\mathbb P^5$) and noticing that $\deg V=2$ in $\mathbb P^5$ . so the square of the exceptional divisor of the blow up of $\mathbb A^6$, which is $-1$-times the hyperplane in $\mathbb P^5$ restricts to $-2L$ on $Y$. Now write $K_Y\sim_{\mathbb Q} \pi^*K_X + aE$ and use the adjunction formula ($Y$ is smooth!) to get
$$
(a+1)E^2= \deg K_{\mathbb P^2} =-3.
$$
Solving for $a$ shows that $a=\dfrac 12$ which shows that $K_X$ cannot be Cartier.
Interesting to note that the same construction does not give a desired example in dimension $2$: The quotient $\mathbb A^2/(x,y)\sim(-x,-y)$ is a cone over a conic which is a surface in $\mathbb P^3$. In particular it is Gorenstein and hence $K_X$ is Cartier.
As for the adjunction formula, it definitely works as long as $K_X+D$ is Cartier and it works up to torsion if it is $\mathbb Q$-Cartier. If it is not $\mathbb Q$-Cartier, it is not clear what the adjunction formula should mean, but even then one can have a sort of adjunction formula involving $\mathscr Ext$'s but this is almost Grotherndieck Duality then.
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edited Feb 15 2011 at 17:53 |
An easy way to produce a $\mathbb Q$-Cartier but not Cartier canonical divisor is by a quotient. For instance for the quotient
$$X=\mathbb A^3/(x,y,z)\sim (-x,-y,-z)$$
$2K_X$ is Cartier, but $K_X$ is not.
I leave it for you to prove that $2K_X$ is Cartier. Here is how to see that $K_X$ is not:
Clearly $X={\rm Spec}k[x^2,y^2,z^2,xy,yz,xz]$ in other words, $X$ is the affine cone over the Veronese surface $\mathbb P^2\simeq V\subset \mathbb P^5$. Blowing up the cone point gives a resolution of singularities $\pi: Y\to X$ with exceptional divisor $E\simeq V$. In fact $E^2\sim -2L$ where $L$ is the class of a line. This follows by considering the blow up as a blow up of the ambient $\mathbb A^6$ (the cone over $\mathbb P^5$) and noticing that $\deg V=2$ in $\mathbb P^5$. Now write $K_Y\sim_{\mathbb Q} \pi^*K_X + aE$ and use the adjunction formula ($Y$ is smooth!) to get
$$
(a+1)E^2= \deg K_{\mathbb P^2} =-2.
-3.
$$
Solving for $a$ shows that $a=\dfrac 12$ which shows that $K_X$ cannot be Cartier.
As for the adjunction formula, it definitely works as long as $K_X+D$ is Cartier and it works up to torsion if it is $\mathbb Q$-Cartier. If it is not $\mathbb Q$-Cartier, it is not clear what the adjunction formula should mean, but even then one can have a sort of adjunction formula involving $\mathscr Ext$'s but this is almost Grotherndieck Duality then.
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edited Feb 15 2011 at 17:18 |
An easy way to produce a $\mathbb Q$-Cartier but not Cartier canonical divisor is by a quotient. For instance for the quotient
$$X=k^3/(x,y,z)\sim $X=\mathbb A^3/(x,y,z)\sim (-x,-y,-z)$$
$2K_X$ is Cartier, but $K_X$ is not.
I leave it for you to prove that $2K_X$ is Cartier. Here is how to see that $K_X$ is not:
Clearly $X={\rm Spec}k[x^2,y^2,z^2,xy,yz,xz]$ in other words, $X$ is the affine cone over the Veronese surface $\mathbb P^2\simeq V\subset \mathbb P^5$. Blowing up the cone point gives a resolution of singularities $\pi: Y\to X$ with exceptional divisor $E\simeq V$. In fact $E^2\sim -2L$ where $L$ is the class of a line. This follows by considering the blow up as a blow up of the ambient $\mathbb A^6$ (the cone over $\mathbb P^5$) and noticing that $\deg V=2$ in $\mathbb P^5$. Now write $K_Y\sim_{\mathbb Q} \pi^*K_X + aE$ and use the adjunction formula ($Y$ is smooth!) to get
$$
(a+1)E^2= \deg K_{\mathbb P^2} =-2.
$$
Solving for $a$ shows that $a=\dfrac 12$ which shows that $K_X$ cannot be Cartier.
