Unramified field extension and elliptic curves
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Let $E/K$ be a elliptic curve over a number field $K$ and let $L/K$ be a finite abelian (Galois) extension. Let $v’$ be a (finite) place of $L$ lying over a place $v$ of $K$. Let $I=I_{v’/v}$ be the inertia group. Suppose $I$ acts trivially on $E(L)$. Why is then $L/K$ unramified at $v’$, where by unramified I mean we have equality $v’=v$ of normalized valuations (or $e=1$ in $mathfrak{p}_vmathcal{O}_L = (mathfrak{p}_{v’})^e cdots$ as prime ideals).
This is basically what Silverman is using on p.212 of Arithmetic of Elliptic Curves. He defines unramified for any Galois module if restriction to the inertia group is trivial. I want to compare this to the usual definition of unramified field extension.
algebraic-number-theory elliptic-curves
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Let $E/K$ be a elliptic curve over a number field $K$ and let $L/K$ be a finite abelian (Galois) extension. Let $v’$ be a (finite) place of $L$ lying over a place $v$ of $K$. Let $I=I_{v’/v}$ be the inertia group. Suppose $I$ acts trivially on $E(L)$. Why is then $L/K$ unramified at $v’$, where by unramified I mean we have equality $v’=v$ of normalized valuations (or $e=1$ in $mathfrak{p}_vmathcal{O}_L = (mathfrak{p}_{v’})^e cdots$ as prime ideals).
This is basically what Silverman is using on p.212 of Arithmetic of Elliptic Curves. He defines unramified for any Galois module if restriction to the inertia group is trivial. I want to compare this to the usual definition of unramified field extension.
algebraic-number-theory elliptic-curves
What is unclear to you in Silverman's proof of the fact that $K' / K$ is unramified at $v'$?
– Watson
Nov 18 at 21:17
It’s all clear except the very last line highlighted in green -why does the fact Q is fixed by inertia imply the field extension is unramified?
– usr0192
Nov 18 at 21:18
add a comment |
up vote
1
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favorite
up vote
1
down vote
favorite
Let $E/K$ be a elliptic curve over a number field $K$ and let $L/K$ be a finite abelian (Galois) extension. Let $v’$ be a (finite) place of $L$ lying over a place $v$ of $K$. Let $I=I_{v’/v}$ be the inertia group. Suppose $I$ acts trivially on $E(L)$. Why is then $L/K$ unramified at $v’$, where by unramified I mean we have equality $v’=v$ of normalized valuations (or $e=1$ in $mathfrak{p}_vmathcal{O}_L = (mathfrak{p}_{v’})^e cdots$ as prime ideals).
This is basically what Silverman is using on p.212 of Arithmetic of Elliptic Curves. He defines unramified for any Galois module if restriction to the inertia group is trivial. I want to compare this to the usual definition of unramified field extension.
algebraic-number-theory elliptic-curves
Let $E/K$ be a elliptic curve over a number field $K$ and let $L/K$ be a finite abelian (Galois) extension. Let $v’$ be a (finite) place of $L$ lying over a place $v$ of $K$. Let $I=I_{v’/v}$ be the inertia group. Suppose $I$ acts trivially on $E(L)$. Why is then $L/K$ unramified at $v’$, where by unramified I mean we have equality $v’=v$ of normalized valuations (or $e=1$ in $mathfrak{p}_vmathcal{O}_L = (mathfrak{p}_{v’})^e cdots$ as prime ideals).
This is basically what Silverman is using on p.212 of Arithmetic of Elliptic Curves. He defines unramified for any Galois module if restriction to the inertia group is trivial. I want to compare this to the usual definition of unramified field extension.
algebraic-number-theory elliptic-curves
algebraic-number-theory elliptic-curves
asked Nov 18 at 20:18
usr0192
1,177412
1,177412
What is unclear to you in Silverman's proof of the fact that $K' / K$ is unramified at $v'$?
– Watson
Nov 18 at 21:17
It’s all clear except the very last line highlighted in green -why does the fact Q is fixed by inertia imply the field extension is unramified?
– usr0192
Nov 18 at 21:18
add a comment |
What is unclear to you in Silverman's proof of the fact that $K' / K$ is unramified at $v'$?
– Watson
Nov 18 at 21:17
It’s all clear except the very last line highlighted in green -why does the fact Q is fixed by inertia imply the field extension is unramified?
– usr0192
Nov 18 at 21:18
What is unclear to you in Silverman's proof of the fact that $K' / K$ is unramified at $v'$?
– Watson
Nov 18 at 21:17
What is unclear to you in Silverman's proof of the fact that $K' / K$ is unramified at $v'$?
– Watson
Nov 18 at 21:17
It’s all clear except the very last line highlighted in green -why does the fact Q is fixed by inertia imply the field extension is unramified?
– usr0192
Nov 18 at 21:18
It’s all clear except the very last line highlighted in green -why does the fact Q is fixed by inertia imply the field extension is unramified?
– usr0192
Nov 18 at 21:18
add a comment |
1 Answer
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1
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We have a number field $K$ and an extension $K' = K(Q)$ where $Q in E(overline K)$ for some elliptic curve $E$. Fix a finite place $v$ of $K$ and a place $v'$ of $K'$ lying above $v$.
