Ramified Field Extensions
$begingroup$
Let $k$ be a field of $char(k)=0$ und we consider an field exension $L/k$ with $[L:k]=n$.
Set $M:= L((t^{1/n}))$ and $F:= k((t))$.
I'm looking for a proof of following two statements:
1) If the residual field $k_F := mathcal{O}_F/pi_F $ equals $k$, then $M/F$ is totally ramified.
2) $char(k) =0$, therefore, $M/F$ is tamely ramified.
I'm not an expert on the field of algebraic number theory/ local field theory so unfortunately I have rather no idea how to show it. I was just curious about the both statements.
Nevertheless for 1) I tried follwing: $M/F$ is totally ramified iff $k_M= mathcal{O}_M/pi_M = mathcal{O}_F/pi_F= k_F$. We know $k_F=k$. Why $k_M=k$?
I know that if $p$ is the unique prime of $F$ then it ramifies via $pmathcal{O}_M= q_1 q_2 ... q_d$ into primes in $mathcal{O}_M$ but from here I'm stuck.
Could anybody give proofs for the two statements.
The main goal is to show that $M/F$ is cyclic extension and I think that one can show this also directly but from didactical point of view (in order to learn some proof techniques from ANT) I would to prefer to go the way through steps 1) and 2).
field-theory algebraic-number-theory local-field ramification
asked Dec 9 '18 at 22:24
KarlPeterKarlPeter
5931315
$endgroup$
|
show 6 more comments
$begingroup$
Let $k$ be a field of $char(k)=0$ und we consider an field exension $L/k$ with $[L:k]=n$.
Set $M:= L((t^{1/n}))$ and $F:= k((t))$.
I'm looking for a proof of following two statements:
1) If the residual field $k_F := mathcal{O}_F/pi_F $ equals $k$, then $M/F$ is totally ramified.
2) $char(k) =0$, therefore, $M/F$ is tamely ramified.
I'm not an expert on the field of algebraic number theory/ local field theory so unfortunately I have rather no idea how to show it. I was just curious about the both statements.
Nevertheless for 1) I tried follwing: $M/F$ is totally ramified iff $k_M= mathcal{O}_M/pi_M = mathcal{O}_F/pi_F= k_F$. We know $k_F=k$. Why $k_M=k$?
I know that if $p$ is the unique prime of $F$ then it ramifies via $pmathcal{O}_M= q_1 q_2 ... q_d$ into primes in $mathcal{O}_M$ but from here I'm stuck.
Could anybody give proofs for the two statements.
The main goal is to show that $M/F$ is cyclic extension and I think that one can show this also directly but from didactical point of view (in order to learn some proof techniques from ANT) I would to prefer to go the way through steps 1) and 2).
field-theory algebraic-number-theory local-field ramification
asked Dec 9 '18 at 22:24
KarlPeterKarlPeter
5931315
$endgroup$
1
$begingroup$
If $k$ is algebraically closed, then $L$ cannot have degree $n$ over $k$, unless $n=1$. Do you mean $k$ any field of characteristic $0$? Also the residue field is not $F/O_F$, it's defined to be (in this case) $k[[t]]$ (a local ring) mod its maximal ideal, which is then just $k$ (you may write this $O_F/m_F$ where $m_F$ is the maximal ideal of $O_F$).
$endgroup$
– Ashwin Iyengar
Dec 9 '18 at 22:58
$begingroup$
Yes, indeed that make no sense. I guess that then $k_F=k$ is an extra assumption.
$endgroup$
– KarlPeter
Dec 9 '18 at 23:02
$begingroup$
And I don't there needs a proof, that the residue fields are the same is the definition of totally ramified extension of local fields (what needs is a proof is that it is related to the ramification in the field they are the completion of). For tamely ramified it means it becomes a Galois extension after an unramified extension.
