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Extension of complete discrete valuation fields and coefficient field
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Let $L$ and $K$ be two complete discrete valuation fields of equal characteristic $0$. Assume that an embedding $Ksubset L$ is fixed and let $F$ be the algebraic closure of $K$ inside $L$. Can you provide a proof or at least an hint for the following proposition?
Proposition: $F$ is a finite extension of $K$ and $Lcong F((t))$.
In particular this means that $F$ is actually the residue field of $L$. why is this true?
Edit: On $L$ we have an additional hypothesis. We know that its residue field $overline L$ is again a complete discrete valuation field (of characteristic $0$). In other words $L$ has "dimension $2$''. For example $L=mathbb Q_p((t))$.
Thank you in advance
abstract-algebra field-theory galois-theory valuation-theory
asked Jul 20 at 20:34
manifold
315213
 |Â
show 2 more comments
up vote
3
down vote
favorite
Let $L$ and $K$ be two complete discrete valuation fields of equal characteristic $0$. Assume that an embedding $Ksubset L$ is fixed and let $F$ be the algebraic closure of $K$ inside $L$. Can you provide a proof or at least an hint for the following proposition?
Proposition: $F$ is a finite extension of $K$ and $Lcong F((t))$.
In particular this means that $F$ is actually the residue field of $L$. why is this true?
Edit: On $L$ we have an additional hypothesis. We know that its residue field $overline L$ is again a complete discrete valuation field (of characteristic $0$). In other words $L$ has "dimension $2$''. For example $L=mathbb Q_p((t))$.
Thank you in advance
abstract-algebra field-theory galois-theory valuation-theory
asked Jul 20 at 20:34
manifold
315213
1
Are you sure this is true as stated? What if $K=L$?
– Torsten Schoeneberg
Jul 20 at 23:32
Seconding Torsten's concern. Consider the case of $K=BbbQ_p$ and $L$ any finite extension. Then $L$ is complete as a finite dimensional vector space over a complete field, and $F=L$.
– Jyrki Lahtonen
Jul 21 at 10:19
1
The statement is taken from the paper "Kato's residue homomorphism and reciprocity laws on arithmetic surfaces - Dongwen Liu". First 3 lines of page 6. I'm puzzled now
– manifold
Jul 21 at 12:34
1
I forgot to add that $L$ is supposed to be two dimensional. It means that the residue field $overline L$ is again a complete discrete valuation field. For example $L=mathbb Q_p((t))$
– manifold
Jul 21 at 12:51
1
I think there are several more conditions given in the paper you cite (I found the statement in the middle of p.7 of the current arxiv version arxiv.org/pdf/1203.6712.pdf). E.g. what you call $K$ (called $k$ in loc.cit.) is a local field of char. 0 with finite residue field, i.e. some finite extension of some $Bbb Q_p$.
– Torsten Schoeneberg
Jul 21 at 20:17
 |Â
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $L$ and $K$ be two complete discrete valuation fields of equal characteristic $0$. Assume that an embedding $Ksubset L$ is fixed and let $F$ be the algebraic closure of $K$ inside $L$. Can you provide a proof or at least an hint for the following proposition?
Proposition: $F$ is a finite extension of $K$ and $Lcong F((t))$.
In particular this means that $F$ is actually the residue field of $L$. why is this true?
Edit: On $L$ we have an additional hypothesis. We know that its residue field $overline L$ is again a complete discrete valuation field (of characteristic $0$). In other words $L$ has "dimension $2$''. For example $L=mathbb Q_p((t))$.
Thank you in advance
abstract-algebra field-theory galois-theory valuation-theory
asked Jul 20 at 20:34
manifold
315213
Let $L$ and $K$ be two complete discrete valuation fields of equal characteristic $0$. Assume that an embedding $Ksubset L$ is fixed and let $F$ be the algebraic closure of $K$ inside $L$. Can you provide a proof or at least an hint for the following proposition?
Proposition: $F$ is a finite extension of $K$ and $Lcong F((t))$.
In particular this means that $F$ is actually the residue field of $L$. why is this true?
Edit: On $L$ we have an additional hypothesis. We know that its residue field $overline L$ is again a complete discrete valuation field (of characteristic $0$). In other words $L$ has "dimension $2$''. For example $L=mathbb Q_p((t))$.
Thank you in advance
abstract-algebra field-theory galois-theory valuation-theory
asked Jul 20 at 20:34
manifold
315213
edited Jul 21 at 13:54
asked Jul 20 at 20:34
manifold
315213
asked Jul 20 at 20:34
manifold
315213
asked Jul 20 at 20:34
manifold
315213
315213
1
Are you sure this is true as stated? What if $K=L$?
– Torsten Schoeneberg
Jul 20 at 23:32
Seconding Torsten's concern. Consider the case of $K=BbbQ_p$ and $L$ any finite extension. Then $L$ is complete as a finite dimensional vector space over a complete field, and $F=L$.
