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Is the pushforward of the sheaf of differentials on an elliptic curve over a scheme necessarily trivial?
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If $f:Erightarrow S$ is an elliptic curve over a scheme $S$ (so $f$ is proper and smooth of relative dimension one with geometrically connected fibers of genus one, equipped with a section $0:Srightarrow E$), then is the sheaf $underlineomega_E/S:=f_*Omega_E/S^1$ actually free of rank one? According to the statement of Grothendieck-Serre duality in Hida's book Geometric Modular Forms and Elliptic Curves, there should be a canonical isomorphism $underlineomega_E/ScongmathcalHom_mathscrO_S(R^1f_*mathscrO_E,mathscrO_S)$, and the first result on elliptic curves in this book is that $R^1f_*mathscrO_EcongmathscrO_S$. It's also not clear to me whether $S$ is being assumed (locally) Noetherian, but if that's necessary for Grothendieck-Serre duality to hold, then I'm fine with assuming it. I can't find another reference with a statement of Grothendieck-Serre duality in this generality which does not use the language of derived categories (which I unfortunately don't understand).
The reason I'm kind of skeptical about this is that in Hida's book, as well as in Katz-Mazur, it is said that $underlineomega_E/S$ is invertible, so that an $mathscrO_S$-basis $omega$ for $underlineomega_E/S$ can be found locally on $S$. If the invertible sheaf in question were really trivial then one would be able to choose a global $mathscrO_S$-basis for $underlineomega_E/S$, and there would be no reason to talk about doing so locally. Hida goes on to say that, choosing an $mathscrO_S$-basis $omega$ locally on $S$ allows one to regard $(Omega_E/S^1,omega)$ as a relative effective Cartier divisor in $E/S$, which also doesn't make complete sense to me because if we can only find $omega$ locally, how are we getting a global section $omegain H^0(E,Omega_E/S^1)=H^0(S,underlineomega_E/S)$ (unless $underlineomega_E/S$ really is trivial)?
algebraic-geometry elliptic-curves
asked Sep 1 '12 at 19:20
Keenan Kidwell
18.9k13069
add a comment |Â
up vote
12
down vote
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If $f:Erightarrow S$ is an elliptic curve over a scheme $S$ (so $f$ is proper and smooth of relative dimension one with geometrically connected fibers of genus one, equipped with a section $0:Srightarrow E$), then is the sheaf $underlineomega_E/S:=f_*Omega_E/S^1$ actually free of rank one? According to the statement of Grothendieck-Serre duality in Hida's book Geometric Modular Forms and Elliptic Curves, there should be a canonical isomorphism $underlineomega_E/ScongmathcalHom_mathscrO_S(R^1f_*mathscrO_E,mathscrO_S)$, and the first result on elliptic curves in this book is that $R^1f_*mathscrO_EcongmathscrO_S$. It's also not clear to me whether $S$ is being assumed (locally) Noetherian, but if that's necessary for Grothendieck-Serre duality to hold, then I'm fine with assuming it. I can't find another reference with a statement of Grothendieck-Serre duality in this generality which does not use the language of derived categories (which I unfortunately don't understand).
The reason I'm kind of skeptical about this is that in Hida's book, as well as in Katz-Mazur, it is said that $underlineomega_E/S$ is invertible, so that an $mathscrO_S$-basis $omega$ for $underlineomega_E/S$ can be found locally on $S$. If the invertible sheaf in question were really trivial then one would be able to choose a global $mathscrO_S$-basis for $underlineomega_E/S$, and there would be no reason to talk about doing so locally. Hida goes on to say that, choosing an $mathscrO_S$-basis $omega$ locally on $S$ allows one to regard $(Omega_E/S^1,omega)$ as a relative effective Cartier divisor in $E/S$, which also doesn't make complete sense to me because if we can only find $omega$ locally, how are we getting a global section $omegain H^0(E,Omega_E/S^1)=H^0(S,underlineomega_E/S)$ (unless $underlineomega_E/S$ really is trivial)?
algebraic-geometry elliptic-curves
asked Sep 1 '12 at 19:20
Keenan Kidwell
18.9k13069
I know I'm resurrecting a long dead post, but I have also encountered this weirdness in Hida's book. As a result I feel like I am less able to rely on his book, if he gets such a basic thing wrong. Do you know if this is justified? Can you remark on how correct the rest of his book is? How does his erroneous assertion that $R^1f_*mathcalO_EcongmathcalO_S$ affect his subsequent results?
