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Pull-back divisor being Cartier
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Let $pi colon X rightarrow Y$ be a projective morphism with connected fibers between normal quasi-projective varieties. Let $N$ be a $mathbbQ$-Cartier divisor on $Y$ so that $pi^*(N)$ is Cartier. Does it follows that $N$ is itself Cartier?
ag.algebraic-geometry birational-geometry divisors
asked 4 hours ago
JoaquÃn Moraga
7921813
 |Â
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Let $pi colon X rightarrow Y$ be a projective morphism with connected fibers between normal quasi-projective varieties. Let $N$ be a $mathbbQ$-Cartier divisor on $Y$ so that $pi^*(N)$ is Cartier. Does it follows that $N$ is itself Cartier?
ag.algebraic-geometry birational-geometry divisors
asked 4 hours ago
JoaquÃn Moraga
7921813
1
Duplicate of math.stackexchange.com/questions/60591/… , I think.
– David E Speyer
4 hours ago
3
Let $Ysubset mathbbP^3$ be a projective cone over a smooth plane conic, i.e., $Y$ is a quadric hypersurface of rank $3$. Let $pi$ be the minimal desingularization. Let $N$ be one of the lines of ruling on $Y$.
– Jason Starr
4 hours ago
4
@JasonStarr Isn't the pullback of $N$ in this case the strict transform plus $frac 1 2 E$, where $E$ is the exceptional curve?
– Stefano
4 hours ago
2
@Stefano. You are correct. I did not read the previous comment about the definition of pullback in this context (I assumed the pullback was inverse image ideal sheaf).
– Jason Starr
4 hours ago
3
What if $Y$ is a smooth curve and $pi $ is a stable family that has a double fiber over $yin Y$. Then isn't $pi ^*((1/2)y)$ Cartier? If we assume $N$ is integral, then we have an inclusion $pi _* O_X(pi ^* N)to O_Y(N)$ where the LHS is torsion free and the RHS is reflexive. This is an isomorphism over the smooth locus of $Y$ (by the projection formula). But it is also surjective since if $fin k(Y)$ satisfies $(f)+Ngeq 0$, then $fcirc pi in k(X)$ satisfies $(fcirc pi)+f^*Ngeq 0$ which shows the inclusion is a surjection.
– Hacon
3 hours ago
 |Â
show 3 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $pi colon X rightarrow Y$ be a projective morphism with connected fibers between normal quasi-projective varieties. Let $N$ be a $mathbbQ$-Cartier divisor on $Y$ so that $pi^*(N)$ is Cartier. Does it follows that $N$ is itself Cartier?
ag.algebraic-geometry birational-geometry divisors
asked 4 hours ago
JoaquÃn Moraga
7921813
Let $pi colon X rightarrow Y$ be a projective morphism with connected fibers between normal quasi-projective varieties. Let $N$ be a $mathbbQ$-Cartier divisor on $Y$ so that $pi^*(N)$ is Cartier. Does it follows that $N$ is itself Cartier?
ag.algebraic-geometry birational-geometry divisors
ag.algebraic-geometry birational-geometry divisors
asked 4 hours ago
JoaquÃn Moraga
7921813
asked 4 hours ago
JoaquÃn Moraga
7921813
asked 4 hours ago
JoaquÃn Moraga
7921813
asked 4 hours ago
JoaquÃn Moraga
7921813
asked 4 hours ago
JoaquÃn Moraga
7921813
7921813
1
Duplicate of math.stackexchange.com/questions/60591/… , I think.
– David E Speyer
4 hours ago
3
Let $Ysubset mathbbP^3$ be a projective cone over a smooth plane conic, i.e., $Y$ is a quadric hypersurface of rank $3$. Let $pi$ be the minimal desingularization. Let $N$ be one of the lines of ruling on $Y$.
– Jason Starr
4 hours ago
4
@JasonStarr Isn't the pullback of $N$ in this case the strict transform plus $frac 1 2 E$, where $E$ is the exceptional curve?
– Stefano
4 hours ago
2
@Stefano. You are correct. I did not read the previous comment about the definition of pullback in this context (I assumed the pullback was inverse image ideal sheaf).