– Jyrki Lahtonen
Jul 21 at 10:19
1
The statement is taken from the paper "Kato's residue homomorphism and reciprocity laws on arithmetic surfaces - Dongwen Liu". First 3 lines of page 6. I'm puzzled now
– manifold
Jul 21 at 12:34
1
I forgot to add that $L$ is supposed to be two dimensional. It means that the residue field $overline L$ is again a complete discrete valuation field. For example $L=mathbb Q_p((t))$
– manifold
Jul 21 at 12:51
1
I think there are several more conditions given in the paper you cite (I found the statement in the middle of p.7 of the current arxiv version arxiv.org/pdf/1203.6712.pdf). E.g. what you call $K$ (called $k$ in loc.cit.) is a local field of char. 0 with finite residue field, i.e. some finite extension of some $Bbb Q_p$.
– Torsten Schoeneberg
Jul 21 at 20:17
 |Â
show 2 more comments
1
Are you sure this is true as stated? What if $K=L$?
– Torsten Schoeneberg
Jul 20 at 23:32
Seconding Torsten's concern. Consider the case of $K=BbbQ_p$ and $L$ any finite extension. Then $L$ is complete as a finite dimensional vector space over a complete field, and $F=L$.
– Jyrki Lahtonen
Jul 21 at 10:19
1
The statement is taken from the paper "Kato's residue homomorphism and reciprocity laws on arithmetic surfaces - Dongwen Liu". First 3 lines of page 6. I'm puzzled now
– manifold
Jul 21 at 12:34
1
I forgot to add that $L$ is supposed to be two dimensional. It means that the residue field $overline L$ is again a complete discrete valuation field. For example $L=mathbb Q_p((t))$
– manifold
Jul 21 at 12:51
1
I think there are several more conditions given in the paper you cite (I found the statement in the middle of p.7 of the current arxiv version arxiv.org/pdf/1203.6712.pdf). E.g. what you call $K$ (called $k$ in loc.cit.) is a local field of char. 0 with finite residue field, i.e. some finite extension of some $Bbb Q_p$.
– Torsten Schoeneberg
Jul 21 at 20:17
1
1
Are you sure this is true as stated? What if $K=L$?
– Torsten Schoeneberg
Jul 20 at 23:32
Are you sure this is true as stated? What if $K=L$?
– Torsten Schoeneberg
Jul 20 at 23:32
Seconding Torsten's concern. Consider the case of $K=BbbQ_p$ and $L$ any finite extension. Then $L$ is complete as a finite dimensional vector space over a complete field, and $F=L$.
– Jyrki Lahtonen
Jul 21 at 10:19
Seconding Torsten's concern. Consider the case of $K=BbbQ_p$ and $L$ any finite extension. Then $L$ is complete as a finite dimensional vector space over a complete field, and $F=L$.
– Jyrki Lahtonen
Jul 21 at 10:19
1
1
The statement is taken from the paper "Kato's residue homomorphism and reciprocity laws on arithmetic surfaces - Dongwen Liu". First 3 lines of page 6. I'm puzzled now
– manifold
Jul 21 at 12:34
The statement is taken from the paper "Kato's residue homomorphism and reciprocity laws on arithmetic surfaces - Dongwen Liu". First 3 lines of page 6. I'm puzzled now
– manifold
Jul 21 at 12:34
1
1
I forgot to add that $L$ is supposed to be two dimensional. It means that the residue field $overline L$ is again a complete discrete valuation field. For example $L=mathbb Q_p((t))$
– manifold
Jul 21 at 12:51
I forgot to add that $L$ is supposed to be two dimensional. It means that the residue field $overline L$ is again a complete discrete valuation field. For example $L=mathbb Q_p((t))$
– manifold
Jul 21 at 12:51
1
1
I think there are several more conditions given in the paper you cite (I found the statement in the middle of p.7 of the current arxiv version arxiv.org/pdf/1203.6712.pdf). E.g. what you call $K$ (called $k$ in loc.cit.) is a local field of char. 0 with finite residue field, i.e. some finite extension of some $Bbb Q_p$.
– Torsten Schoeneberg
Jul 21 at 20:17
I think there are several more conditions given in the paper you cite (I found the statement in the middle of p.7 of the current arxiv version arxiv.org/pdf/1203.6712.pdf). E.g. what you call $K$ (called $k$ in loc.cit.) is a local field of char. 0 with finite residue field, i.e. some finite extension of some $Bbb Q_p$.
– Torsten Schoeneberg
Jul 21 at 20:17
 |Â
show 2 more comments
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1
Are you sure this is true as stated? What if $K=L$?
– Torsten Schoeneberg
Jul 20 at 23:32
Seconding Torsten's concern. Consider the case of $K=BbbQ_p$ and $L$ any finite extension. Then $L$ is complete as a finite dimensional vector space over a complete field, and $F=L$.
– Jyrki Lahtonen
Jul 21 at 10:19
1
The statement is taken from the paper "Kato's residue homomorphism and reciprocity laws on arithmetic surfaces - Dongwen Liu". First 3 lines of page 6. I'm puzzled now
– manifold
Jul 21 at 12:34
1
I forgot to add that $L$ is supposed to be two dimensional. It means that the residue field $overline L$ is again a complete discrete valuation field. For example $L=mathbb Q_p((t))$
– manifold
Jul 21 at 12:51
1
I think there are several more conditions given in the paper you cite (I found the statement in the middle of p.7 of the current arxiv version arxiv.org/pdf/1203.6712.pdf). E.g. what you call $K$ (called $k$ in loc.cit.) is a local field of char. 0 with finite residue field, i.e. some finite extension of some $Bbb Q_p$.
– Torsten Schoeneberg
Jul 21 at 20:17