– oxeimon
Dec 21 '15 at 20:21
add a comment |Â
up vote
12
down vote
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up vote
12
down vote
favorite
If $f:Erightarrow S$ is an elliptic curve over a scheme $S$ (so $f$ is proper and smooth of relative dimension one with geometrically connected fibers of genus one, equipped with a section $0:Srightarrow E$), then is the sheaf $underlineomega_E/S:=f_*Omega_E/S^1$ actually free of rank one? According to the statement of Grothendieck-Serre duality in Hida's book Geometric Modular Forms and Elliptic Curves, there should be a canonical isomorphism $underlineomega_E/ScongmathcalHom_mathscrO_S(R^1f_*mathscrO_E,mathscrO_S)$, and the first result on elliptic curves in this book is that $R^1f_*mathscrO_EcongmathscrO_S$. It's also not clear to me whether $S$ is being assumed (locally) Noetherian, but if that's necessary for Grothendieck-Serre duality to hold, then I'm fine with assuming it. I can't find another reference with a statement of Grothendieck-Serre duality in this generality which does not use the language of derived categories (which I unfortunately don't understand).
The reason I'm kind of skeptical about this is that in Hida's book, as well as in Katz-Mazur, it is said that $underlineomega_E/S$ is invertible, so that an $mathscrO_S$-basis $omega$ for $underlineomega_E/S$ can be found locally on $S$. If the invertible sheaf in question were really trivial then one would be able to choose a global $mathscrO_S$-basis for $underlineomega_E/S$, and there would be no reason to talk about doing so locally. Hida goes on to say that, choosing an $mathscrO_S$-basis $omega$ locally on $S$ allows one to regard $(Omega_E/S^1,omega)$ as a relative effective Cartier divisor in $E/S$, which also doesn't make complete sense to me because if we can only find $omega$ locally, how are we getting a global section $omegain H^0(E,Omega_E/S^1)=H^0(S,underlineomega_E/S)$ (unless $underlineomega_E/S$ really is trivial)?
algebraic-geometry elliptic-curves
asked Sep 1 '12 at 19:20
Keenan Kidwell
18.9k13069
If $f:Erightarrow S$ is an elliptic curve over a scheme $S$ (so $f$ is proper and smooth of relative dimension one with geometrically connected fibers of genus one, equipped with a section $0:Srightarrow E$), then is the sheaf $underlineomega_E/S:=f_*Omega_E/S^1$ actually free of rank one? According to the statement of Grothendieck-Serre duality in Hida's book Geometric Modular Forms and Elliptic Curves, there should be a canonical isomorphism $underlineomega_E/ScongmathcalHom_mathscrO_S(R^1f_*mathscrO_E,mathscrO_S)$, and the first result on elliptic curves in this book is that $R^1f_*mathscrO_EcongmathscrO_S$. It's also not clear to me whether $S$ is being assumed (locally) Noetherian, but if that's necessary for Grothendieck-Serre duality to hold, then I'm fine with assuming it. I can't find another reference with a statement of Grothendieck-Serre duality in this generality which does not use the language of derived categories (which I unfortunately don't understand).
The reason I'm kind of skeptical about this is that in Hida's book, as well as in Katz-Mazur, it is said that $underlineomega_E/S$ is invertible, so that an $mathscrO_S$-basis $omega$ for $underlineomega_E/S$ can be found locally on $S$. If the invertible sheaf in question were really trivial then one would be able to choose a global $mathscrO_S$-basis for $underlineomega_E/S$, and there would be no reason to talk about doing so locally. Hida goes on to say that, choosing an $mathscrO_S$-basis $omega$ locally on $S$ allows one to regard $(Omega_E/S^1,omega)$ as a relative effective Cartier divisor in $E/S$, which also doesn't make complete sense to me because if we can only find $omega$ locally, how are we getting a global section $omegain H^0(E,Omega_E/S^1)=H^0(S,underlineomega_E/S)$ (unless $underlineomega_E/S$ really is trivial)?