– Jason Starr
4 hours ago
3
What if $Y$ is a smooth curve and $pi $ is a stable family that has a double fiber over $yin Y$. Then isn't $pi ^*((1/2)y)$ Cartier? If we assume $N$ is integral, then we have an inclusion $pi _* O_X(pi ^* N)to O_Y(N)$ where the LHS is torsion free and the RHS is reflexive. This is an isomorphism over the smooth locus of $Y$ (by the projection formula). But it is also surjective since if $fin k(Y)$ satisfies $(f)+Ngeq 0$, then $fcirc pi in k(X)$ satisfies $(fcirc pi)+f^*Ngeq 0$ which shows the inclusion is a surjection.
– Hacon
3 hours ago
 |Â
show 3 more comments
1
Duplicate of math.stackexchange.com/questions/60591/… , I think.
– David E Speyer
4 hours ago
3
Let $Ysubset mathbbP^3$ be a projective cone over a smooth plane conic, i.e., $Y$ is a quadric hypersurface of rank $3$. Let $pi$ be the minimal desingularization. Let $N$ be one of the lines of ruling on $Y$.
– Jason Starr
4 hours ago
4
@JasonStarr Isn't the pullback of $N$ in this case the strict transform plus $frac 1 2 E$, where $E$ is the exceptional curve?
– Stefano
4 hours ago
2
@Stefano. You are correct. I did not read the previous comment about the definition of pullback in this context (I assumed the pullback was inverse image ideal sheaf).
– Jason Starr
4 hours ago
3
What if $Y$ is a smooth curve and $pi $ is a stable family that has a double fiber over $yin Y$. Then isn't $pi ^*((1/2)y)$ Cartier? If we assume $N$ is integral, then we have an inclusion $pi _* O_X(pi ^* N)to O_Y(N)$ where the LHS is torsion free and the RHS is reflexive. This is an isomorphism over the smooth locus of $Y$ (by the projection formula). But it is also surjective since if $fin k(Y)$ satisfies $(f)+Ngeq 0$, then $fcirc pi in k(X)$ satisfies $(fcirc pi)+f^*Ngeq 0$ which shows the inclusion is a surjection.
– Hacon
3 hours ago
1
1
Duplicate of math.stackexchange.com/questions/60591/… , I think.
– David E Speyer
4 hours ago
Duplicate of math.stackexchange.com/questions/60591/… , I think.
– David E Speyer
4 hours ago
3
3
Let $Ysubset mathbbP^3$ be a projective cone over a smooth plane conic, i.e., $Y$ is a quadric hypersurface of rank $3$. Let $pi$ be the minimal desingularization. Let $N$ be one of the lines of ruling on $Y$.
– Jason Starr
4 hours ago
Let $Ysubset mathbbP^3$ be a projective cone over a smooth plane conic, i.e., $Y$ is a quadric hypersurface of rank $3$. Let $pi$ be the minimal desingularization. Let $N$ be one of the lines of ruling on $Y$.
– Jason Starr
4 hours ago
4
4
@JasonStarr Isn't the pullback of $N$ in this case the strict transform plus $frac 1 2 E$, where $E$ is the exceptional curve?
– Stefano
4 hours ago
@JasonStarr Isn't the pullback of $N$ in this case the strict transform plus $frac 1 2 E$, where $E$ is the exceptional curve?
– Stefano
4 hours ago
2
2
@Stefano. You are correct. I did not read the previous comment about the definition of pullback in this context (I assumed the pullback was inverse image ideal sheaf).
– Jason Starr
4 hours ago
@Stefano. You are correct. I did not read the previous comment about the definition of pullback in this context (I assumed the pullback was inverse image ideal sheaf).
– Jason Starr
4 hours ago
3
3
What if $Y$ is a smooth curve and $pi $ is a stable family that has a double fiber over $yin Y$. Then isn't $pi ^*((1/2)y)$ Cartier? If we assume $N$ is integral, then we have an inclusion $pi _* O_X(pi ^* N)to O_Y(N)$ where the LHS is torsion free and the RHS is reflexive. This is an isomorphism over the smooth locus of $Y$ (by the projection formula). But it is also surjective since if $fin k(Y)$ satisfies $(f)+Ngeq 0$, then $fcirc pi in k(X)$ satisfies $(fcirc pi)+f^*Ngeq 0$ which shows the inclusion is a surjection.