Assume that every element of the inertia group $I_{v'/v}$ fixes $Q$. We want to deduce that $K' = K(Q)$ is unramified over $K$ at $v'$. By definition, it means that the ramification index $e(v'/v)$ equals $1$.
Recall that the inertia group $I_{v'/v}$ has order $e(v'/v)$. Moreover, any $g in I_{v'/v}$ fixes $Q$, and also fix $K$, since $I_{v'/v} subset mathrm{Gal}(K'/K)$ (I think on p. 209, Silverman assumes $E[m] subset E(K)$ so that $K'/K$ is Galois). Thus $g$ fixes $K(Q)$, that is $g = mathrm{id}_{K'}$. This shows that $I_{v'/v}$ is the trivial group, of order $e(v'/v)=1$.
This shows that $K'/K$ is unramified at $v'$, as claimed.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We have a number field $K$ and an extension $K' = K(Q)$ where $Q in E(overline K)$ for some elliptic curve $E$. Fix a finite place $v$ of $K$ and a place $v'$ of $K'$ lying above $v$.
Assume that every element of the inertia group $I_{v'/v}$ fixes $Q$. We want to deduce that $K' = K(Q)$ is unramified over $K$ at $v'$. By definition, it means that the ramification index $e(v'/v)$ equals $1$.
Recall that the inertia group $I_{v'/v}$ has order $e(v'/v)$. Moreover, any $g in I_{v'/v}$ fixes $Q$, and also fix $K$, since $I_{v'/v} subset mathrm{Gal}(K'/K)$ (I think on p. 209, Silverman assumes $E[m] subset E(K)$ so that $K'/K$ is Galois). Thus $g$ fixes $K(Q)$, that is $g = mathrm{id}_{K'}$. This shows that $I_{v'/v}$ is the trivial group, of order $e(v'/v)=1$.
This shows that $K'/K$ is unramified at $v'$, as claimed.
add a comment |
up vote
1
down vote
accepted
We have a number field $K$ and an extension $K' = K(Q)$ where $Q in E(overline K)$ for some elliptic curve $E$. Fix a finite place $v$ of $K$ and a place $v'$ of $K'$ lying above $v$.
Assume that every element of the inertia group $I_{v'/v}$ fixes $Q$. We want to deduce that $K' = K(Q)$ is unramified over $K$ at $v'$. By definition, it means that the ramification index $e(v'/v)$ equals $1$.
Recall that the inertia group $I_{v'/v}$ has order $e(v'/v)$. Moreover, any $g in I_{v'/v}$ fixes $Q$, and also fix $K$, since $I_{v'/v} subset mathrm{Gal}(K'/K)$ (I think on p. 209, Silverman assumes $E[m] subset E(K)$ so that $K'/K$ is Galois). Thus $g$ fixes $K(Q)$, that is $g = mathrm{id}_{K'}$. This shows that $I_{v'/v}$ is the trivial group, of order $e(v'/v)=1$.
This shows that $K'/K$ is unramified at $v'$, as claimed.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We have a number field $K$ and an extension $K' = K(Q)$ where $Q in E(overline K)$ for some elliptic curve $E$. Fix a finite place $v$ of $K$ and a place $v'$ of $K'$ lying above $v$.
Assume that every element of the inertia group $I_{v'/v}$ fixes $Q$. We want to deduce that $K' = K(Q)$ is unramified over $K$ at $v'$. By definition, it means that the ramification index $e(v'/v)$ equals $1$.
Recall that the inertia group $I_{v'/v}$ has order $e(v'/v)$. Moreover, any $g in I_{v'/v}$ fixes $Q$, and also fix $K$, since $I_{v'/v} subset mathrm{Gal}(K'/K)$ (I think on p. 209, Silverman assumes $E[m] subset E(K)$ so that $K'/K$ is Galois). Thus $g$ fixes $K(Q)$, that is $g = mathrm{id}_{K'}$. This shows that $I_{v'/v}$ is the trivial group, of order $e(v'/v)=1$.
This shows that $K'/K$ is unramified at $v'$, as claimed.
We have a number field $K$ and an extension $K' = K(Q)$ where $Q in E(overline K)$ for some elliptic curve $E$. Fix a finite place $v$ of $K$ and a place $v'$ of $K'$ lying above $v$.
Assume that every element of the inertia group $I_{v'/v}$ fixes $Q$. We want to deduce that $K' = K(Q)$ is unramified over $K$ at $v'$. By definition, it means that the ramification index $e(v'/v)$ equals $1$.
Recall that the inertia group $I_{v'/v}$ has order $e(v'/v)$. Moreover, any $g in I_{v'/v}$ fixes $Q$, and also fix $K$, since $I_{v'/v} subset mathrm{Gal}(K'/K)$ (I think on p. 209, Silverman assumes $E[m] subset E(K)$ so that $K'/K$ is Galois). Thus $g$ fixes $K(Q)$, that is $g = mathrm{id}_{K'}$. This shows that $I_{v'/v}$ is the trivial group, of order $e(v'/v)=1$.
This shows that $K'/K$ is unramified at $v'$, as claimed.
answered Nov 18 at 21:28
Watson
15.7k92870
15.7k92870
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What is unclear to you in Silverman's proof of the fact that $K' / K$ is unramified at $v'$?
– Watson
Nov 18 at 21:17
It’s all clear except the very last line highlighted in green -why does the fact Q is fixed by inertia imply the field extension is unramified?
– usr0192
Nov 18 at 21:18