Mixing
Proof explanation of Katz and Mazur's “autoduality theorem†for elliptic curves
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I am currently studying Katz and Mazur's book "Arithmetic Moduli of Elliptic Curves" and I face difficulties understand a part of the proof of theorem 2.5.1, page 77 in the book (available here, page 44 in this pdf). The statement I would like to show is the following.
Let $S$ be a scheme and $E/S$ an elliptic curve. The structure of $S$-group-scheme on $E/S$ given by Abel's theorem (cf. 2.1.2, p.63 in the book, p.37 in the pdf) is the unique structure of $S$-group-scheme for which the section $"0"in E(S)$ is the identity.
To show this, we consider $K:Etimes_SE rightarrow E$ a structure of $S$-group-scheme on $E/S$ having $"0"$ as the identity. We want to show that for any $S$-scheme $T$, for any $P,Qin E(T)$, we have $$K(P,Q)=P+Q$$
To do this, we reduce to the case $T=S$ by base change. Then, we fix $Pin E(S)$ and define the $S$-morphism $f_P:Erightarrow E$ by $$f_P(Q)=K(P,Q)-P$$ and we want to show that this is the identity on $E$.
After some arguing (cf. the book), we prove that $f_P$ actually is an automorphism of $E$ for its "Abel's theorem" structure. Then, the last part of the proof goes like this:
Viewing $P$ as a variable, ie making the f.p.p.f. base change $Erightarrow S$, we may apply the rigidity theorem (cf. 2.4.2, p.76 in the book, p.44 in the pdf) to the homomorphism $$f_P-operatornameid:E_Erightarrow E_E$$ of elliptic curves over the base $E$. This homomorphism is zero for $P=0$, ie over the zero-section of the base $E$. As the zero section meets every connected component of $E$, this implies $f_P=operatornameid$ over all of $E$, in particular for any $Pin E(S)$. Therefore, $K(P+Q)=P+Q$ as required.
This conclusion sounds to me somewhat tricky and I can't get a clear mind about what is going on here. Actually, I find this base change quite confusing. I do not understand to what extent it makes $P$ the variable, as stated. I do not get what object is now refered to as the "zero-section of the base $E$" (is it $0in E(S)$? Or in $E(E)=E_E(E)$ after base change?). And lastly, I absolutely do not see why the zero section (in $E(S)$ ?) meets every connected component.
Could someone please provide some details about what is happening here? I would be truly thankful.
algebraic-geometry proof-explanation elliptic-curves
asked Jul 16 at 11:06
Suzet
2,216527
add a comment |Â
up vote
1
down vote
favorite
I am currently studying Katz and Mazur's book "Arithmetic Moduli of Elliptic Curves" and I face difficulties understand a part of the proof of theorem 2.5.1, page 77 in the book (available here, page 44 in this pdf). The statement I would like to show is the following.
Let $S$ be a scheme and $E/S$ an elliptic curve. The structure of $S$-group-scheme on $E/S$ given by Abel's theorem (cf. 2.1.2, p.63 in the book, p.37 in the pdf) is the unique structure of $S$-group-scheme for which the section $"0"in E(S)$ is the identity.
To show this, we consider $K:Etimes_SE rightarrow E$ a structure of $S$-group-scheme on $E/S$ having $"0"$ as the identity. We want to show that for any $S$-scheme $T$, for any $P,Qin E(T)$, we have $$K(P,Q)=P+Q$$
To do this, we reduce to the case $T=S$ by base change. Then, we fix $Pin E(S)$ and define the $S$-morphism $f_P:Erightarrow E$ by $$f_P(Q)=K(P,Q)-P$$ and we want to show that this is the identity on $E$.
After some arguing (cf. the book), we prove that $f_P$ actually is an automorphism of $E$ for its "Abel's theorem" structure. Then, the last part of the proof goes like this:
Viewing $P$ as a variable, ie making the f.p.p.f. base change $Erightarrow S$, we may apply the rigidity theorem (cf. 2.4.2, p.76 in the book, p.44 in the pdf) to the homomorphism $$f_P-operatornameid:E_Erightarrow E_E$$ of elliptic curves over the base $E$. This homomorphism is zero for $P=0$, ie over the zero-section of the base $E$. As the zero section meets every connected component of $E$, this implies $f_P=operatornameid$ over all of $E$, in particular for any $Pin E(S)$. Therefore, $K(P+Q)=P+Q$ as required.
This conclusion sounds to me somewhat tricky and I can't get a clear mind about what is going on here. Actually, I find this base change quite confusing. I do not understand to what extent it makes $P$ the variable, as stated. I do not get what object is now refered to as the "zero-section of the base $E$" (is it $0in E(S)$? Or in $E(E)=E_E(E)$ after base change?). And lastly, I absolutely do not see why the zero section (in $E(S)$ ?) meets every connected component.
Could someone please provide some details about what is happening here? I would be truly thankful.
algebraic-geometry proof-explanation elliptic-curves
asked Jul 16 at 11:06
Suzet
2,216527
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am currently studying Katz and Mazur's book "Arithmetic Moduli of Elliptic Curves" and I face difficulties understand a part of the proof of theorem 2.5.1, page 77 in the book (available here, page 44 in this pdf). The statement I would like to show is the following.