$endgroup$
– reuns
Dec 9 '18 at 23:31
$begingroup$
@reuns:Possibly I missunderstand you but is this what is to proof, insn't it? As you pointed out, a totally ramified extension of lfs is characterized by property that the residue fields for both fields coinside. We know that $k_F=k$. But is it obvoiusly that also $k_M=k$? Which completion wrt which valuation do you mean?
$endgroup$
– KarlPeter
Dec 9 '18 at 23:44
$begingroup$
Sorry for the maybe stupid question but now I have a huge lack of knowledge on this topic
$endgroup$
– KarlPeter
Dec 9 '18 at 23:46
|
show 6 more comments
$begingroup$
Let $k$ be a field of $char(k)=0$ und we consider an field exension $L/k$ with $[L:k]=n$.
Set $M:= L((t^{1/n}))$ and $F:= k((t))$.
I'm looking for a proof of following two statements:
1) If the residual field $k_F := mathcal{O}_F/pi_F $ equals $k$, then $M/F$ is totally ramified.
2) $char(k) =0$, therefore, $M/F$ is tamely ramified.
I'm not an expert on the field of algebraic number theory/ local field theory so unfortunately I have rather no idea how to show it. I was just curious about the both statements.
Nevertheless for 1) I tried follwing: $M/F$ is totally ramified iff $k_M= mathcal{O}_M/pi_M = mathcal{O}_F/pi_F= k_F$. We know $k_F=k$. Why $k_M=k$?
I know that if $p$ is the unique prime of $F$ then it ramifies via $pmathcal{O}_M= q_1 q_2 ... q_d$ into primes in $mathcal{O}_M$ but from here I'm stuck.
Could anybody give proofs for the two statements.
The main goal is to show that $M/F$ is cyclic extension and I think that one can show this also directly but from didactical point of view (in order to learn some proof techniques from ANT) I would to prefer to go the way through steps 1) and 2).
field-theory algebraic-number-theory local-field ramification
asked Dec 9 '18 at 22:24
KarlPeterKarlPeter
5931315
$endgroup$
Let $k$ be a field of $char(k)=0$ und we consider an field exension $L/k$ with $[L:k]=n$.
Set $M:= L((t^{1/n}))$ and $F:= k((t))$.
I'm looking for a proof of following two statements:
1) If the residual field $k_F := mathcal{O}_F/pi_F $ equals $k$, then $M/F$ is totally ramified.
2) $char(k) =0$, therefore, $M/F$ is tamely ramified.
I'm not an expert on the field of algebraic number theory/ local field theory so unfortunately I have rather no idea how to show it. I was just curious about the both statements.
Nevertheless for 1) I tried follwing: $M/F$ is totally ramified iff $k_M= mathcal{O}_M/pi_M = mathcal{O}_F/pi_F= k_F$. We know $k_F=k$. Why $k_M=k$?
I know that if $p$ is the unique prime of $F$ then it ramifies via $pmathcal{O}_M= q_1 q_2 ... q_d$ into primes in $mathcal{O}_M$ but from here I'm stuck.
Could anybody give proofs for the two statements.
The main goal is to show that $M/F$ is cyclic extension and I think that one can show this also directly but from didactical point of view (in order to learn some proof techniques from ANT) I would to prefer to go the way through steps 1) and 2).
field-theory algebraic-number-theory local-field ramification
field-theory algebraic-number-theory local-field ramification
asked Dec 9 '18 at 22:24
KarlPeterKarlPeter
5931315
asked Dec 9 '18 at 22:24
KarlPeterKarlPeter
5931315
edited Dec 9 '18 at 23:58
KarlPeter
asked Dec 9 '18 at 22:24
KarlPeterKarlPeter
5931315
asked Dec 9 '18 at 22:24
KarlPeterKarlPeter
5931315
asked Dec 9 '18 at 22:24
KarlPeterKarlPeter
5931315
5931315
1
$begingroup$
If $k$ is algebraically closed, then $L$ cannot have degree $n$ over $k$, unless $n=1$. Do you mean $k$ any field of characteristic $0$? Also the residue field is not $F/O_F$, it's defined to be (in this case) $k[[t]]$ (a local ring) mod its maximal ideal, which is then just $k$ (you may write this $O_F/m_F$ where $m_F$ is the maximal ideal of $O_F$).