algebraic-geometry elliptic-curves
asked Sep 1 '12 at 19:20
Keenan Kidwell
18.9k13069
asked Sep 1 '12 at 19:20
Keenan Kidwell
18.9k13069
asked Sep 1 '12 at 19:20
Keenan Kidwell
18.9k13069
asked Sep 1 '12 at 19:20
Keenan Kidwell
18.9k13069
18.9k13069
I know I'm resurrecting a long dead post, but I have also encountered this weirdness in Hida's book. As a result I feel like I am less able to rely on his book, if he gets such a basic thing wrong. Do you know if this is justified? Can you remark on how correct the rest of his book is? How does his erroneous assertion that $R^1f_*mathcalO_EcongmathcalO_S$ affect his subsequent results?
– oxeimon
Dec 21 '15 at 20:21
add a comment |Â
I know I'm resurrecting a long dead post, but I have also encountered this weirdness in Hida's book. As a result I feel like I am less able to rely on his book, if he gets such a basic thing wrong. Do you know if this is justified? Can you remark on how correct the rest of his book is? How does his erroneous assertion that $R^1f_*mathcalO_EcongmathcalO_S$ affect his subsequent results?
– oxeimon
Dec 21 '15 at 20:21
I know I'm resurrecting a long dead post, but I have also encountered this weirdness in Hida's book. As a result I feel like I am less able to rely on his book, if he gets such a basic thing wrong. Do you know if this is justified? Can you remark on how correct the rest of his book is? How does his erroneous assertion that $R^1f_*mathcalO_EcongmathcalO_S$ affect his subsequent results?
– oxeimon
Dec 21 '15 at 20:21
I know I'm resurrecting a long dead post, but I have also encountered this weirdness in Hida's book. As a result I feel like I am less able to rely on his book, if he gets such a basic thing wrong. Do you know if this is justified? Can you remark on how correct the rest of his book is? How does his erroneous assertion that $R^1f_*mathcalO_EcongmathcalO_S$ affect his subsequent results?
– oxeimon
Dec 21 '15 at 20:21
add a comment |Â
1 Answer
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The isomorphism coming from duality is correct, but it is not true that the invertible sheaf $R^1f_*O_E$ on $S$ is trivial in general.
Here is a down-to-earth construction of an example where $f_*Omega_E/S$ is not free. Let $S=mathrmSpec(R)$ be a domain (regular of dimension $1$ if you want) such that $2in R^*$ and $mathrmPic(S)=H^1(S, O_S^*)$ has an element of order $4$. Represent this element by a $1$-cocycle $(u_ij)_ij$ with $u_ijin O_S(U_icap U_j)^*$ where $ U_i _i$ is some open affine covering of $S$. By hypothesis of $4$-torsion, there exist $v_iin O_S(U_i)^*$ such that $u_ij^4=v_iv_j^-1$ for all $i,j$.
Let $E_i/U_i$ be the elliptic curve defined by the affine equation
$$ y_i^2=x_i^3+ v_i x_i.$$
Over $U_icap U_j$, the elliptic curves $E_i, E_j$ are isomorphic by the change of variables
$$y_j=u_ij^3y_i, quad x_j=u_ij^2x_i.$$
The cocycle condition on the $u_ij$'s insures that the $E_i$'s glue together and gives rise to an elliptic curve $E/S$.
Now let's compute the differential sheaf on $E$. On $E_i$, a canonical basis of $Omega_E_i/U_i$ is $omega_i:=dx_i/y_i$ and we have $omega_j=u_ij^-1omega_i$ above $U_icap U_j$. If $omega_E/S$ is free and generated by $omega$, then writing $omega_i=u_iomega|_E_i$ with $u_iin O(U_i)^*$, we get $u_ij=u_iu_j^-1$. Hence the class of $(u_ij)$ in $H^1(S, O_S^*)$ is trivial, contradiction.
One can show that over a Dedekind domain (or any ring I think), $f_*Omega_E/S$ is free if and only $E/S$ has a global smooth Weierstrass equation. It is well known (?) that this doesn't hold in general over number fields.
Dear QiL, the phrase " ...has an element of order $4$ and $2in R^*$ " is difficult to scan : it seems to imply that there are two orders involved and that they are in $R^*$ ! Might I humbly suggest that you write $2in R^*$ somewhere earlier in the sentence? (I have upvoted your answer , needless to say).