– Hacon
3 hours ago
What if $Y$ is a smooth curve and $pi $ is a stable family that has a double fiber over $yin Y$. Then isn't $pi ^*((1/2)y)$ Cartier? If we assume $N$ is integral, then we have an inclusion $pi _* O_X(pi ^* N)to O_Y(N)$ where the LHS is torsion free and the RHS is reflexive. This is an isomorphism over the smooth locus of $Y$ (by the projection formula). But it is also surjective since if $fin k(Y)$ satisfies $(f)+Ngeq 0$, then $fcirc pi in k(X)$ satisfies $(fcirc pi)+f^*Ngeq 0$ which shows the inclusion is a surjection.
– Hacon
3 hours ago
 |Â
show 3 more comments
1 Answer
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3
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Following the clarification in the comments, I am interpreting the question as follows.
Question. For an effective Weil divisor $N$ on $Y$, for an effective Cartier divisor $A$ on $X$, for a positive integer $ell$ such that the effective Weil divisor $ell N$ is Cartier and such that the pullback effective Cartier divisor $pi^*(ell N)$ equals $ell A$ as effective Cartier divisors, is $N$ Cartier?
The answer to that question is no. The following example is a modification of the example in my comment avoiding the mistake identified by Stefano.
Let $Csubset mathbbP^2$ be a smooth, plane cubic (a genus $1$ curve). Let $Y$ be the projective cone in $mathbbP^3$ over $C$. The minimal desingularization of $Y$, $$pi:Xto Y,$$ is the closure in $Ytimes C$ of the graph of the linear projection from $Y$ to $C$. Denote by $rho$ the projection, $$ rho:Xto C.$$ This morphism is a $mathbbP^1$-bundle. The morphism $rho$ maps the exceptional locus $E$ of $pi$ isomorphically to $C$, and $E$ is a relative hyperplane class for $rho$. The normal sheaf $mathcalO_X(underlineE)|_E$ is isomorphic to $mathcalO_mathbbP^2(-1)|_C$.
Let $Hsubset C$ be the restriction to $C$ of a general hyperplane in $mathbbP^2$. Let $Dsubset C$ be a degree $3$ effective divisor such that the divisor $H-D$ has finite order $ell>1$ in the Picard group of $C$, i.e., $ell H - ell D$ is the divisor of a rational function $f$ on $C$.
Let $Msubset Y$, resp. $N subset Y$, be the cone over $H$, resp. $D$. Note that $ell N$ and $ell M$ are linearly equivalent Cartier divisors on $Y$ in the linear equivalence class of $mathcalO_mathbbP^3(ell)|_Y$. In fact, $ell$ is the smallest positive integer such that $ell N$ is a Cartier divisor on $Y$.
Denote by $widetildeMsubset X$, resp. by $widetildeNsubset X$, the strict transform under $pi$ of $M$, resp. of $N$. Note that $ell widetildeM-ellwidetildeN$ equals the Cartier divisor of $fcirc rho$, as does the total pullback under $pi$ of the Cartier divisor $ell N-ell M$. Thus, the coefficient of $E$ in the pullback of $ell M$ equals the coefficient of $E$ in the pullback of $ell N$.
Already $M$ is an effective Cartier divisor on $Y$, and the pullback of $M$ is the strict transform $widetildeM$ plus the exceptional divisor $E$. One way to see this is to deform $M$ to a hyperplane section of $Y$ that is disjoint from the vertex of the cone. Since the intersection number of the total transform of $M$ with $E$ equals $0$, it follows that the coefficient of $E$ equals $1$. Thus, the total pullback of $ell M$ equals $ell widetildeM + ell E$.
Therefore also $ell N$ equals t$ell widetildeN +ell E$. Although $N$ is not Cartier on $Y$, the pullback of $ell N$ equals $ell A$ on $X$ for the effective Cartier divisor $A=widetildeN+E$.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Following the clarification in the comments, I am interpreting the question as follows.
Question. For an effective Weil divisor $N$ on $Y$, for an effective Cartier divisor $A$ on $X$, for a positive integer $ell$ such that the effective Weil divisor $ell N$ is Cartier and such that the pullback effective Cartier divisor $pi^*(ell N)$ equals $ell A$ as effective Cartier divisors, is $N$ Cartier?
The answer to that question is no. The following example is a modification of the example in my comment avoiding the mistake identified by Stefano.