$endgroup$
– Ashwin Iyengar
Dec 9 '18 at 22:58
$begingroup$
Yes, indeed that make no sense. I guess that then $k_F=k$ is an extra assumption.
$endgroup$
– KarlPeter
Dec 9 '18 at 23:02
$begingroup$
And I don't there needs a proof, that the residue fields are the same is the definition of totally ramified extension of local fields (what needs is a proof is that it is related to the ramification in the field they are the completion of). For tamely ramified it means it becomes a Galois extension after an unramified extension.
$endgroup$
– reuns
Dec 9 '18 at 23:31
$begingroup$
@reuns:Possibly I missunderstand you but is this what is to proof, insn't it? As you pointed out, a totally ramified extension of lfs is characterized by property that the residue fields for both fields coinside. We know that $k_F=k$. But is it obvoiusly that also $k_M=k$? Which completion wrt which valuation do you mean?
$endgroup$
– KarlPeter
Dec 9 '18 at 23:44
$begingroup$
Sorry for the maybe stupid question but now I have a huge lack of knowledge on this topic
$endgroup$
– KarlPeter
Dec 9 '18 at 23:46
|
show 6 more comments
1
$begingroup$
If $k$ is algebraically closed, then $L$ cannot have degree $n$ over $k$, unless $n=1$. Do you mean $k$ any field of characteristic $0$? Also the residue field is not $F/O_F$, it's defined to be (in this case) $k[[t]]$ (a local ring) mod its maximal ideal, which is then just $k$ (you may write this $O_F/m_F$ where $m_F$ is the maximal ideal of $O_F$).
$endgroup$
– Ashwin Iyengar
Dec 9 '18 at 22:58
$begingroup$
Yes, indeed that make no sense. I guess that then $k_F=k$ is an extra assumption.
$endgroup$
– KarlPeter
Dec 9 '18 at 23:02
$begingroup$
And I don't there needs a proof, that the residue fields are the same is the definition of totally ramified extension of local fields (what needs is a proof is that it is related to the ramification in the field they are the completion of). For tamely ramified it means it becomes a Galois extension after an unramified extension.
$endgroup$
– reuns
Dec 9 '18 at 23:31
$begingroup$
@reuns:Possibly I missunderstand you but is this what is to proof, insn't it? As you pointed out, a totally ramified extension of lfs is characterized by property that the residue fields for both fields coinside. We know that $k_F=k$. But is it obvoiusly that also $k_M=k$? Which completion wrt which valuation do you mean?
$endgroup$
– KarlPeter
Dec 9 '18 at 23:44
$begingroup$
Sorry for the maybe stupid question but now I have a huge lack of knowledge on this topic
$endgroup$
– KarlPeter
Dec 9 '18 at 23:46
1
1
$begingroup$
If $k$ is algebraically closed, then $L$ cannot have degree $n$ over $k$, unless $n=1$. Do you mean $k$ any field of characteristic $0$? Also the residue field is not $F/O_F$, it's defined to be (in this case) $k[[t]]$ (a local ring) mod its maximal ideal, which is then just $k$ (you may write this $O_F/m_F$ where $m_F$ is the maximal ideal of $O_F$).
$endgroup$
– Ashwin Iyengar
Dec 9 '18 at 22:58
$begingroup$
If $k$ is algebraically closed, then $L$ cannot have degree $n$ over $k$, unless $n=1$. Do you mean $k$ any field of characteristic $0$? Also the residue field is not $F/O_F$, it's defined to be (in this case) $k[[t]]$ (a local ring) mod its maximal ideal, which is then just $k$ (you may write this $O_F/m_F$ where $m_F$ is the maximal ideal of $O_F$).