– Georges Elencwajg
Sep 2 '12 at 12:33
Dear @GeorgesElencwajg, you are right. The second condition was added while writing the example. Thanks !
– user18119
Sep 2 '12 at 13:49
Thanks for the very clear, detailed answer, @QiL!
– Keenan Kidwell
Sep 2 '12 at 14:05
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
The isomorphism coming from duality is correct, but it is not true that the invertible sheaf $R^1f_*O_E$ on $S$ is trivial in general.
Here is a down-to-earth construction of an example where $f_*Omega_E/S$ is not free. Let $S=mathrmSpec(R)$ be a domain (regular of dimension $1$ if you want) such that $2in R^*$ and $mathrmPic(S)=H^1(S, O_S^*)$ has an element of order $4$. Represent this element by a $1$-cocycle $(u_ij)_ij$ with $u_ijin O_S(U_icap U_j)^*$ where $ U_i _i$ is some open affine covering of $S$. By hypothesis of $4$-torsion, there exist $v_iin O_S(U_i)^*$ such that $u_ij^4=v_iv_j^-1$ for all $i,j$.
Let $E_i/U_i$ be the elliptic curve defined by the affine equation
$$ y_i^2=x_i^3+ v_i x_i.$$
Over $U_icap U_j$, the elliptic curves $E_i, E_j$ are isomorphic by the change of variables
$$y_j=u_ij^3y_i, quad x_j=u_ij^2x_i.$$
The cocycle condition on the $u_ij$'s insures that the $E_i$'s glue together and gives rise to an elliptic curve $E/S$.
Now let's compute the differential sheaf on $E$. On $E_i$, a canonical basis of $Omega_E_i/U_i$ is $omega_i:=dx_i/y_i$ and we have $omega_j=u_ij^-1omega_i$ above $U_icap U_j$. If $omega_E/S$ is free and generated by $omega$, then writing $omega_i=u_iomega|_E_i$ with $u_iin O(U_i)^*$, we get $u_ij=u_iu_j^-1$. Hence the class of $(u_ij)$ in $H^1(S, O_S^*)$ is trivial, contradiction.
One can show that over a Dedekind domain (or any ring I think), $f_*Omega_E/S$ is free if and only $E/S$ has a global smooth Weierstrass equation. It is well known (?) that this doesn't hold in general over number fields.
Dear QiL, the phrase " ...has an element of order $4$ and $2in R^*$ " is difficult to scan : it seems to imply that there are two orders involved and that they are in $R^*$ ! Might I humbly suggest that you write $2in R^*$ somewhere earlier in the sentence? (I have upvoted your answer , needless to say).
– Georges Elencwajg
Sep 2 '12 at 12:33
Dear @GeorgesElencwajg, you are right. The second condition was added while writing the example. Thanks !
– user18119
Sep 2 '12 at 13:49
Thanks for the very clear, detailed answer, @QiL!
– Keenan Kidwell
Sep 2 '12 at 14:05
add a comment |Â
up vote
12
down vote
accepted
The isomorphism coming from duality is correct, but it is not true that the invertible sheaf $R^1f_*O_E$ on $S$ is trivial in general.
Here is a down-to-earth construction of an example where $f_*Omega_E/S$ is not free. Let $S=mathrmSpec(R)$ be a domain (regular of dimension $1$ if you want) such that $2in R^*$ and $mathrmPic(S)=H^1(S, O_S^*)$ has an element of order $4$. Represent this element by a $1$-cocycle $(u_ij)_ij$ with $u_ijin O_S(U_icap U_j)^*$ where $ U_i _i$ is some open affine covering of $S$. By hypothesis of $4$-torsion, there exist $v_iin O_S(U_i)^*$ such that $u_ij^4=v_iv_j^-1$ for all $i,j$.
Let $E_i/U_i$ be the elliptic curve defined by the affine equation
$$ y_i^2=x_i^3+ v_i x_i.$$
Over $U_icap U_j$, the elliptic curves $E_i, E_j$ are isomorphic by the change of variables
$$y_j=u_ij^3y_i, quad x_j=u_ij^2x_i.$$
The cocycle condition on the $u_ij$'s insures that the $E_i$'s glue together and gives rise to an elliptic curve $E/S$.