Let $Csubset mathbbP^2$ be a smooth, plane cubic (a genus $1$ curve). Let $Y$ be the projective cone in $mathbbP^3$ over $C$. The minimal desingularization of $Y$, $$pi:Xto Y,$$ is the closure in $Ytimes C$ of the graph of the linear projection from $Y$ to $C$. Denote by $rho$ the projection, $$ rho:Xto C.$$ This morphism is a $mathbbP^1$-bundle. The morphism $rho$ maps the exceptional locus $E$ of $pi$ isomorphically to $C$, and $E$ is a relative hyperplane class for $rho$. The normal sheaf $mathcalO_X(underlineE)|_E$ is isomorphic to $mathcalO_mathbbP^2(-1)|_C$.
Let $Hsubset C$ be the restriction to $C$ of a general hyperplane in $mathbbP^2$. Let $Dsubset C$ be a degree $3$ effective divisor such that the divisor $H-D$ has finite order $ell>1$ in the Picard group of $C$, i.e., $ell H - ell D$ is the divisor of a rational function $f$ on $C$.
Let $Msubset Y$, resp. $N subset Y$, be the cone over $H$, resp. $D$. Note that $ell N$ and $ell M$ are linearly equivalent Cartier divisors on $Y$ in the linear equivalence class of $mathcalO_mathbbP^3(ell)|_Y$. In fact, $ell$ is the smallest positive integer such that $ell N$ is a Cartier divisor on $Y$.
Denote by $widetildeMsubset X$, resp. by $widetildeNsubset X$, the strict transform under $pi$ of $M$, resp. of $N$. Note that $ell widetildeM-ellwidetildeN$ equals the Cartier divisor of $fcirc rho$, as does the total pullback under $pi$ of the Cartier divisor $ell N-ell M$. Thus, the coefficient of $E$ in the pullback of $ell M$ equals the coefficient of $E$ in the pullback of $ell N$.
Already $M$ is an effective Cartier divisor on $Y$, and the pullback of $M$ is the strict transform $widetildeM$ plus the exceptional divisor $E$. One way to see this is to deform $M$ to a hyperplane section of $Y$ that is disjoint from the vertex of the cone. Since the intersection number of the total transform of $M$ with $E$ equals $0$, it follows that the coefficient of $E$ equals $1$. Thus, the total pullback of $ell M$ equals $ell widetildeM + ell E$.
Therefore also $ell N$ equals t$ell widetildeN +ell E$. Although $N$ is not Cartier on $Y$, the pullback of $ell N$ equals $ell A$ on $X$ for the effective Cartier divisor $A=widetildeN+E$.
add a comment |Â
up vote
3
down vote
accepted
Following the clarification in the comments, I am interpreting the question as follows.
Question. For an effective Weil divisor $N$ on $Y$, for an effective Cartier divisor $A$ on $X$, for a positive integer $ell$ such that the effective Weil divisor $ell N$ is Cartier and such that the pullback effective Cartier divisor $pi^*(ell N)$ equals $ell A$ as effective Cartier divisors, is $N$ Cartier?
The answer to that question is no. The following example is a modification of the example in my comment avoiding the mistake identified by Stefano.
Let $Csubset mathbbP^2$ be a smooth, plane cubic (a genus $1$ curve). Let $Y$ be the projective cone in $mathbbP^3$ over $C$. The minimal desingularization of $Y$, $$pi:Xto Y,$$ is the closure in $Ytimes C$ of the graph of the linear projection from $Y$ to $C$. Denote by $rho$ the projection, $$ rho:Xto C.$$ This morphism is a $mathbbP^1$-bundle. The morphism $rho$ maps the exceptional locus $E$ of $pi$ isomorphically to $C$, and $E$ is a relative hyperplane class for $rho$. The normal sheaf $mathcalO_X(underlineE)|_E$ is isomorphic to $mathcalO_mathbbP^2(-1)|_C$.
Let $Hsubset C$ be the restriction to $C$ of a general hyperplane in $mathbbP^2$. Let $Dsubset C$ be a degree $3$ effective divisor such that the divisor $H-D$ has finite order $ell>1$ in the Picard group of $C$, i.e., $ell H - ell D$ is the divisor of a rational function $f$ on $C$.
Let $Msubset Y$, resp. $N subset Y$, be the cone over $H$, resp. $D$. Note that $ell N$ and $ell M$ are linearly equivalent Cartier divisors on $Y$ in the linear equivalence class of $mathcalO_mathbbP^3(ell)|_Y$. In fact, $ell$ is the smallest positive integer such that $ell N$ is a Cartier divisor on $Y$.