$endgroup$
– Ashwin Iyengar
Dec 9 '18 at 22:58
$begingroup$
Yes, indeed that make no sense. I guess that then $k_F=k$ is an extra assumption.
$endgroup$
– KarlPeter
Dec 9 '18 at 23:02
$begingroup$
Yes, indeed that make no sense. I guess that then $k_F=k$ is an extra assumption.
$endgroup$
– KarlPeter
Dec 9 '18 at 23:02
$begingroup$
And I don't there needs a proof, that the residue fields are the same is the definition of totally ramified extension of local fields (what needs is a proof is that it is related to the ramification in the field they are the completion of). For tamely ramified it means it becomes a Galois extension after an unramified extension.
$endgroup$
– reuns
Dec 9 '18 at 23:31
$begingroup$
And I don't there needs a proof, that the residue fields are the same is the definition of totally ramified extension of local fields (what needs is a proof is that it is related to the ramification in the field they are the completion of). For tamely ramified it means it becomes a Galois extension after an unramified extension.
$endgroup$
– reuns
Dec 9 '18 at 23:31
$begingroup$
@reuns:Possibly I missunderstand you but is this what is to proof, insn't it? As you pointed out, a totally ramified extension of lfs is characterized by property that the residue fields for both fields coinside. We know that $k_F=k$. But is it obvoiusly that also $k_M=k$? Which completion wrt which valuation do you mean?
$endgroup$
– KarlPeter
Dec 9 '18 at 23:44
$begingroup$
@reuns:Possibly I missunderstand you but is this what is to proof, insn't it? As you pointed out, a totally ramified extension of lfs is characterized by property that the residue fields for both fields coinside. We know that $k_F=k$. But is it obvoiusly that also $k_M=k$? Which completion wrt which valuation do you mean?
$endgroup$
– KarlPeter
Dec 9 '18 at 23:44
$begingroup$
Sorry for the maybe stupid question but now I have a huge lack of knowledge on this topic
$endgroup$
– KarlPeter
Dec 9 '18 at 23:46
$begingroup$
Sorry for the maybe stupid question but now I have a huge lack of knowledge on this topic
$endgroup$
– KarlPeter
Dec 9 '18 at 23:46
|
show 6 more comments
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1
$begingroup$
If $k$ is algebraically closed, then $L$ cannot have degree $n$ over $k$, unless $n=1$. Do you mean $k$ any field of characteristic $0$? Also the residue field is not $F/O_F$, it's defined to be (in this case) $k[[t]]$ (a local ring) mod its maximal ideal, which is then just $k$ (you may write this $O_F/m_F$ where $m_F$ is the maximal ideal of $O_F$).
$endgroup$
– Ashwin Iyengar
Dec 9 '18 at 22:58
$begingroup$
Yes, indeed that make no sense. I guess that then $k_F=k$ is an extra assumption.
$endgroup$
– KarlPeter
Dec 9 '18 at 23:02
$begingroup$
And I don't there needs a proof, that the residue fields are the same is the definition of totally ramified extension of local fields (what needs is a proof is that it is related to the ramification in the field they are the completion of). For tamely ramified it means it becomes a Galois extension after an unramified extension.
$endgroup$
– reuns
Dec 9 '18 at 23:31
$begingroup$
@reuns:Possibly I missunderstand you but is this what is to proof, insn't it? As you pointed out, a totally ramified extension of lfs is characterized by property that the residue fields for both fields coinside. We know that $k_F=k$. But is it obvoiusly that also $k_M=k$? Which completion wrt which valuation do you mean?
$endgroup$
– KarlPeter
Dec 9 '18 at 23:44
$begingroup$
Sorry for the maybe stupid question but now I have a huge lack of knowledge on this topic
$endgroup$
– KarlPeter
Dec 9 '18 at 23:46