Now let's compute the differential sheaf on $E$. On $E_i$, a canonical basis of $Omega_E_i/U_i$ is $omega_i:=dx_i/y_i$ and we have $omega_j=u_ij^-1omega_i$ above $U_icap U_j$. If $omega_E/S$ is free and generated by $omega$, then writing $omega_i=u_iomega|_E_i$ with $u_iin O(U_i)^*$, we get $u_ij=u_iu_j^-1$. Hence the class of $(u_ij)$ in $H^1(S, O_S^*)$ is trivial, contradiction.
One can show that over a Dedekind domain (or any ring I think), $f_*Omega_E/S$ is free if and only $E/S$ has a global smooth Weierstrass equation. It is well known (?) that this doesn't hold in general over number fields.
Dear QiL, the phrase " ...has an element of order $4$ and $2in R^*$ " is difficult to scan : it seems to imply that there are two orders involved and that they are in $R^*$ ! Might I humbly suggest that you write $2in R^*$ somewhere earlier in the sentence? (I have upvoted your answer , needless to say).
– Georges Elencwajg
Sep 2 '12 at 12:33
Dear @GeorgesElencwajg, you are right. The second condition was added while writing the example. Thanks !
– user18119
Sep 2 '12 at 13:49
Thanks for the very clear, detailed answer, @QiL!
– Keenan Kidwell
Sep 2 '12 at 14:05
add a comment |Â
up vote
12
down vote
accepted
up vote
12
down vote
accepted
The isomorphism coming from duality is correct, but it is not true that the invertible sheaf $R^1f_*O_E$ on $S$ is trivial in general.
Here is a down-to-earth construction of an example where $f_*Omega_E/S$ is not free. Let $S=mathrmSpec(R)$ be a domain (regular of dimension $1$ if you want) such that $2in R^*$ and $mathrmPic(S)=H^1(S, O_S^*)$ has an element of order $4$. Represent this element by a $1$-cocycle $(u_ij)_ij$ with $u_ijin O_S(U_icap U_j)^*$ where $ U_i _i$ is some open affine covering of $S$. By hypothesis of $4$-torsion, there exist $v_iin O_S(U_i)^*$ such that $u_ij^4=v_iv_j^-1$ for all $i,j$.
Let $E_i/U_i$ be the elliptic curve defined by the affine equation
$$ y_i^2=x_i^3+ v_i x_i.$$
Over $U_icap U_j$, the elliptic curves $E_i, E_j$ are isomorphic by the change of variables
$$y_j=u_ij^3y_i, quad x_j=u_ij^2x_i.$$
The cocycle condition on the $u_ij$'s insures that the $E_i$'s glue together and gives rise to an elliptic curve $E/S$.
Now let's compute the differential sheaf on $E$. On $E_i$, a canonical basis of $Omega_E_i/U_i$ is $omega_i:=dx_i/y_i$ and we have $omega_j=u_ij^-1omega_i$ above $U_icap U_j$. If $omega_E/S$ is free and generated by $omega$, then writing $omega_i=u_iomega|_E_i$ with $u_iin O(U_i)^*$, we get $u_ij=u_iu_j^-1$. Hence the class of $(u_ij)$ in $H^1(S, O_S^*)$ is trivial, contradiction.
One can show that over a Dedekind domain (or any ring I think), $f_*Omega_E/S$ is free if and only $E/S$ has a global smooth Weierstrass equation. It is well known (?) that this doesn't hold in general over number fields.
The isomorphism coming from duality is correct, but it is not true that the invertible sheaf $R^1f_*O_E$ on $S$ is trivial in general.
Here is a down-to-earth construction of an example where $f_*Omega_E/S$ is not free. Let $S=mathrmSpec(R)$ be a domain (regular of dimension $1$ if you want) such that $2in R^*$ and $mathrmPic(S)=H^1(S, O_S^*)$ has an element of order $4$. Represent this element by a $1$-cocycle $(u_ij)_ij$ with $u_ijin O_S(U_icap U_j)^*$ where $ U_i _i$ is some open affine covering of $S$. By hypothesis of $4$-torsion, there exist $v_iin O_S(U_i)^*$ such that $u_ij^4=v_iv_j^-1$ for all $i,j$.