Denote by $widetildeMsubset X$, resp. by $widetildeNsubset X$, the strict transform under $pi$ of $M$, resp. of $N$. Note that $ell widetildeM-ellwidetildeN$ equals the Cartier divisor of $fcirc rho$, as does the total pullback under $pi$ of the Cartier divisor $ell N-ell M$. Thus, the coefficient of $E$ in the pullback of $ell M$ equals the coefficient of $E$ in the pullback of $ell N$.
Already $M$ is an effective Cartier divisor on $Y$, and the pullback of $M$ is the strict transform $widetildeM$ plus the exceptional divisor $E$. One way to see this is to deform $M$ to a hyperplane section of $Y$ that is disjoint from the vertex of the cone. Since the intersection number of the total transform of $M$ with $E$ equals $0$, it follows that the coefficient of $E$ equals $1$. Thus, the total pullback of $ell M$ equals $ell widetildeM + ell E$.
Therefore also $ell N$ equals t$ell widetildeN +ell E$. Although $N$ is not Cartier on $Y$, the pullback of $ell N$ equals $ell A$ on $X$ for the effective Cartier divisor $A=widetildeN+E$.
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Following the clarification in the comments, I am interpreting the question as follows.
Question. For an effective Weil divisor $N$ on $Y$, for an effective Cartier divisor $A$ on $X$, for a positive integer $ell$ such that the effective Weil divisor $ell N$ is Cartier and such that the pullback effective Cartier divisor $pi^*(ell N)$ equals $ell A$ as effective Cartier divisors, is $N$ Cartier?
The answer to that question is no. The following example is a modification of the example in my comment avoiding the mistake identified by Stefano.
Let $Csubset mathbbP^2$ be a smooth, plane cubic (a genus $1$ curve). Let $Y$ be the projective cone in $mathbbP^3$ over $C$. The minimal desingularization of $Y$, $$pi:Xto Y,$$ is the closure in $Ytimes C$ of the graph of the linear projection from $Y$ to $C$. Denote by $rho$ the projection, $$ rho:Xto C.$$ This morphism is a $mathbbP^1$-bundle. The morphism $rho$ maps the exceptional locus $E$ of $pi$ isomorphically to $C$, and $E$ is a relative hyperplane class for $rho$. The normal sheaf $mathcalO_X(underlineE)|_E$ is isomorphic to $mathcalO_mathbbP^2(-1)|_C$.
Let $Hsubset C$ be the restriction to $C$ of a general hyperplane in $mathbbP^2$. Let $Dsubset C$ be a degree $3$ effective divisor such that the divisor $H-D$ has finite order $ell>1$ in the Picard group of $C$, i.e., $ell H - ell D$ is the divisor of a rational function $f$ on $C$.
Let $Msubset Y$, resp. $N subset Y$, be the cone over $H$, resp. $D$. Note that $ell N$ and $ell M$ are linearly equivalent Cartier divisors on $Y$ in the linear equivalence class of $mathcalO_mathbbP^3(ell)|_Y$. In fact, $ell$ is the smallest positive integer such that $ell N$ is a Cartier divisor on $Y$.
Denote by $widetildeMsubset X$, resp. by $widetildeNsubset X$, the strict transform under $pi$ of $M$, resp. of $N$. Note that $ell widetildeM-ellwidetildeN$ equals the Cartier divisor of $fcirc rho$, as does the total pullback under $pi$ of the Cartier divisor $ell N-ell M$. Thus, the coefficient of $E$ in the pullback of $ell M$ equals the coefficient of $E$ in the pullback of $ell N$.
Already $M$ is an effective Cartier divisor on $Y$, and the pullback of $M$ is the strict transform $widetildeM$ plus the exceptional divisor $E$. One way to see this is to deform $M$ to a hyperplane section of $Y$ that is disjoint from the vertex of the cone. Since the intersection number of the total transform of $M$ with $E$ equals $0$, it follows that the coefficient of $E$ equals $1$. Thus, the total pullback of $ell M$ equals $ell widetildeM + ell E$.
Therefore also $ell N$ equals t$ell widetildeN +ell E$. Although $N$ is not Cartier on $Y$, the pullback of $ell N$ equals $ell A$ on $X$ for the effective Cartier divisor $A=widetildeN+E$.
Following the clarification in the comments, I am interpreting the question as follows.