As for the adjunction formula, it definitely works as long as $K_X+D$ is Cartier and it works up to torsion if it is $\mathbb Q$-Cartier. If it is not $\mathbb Q$-Cartier, it is not clear what the adjunction formula should mean, but even then one can have a sort of adjunction formula involving $\mathscr Ext$'s but this is almost Grotherndieck Duality then.
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edited Feb 15 2011 at 16:54 |
An easy way to produce a $\mathbb Q$-Cartier but not Cartier canonical divisor is by a quotient. For instance for the quotient
$$X=k^3/(x,y,z)\sim (-x,-y,-z)$$
$2K_X$ is Cartier, but $K_X$ is not.
I leave it for you to prove that $2K_X$ is Cartier. Here is how to see that $K_X$ is not:
Clearly $X={\rm Spec}k[x^2,y^2,z^2,xy,yz,xz]$ in other words, $X$ is the affine cone over the Veronese surface $\mathbb P^2\simeq V\subset \mathbb P^5$. Blowing up the cone point gives a resolution of singularities $\pi: Y\to X$ with exceptional divisor $E\simeq V$. In fact $E^2\sim -2L$ where $L$ is the class of a line. This follows by considering the blow up as a blow up of the ambient $\mathbb A^6$ (the cone over $\mathbb P^5$) and noticing that $\deg V=2$ in $\mathbb P^5$. Now write $K_Y\sim_{\mathbb Q} \pi^*K_X + aE$ and use the adjunction formula (!) $Y$ is smooth!) to get
$$
(a+1)E^2= \deg K_{\mathbb P^2} =-2.
$$
Solving for $a$ shows that $a=\dfrac 12$ which shows that $K_X$ cannot be Cartier.
As for the adjunction formula, it definitely works as long as $K_X$ K_X+D$ is Cartier and it works up to torsion if it is $\mathbb Q$-Cartier. If it is not $\mathbb Q$-Cartier, it is not clear what the adjunction formula should mean, but even then one can have a sort of adjunction formula involving $\mathscr Ext$'s but this is almost Grotherndieck Duality then.
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edited Feb 15 2011 at 16:37 |
An easy way to produce a $\mathbb Q$-Cartier but not Cartier canonical divisor is by a quotient. For instance for the quotient
$$X=\mathbb C^3/(x,y,z)\sim $X=k^3/(x,y,z)\sim (-x,-y,-z)$$
$2K_X$ is Cartier, but $K_X$ is not.
I leave it for you to prove that $2K_X$ is Cartier. Here is how to see that $K_X$ is not:
Clearly $X={\rm Spec}k[x^2,y^2,z^2,xy,yz,xz]$ in other words, $X$ is the affine cone over the Veronese surface $\mathbb P^2\simeq V\subset \mathbb P^5$. Blowing up the cone point gives a resolution of singularities $\pi: Y\to X$ with exceptional divisor $E\simeq V$. In fact $E^2\sim -2L$ where $L$ is the class of a line. This follows by considering the blow up as a blow up of the ambient $\mathbb A^6$ (the cone over $\mathbb P^5$) and noticing that $\deg V=2$ in $\mathbb P^5$. Now write $K_Y\sim_{\mathbb Q} \pi^*K_X + aE$ and use the adjunction formula (!) to get
$$
(a+1)E^2= \deg K_{\mathbb P^2} =-2.
$$
Solving for $a$ shows that $a=\dfrac 12$ which shows that $K_X$ cannot be Cartier.
As for the adjunction formula, it definitely works as long as $K_X$ is Cartier and it works up to torsion if it is $\mathbb Q$-Cartier. If it is not $\mathbb Q$-Cartier, it is not clear what the adjunction formula should mean, but even then one can have a sort of adjunction formula involving $\mathscr Ext$'s but this is almost Grotherndieck Duality then.
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answered Feb 15 2011 at 16:26 |
An easy way to produce a $\mathbb Q$-Cartier but not Cartier canonical divisor is by a quotient. For instance for the quotient
$$X=\mathbb C^3/(x,y,z)\sim (-x,-y,-z)$$
$2K_X$ is Cartier, but $K_X$ is not.
As for the adjunction formula, it definitely works as long as $K_X$ is Cartier and it works up to torsion if it is $\mathbb Q$-Cartier. If it is not $\mathbb Q$-Cartier, it is not clear what the adjunction formula should mean, but even then one can have a sort of adjunction formula involving $\mathscr Ext$'s but this is almost Grotherndieck Duality then.
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