Let $S$ be a scheme and $E/S$ an elliptic curve. The structure of $S$-group-scheme on $E/S$ given by Abel's theorem (cf. 2.1.2, p.63 in the book, p.37 in the pdf) is the unique structure of $S$-group-scheme for which the section $"0"in E(S)$ is the identity.
To show this, we consider $K:Etimes_SE rightarrow E$ a structure of $S$-group-scheme on $E/S$ having $"0"$ as the identity. We want to show that for any $S$-scheme $T$, for any $P,Qin E(T)$, we have $$K(P,Q)=P+Q$$
To do this, we reduce to the case $T=S$ by base change. Then, we fix $Pin E(S)$ and define the $S$-morphism $f_P:Erightarrow E$ by $$f_P(Q)=K(P,Q)-P$$ and we want to show that this is the identity on $E$.
After some arguing (cf. the book), we prove that $f_P$ actually is an automorphism of $E$ for its "Abel's theorem" structure. Then, the last part of the proof goes like this:
Viewing $P$ as a variable, ie making the f.p.p.f. base change $Erightarrow S$, we may apply the rigidity theorem (cf. 2.4.2, p.76 in the book, p.44 in the pdf) to the homomorphism $$f_P-operatornameid:E_Erightarrow E_E$$ of elliptic curves over the base $E$. This homomorphism is zero for $P=0$, ie over the zero-section of the base $E$. As the zero section meets every connected component of $E$, this implies $f_P=operatornameid$ over all of $E$, in particular for any $Pin E(S)$. Therefore, $K(P+Q)=P+Q$ as required.
This conclusion sounds to me somewhat tricky and I can't get a clear mind about what is going on here. Actually, I find this base change quite confusing. I do not understand to what extent it makes $P$ the variable, as stated. I do not get what object is now refered to as the "zero-section of the base $E$" (is it $0in E(S)$? Or in $E(E)=E_E(E)$ after base change?). And lastly, I absolutely do not see why the zero section (in $E(S)$ ?) meets every connected component.
Could someone please provide some details about what is happening here? I would be truly thankful.
algebraic-geometry proof-explanation elliptic-curves
asked Jul 16 at 11:06
Suzet
2,216527
I am currently studying Katz and Mazur's book "Arithmetic Moduli of Elliptic Curves" and I face difficulties understand a part of the proof of theorem 2.5.1, page 77 in the book (available here, page 44 in this pdf). The statement I would like to show is the following.
Let $S$ be a scheme and $E/S$ an elliptic curve. The structure of $S$-group-scheme on $E/S$ given by Abel's theorem (cf. 2.1.2, p.63 in the book, p.37 in the pdf) is the unique structure of $S$-group-scheme for which the section $"0"in E(S)$ is the identity.
To show this, we consider $K:Etimes_SE rightarrow E$ a structure of $S$-group-scheme on $E/S$ having $"0"$ as the identity. We want to show that for any $S$-scheme $T$, for any $P,Qin E(T)$, we have $$K(P,Q)=P+Q$$
To do this, we reduce to the case $T=S$ by base change. Then, we fix $Pin E(S)$ and define the $S$-morphism $f_P:Erightarrow E$ by $$f_P(Q)=K(P,Q)-P$$ and we want to show that this is the identity on $E$.
After some arguing (cf. the book), we prove that $f_P$ actually is an automorphism of $E$ for its "Abel's theorem" structure. Then, the last part of the proof goes like this:
Viewing $P$ as a variable, ie making the f.p.p.f. base change $Erightarrow S$, we may apply the rigidity theorem (cf. 2.4.2, p.76 in the book, p.44 in the pdf) to the homomorphism $$f_P-operatornameid:E_Erightarrow E_E$$ of elliptic curves over the base $E$. This homomorphism is zero for $P=0$, ie over the zero-section of the base $E$. As the zero section meets every connected component of $E$, this implies $f_P=operatornameid$ over all of $E$, in particular for any $Pin E(S)$. Therefore, $K(P+Q)=P+Q$ as required.
This conclusion sounds to me somewhat tricky and I can't get a clear mind about what is going on here. Actually, I find this base change quite confusing. I do not understand to what extent it makes $P$ the variable, as stated. I do not get what object is now refered to as the "zero-section of the base $E$" (is it $0in E(S)$? Or in $E(E)=E_E(E)$ after base change?). And lastly, I absolutely do not see why the zero section (in $E(S)$ ?) meets every connected component.
Could someone please provide some details about what is happening here? I would be truly thankful.
algebraic-geometry proof-explanation elliptic-curves
asked Jul 16 at 11:06
Suzet
2,216527
asked Jul 16 at 11:06
Suzet
2,216527
asked Jul 16 at 11:06
Suzet
2,216527
asked Jul 16 at 11:06
Suzet
2,216527
2,216527
add a comment |Â
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853319%2fproof-explanation-of-katz-and-mazurs-autoduality-theorem-for-elliptic-curves%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password