Let $E_i/U_i$ be the elliptic curve defined by the affine equation
$$ y_i^2=x_i^3+ v_i x_i.$$
Over $U_icap U_j$, the elliptic curves $E_i, E_j$ are isomorphic by the change of variables
$$y_j=u_ij^3y_i, quad x_j=u_ij^2x_i.$$
The cocycle condition on the $u_ij$'s insures that the $E_i$'s glue together and gives rise to an elliptic curve $E/S$.
Now let's compute the differential sheaf on $E$. On $E_i$, a canonical basis of $Omega_E_i/U_i$ is $omega_i:=dx_i/y_i$ and we have $omega_j=u_ij^-1omega_i$ above $U_icap U_j$. If $omega_E/S$ is free and generated by $omega$, then writing $omega_i=u_iomega|_E_i$ with $u_iin O(U_i)^*$, we get $u_ij=u_iu_j^-1$. Hence the class of $(u_ij)$ in $H^1(S, O_S^*)$ is trivial, contradiction.
One can show that over a Dedekind domain (or any ring I think), $f_*Omega_E/S$ is free if and only $E/S$ has a global smooth Weierstrass equation. It is well known (?) that this doesn't hold in general over number fields.
edited Sep 2 '12 at 13:48
answered Sep 2 '12 at 10:43
user18119
Dear QiL, the phrase " ...has an element of order $4$ and $2in R^*$ " is difficult to scan : it seems to imply that there are two orders involved and that they are in $R^*$ ! Might I humbly suggest that you write $2in R^*$ somewhere earlier in the sentence? (I have upvoted your answer , needless to say).
– Georges Elencwajg
Sep 2 '12 at 12:33
Dear @GeorgesElencwajg, you are right. The second condition was added while writing the example. Thanks !
– user18119
Sep 2 '12 at 13:49
Thanks for the very clear, detailed answer, @QiL!
– Keenan Kidwell
Sep 2 '12 at 14:05
add a comment |Â
Dear QiL, the phrase " ...has an element of order $4$ and $2in R^*$ " is difficult to scan : it seems to imply that there are two orders involved and that they are in $R^*$ ! Might I humbly suggest that you write $2in R^*$ somewhere earlier in the sentence? (I have upvoted your answer , needless to say).
– Georges Elencwajg
Sep 2 '12 at 12:33
Dear @GeorgesElencwajg, you are right. The second condition was added while writing the example. Thanks !
– user18119
Sep 2 '12 at 13:49
Thanks for the very clear, detailed answer, @QiL!
– Keenan Kidwell
Sep 2 '12 at 14:05
Dear QiL, the phrase " ...has an element of order $4$ and $2in R^*$ " is difficult to scan : it seems to imply that there are two orders involved and that they are in $R^*$ ! Might I humbly suggest that you write $2in R^*$ somewhere earlier in the sentence? (I have upvoted your answer , needless to say).
– Georges Elencwajg
Sep 2 '12 at 12:33
Dear QiL, the phrase " ...has an element of order $4$ and $2in R^*$ " is difficult to scan : it seems to imply that there are two orders involved and that they are in $R^*$ ! Might I humbly suggest that you write $2in R^*$ somewhere earlier in the sentence? (I have upvoted your answer , needless to say).
– Georges Elencwajg
Sep 2 '12 at 12:33
Dear @GeorgesElencwajg, you are right. The second condition was added while writing the example. Thanks !
– user18119
Sep 2 '12 at 13:49
Dear @GeorgesElencwajg, you are right. The second condition was added while writing the example. Thanks !
– user18119
Sep 2 '12 at 13:49
Thanks for the very clear, detailed answer, @QiL!
– Keenan Kidwell
Sep 2 '12 at 14:05
Thanks for the very clear, detailed answer, @QiL!
– Keenan Kidwell
Sep 2 '12 at 14:05
add a comment |Â
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I know I'm resurrecting a long dead post, but I have also encountered this weirdness in Hida's book. As a result I feel like I am less able to rely on his book, if he gets such a basic thing wrong. Do you know if this is justified? Can you remark on how correct the rest of his book is? How does his erroneous assertion that $R^1f_*mathcalO_EcongmathcalO_S$ affect his subsequent results?
– oxeimon
Dec 21 '15 at 20:21