Question. For an effective Weil divisor $N$ on $Y$, for an effective Cartier divisor $A$ on $X$, for a positive integer $ell$ such that the effective Weil divisor $ell N$ is Cartier and such that the pullback effective Cartier divisor $pi^*(ell N)$ equals $ell A$ as effective Cartier divisors, is $N$ Cartier?
The answer to that question is no. The following example is a modification of the example in my comment avoiding the mistake identified by Stefano.
Let $Csubset mathbbP^2$ be a smooth, plane cubic (a genus $1$ curve). Let $Y$ be the projective cone in $mathbbP^3$ over $C$. The minimal desingularization of $Y$, $$pi:Xto Y,$$ is the closure in $Ytimes C$ of the graph of the linear projection from $Y$ to $C$. Denote by $rho$ the projection, $$ rho:Xto C.$$ This morphism is a $mathbbP^1$-bundle. The morphism $rho$ maps the exceptional locus $E$ of $pi$ isomorphically to $C$, and $E$ is a relative hyperplane class for $rho$. The normal sheaf $mathcalO_X(underlineE)|_E$ is isomorphic to $mathcalO_mathbbP^2(-1)|_C$.
Let $Hsubset C$ be the restriction to $C$ of a general hyperplane in $mathbbP^2$. Let $Dsubset C$ be a degree $3$ effective divisor such that the divisor $H-D$ has finite order $ell>1$ in the Picard group of $C$, i.e., $ell H - ell D$ is the divisor of a rational function $f$ on $C$.
Let $Msubset Y$, resp. $N subset Y$, be the cone over $H$, resp. $D$. Note that $ell N$ and $ell M$ are linearly equivalent Cartier divisors on $Y$ in the linear equivalence class of $mathcalO_mathbbP^3(ell)|_Y$. In fact, $ell$ is the smallest positive integer such that $ell N$ is a Cartier divisor on $Y$.
Denote by $widetildeMsubset X$, resp. by $widetildeNsubset X$, the strict transform under $pi$ of $M$, resp. of $N$. Note that $ell widetildeM-ellwidetildeN$ equals the Cartier divisor of $fcirc rho$, as does the total pullback under $pi$ of the Cartier divisor $ell N-ell M$. Thus, the coefficient of $E$ in the pullback of $ell M$ equals the coefficient of $E$ in the pullback of $ell N$.
Already $M$ is an effective Cartier divisor on $Y$, and the pullback of $M$ is the strict transform $widetildeM$ plus the exceptional divisor $E$. One way to see this is to deform $M$ to a hyperplane section of $Y$ that is disjoint from the vertex of the cone. Since the intersection number of the total transform of $M$ with $E$ equals $0$, it follows that the coefficient of $E$ equals $1$. Thus, the total pullback of $ell M$ equals $ell widetildeM + ell E$.
Therefore also $ell N$ equals t$ell widetildeN +ell E$. Although $N$ is not Cartier on $Y$, the pullback of $ell N$ equals $ell A$ on $X$ for the effective Cartier divisor $A=widetildeN+E$.
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1
Duplicate of math.stackexchange.com/questions/60591/… , I think.
– David E Speyer
4 hours ago
3
Let $Ysubset mathbbP^3$ be a projective cone over a smooth plane conic, i.e., $Y$ is a quadric hypersurface of rank $3$. Let $pi$ be the minimal desingularization. Let $N$ be one of the lines of ruling on $Y$.
– Jason Starr
4 hours ago
4
@JasonStarr Isn't the pullback of $N$ in this case the strict transform plus $frac 1 2 E$, where $E$ is the exceptional curve?
– Stefano
4 hours ago
2
@Stefano. You are correct. I did not read the previous comment about the definition of pullback in this context (I assumed the pullback was inverse image ideal sheaf).
– Jason Starr
4 hours ago
3
What if $Y$ is a smooth curve and $pi $ is a stable family that has a double fiber over $yin Y$. Then isn't $pi ^*((1/2)y)$ Cartier? If we assume $N$ is integral, then we have an inclusion $pi _* O_X(pi ^* N)to O_Y(N)$ where the LHS is torsion free and the RHS is reflexive. This is an isomorphism over the smooth locus of $Y$ (by the projection formula). But it is also surjective since if $fin k(Y)$ satisfies $(f)+Ngeq 0$, then $fcirc pi in k(X)$ satisfies $(fcirc pi)+f^*Ngeq 0$ which shows the inclusion is a surjection.
– Hacon
3 hours ago