Tame ramification for non-henselian fields
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I have a question concerning the remark after defintion (7.6) in Chapter 2 of Neukirch’s Algebraic number theory. This is the defintion:

($T$ denotes the maximal unramified subextension of $L$)
And this is the remark:
Namely, he writes that we can use this definition for non-henselian fields as well. But the existence of $T$ relies on henselianity (at least, the proof in book uses it). So what does he really mean and what is the „correct“ definition?
number-theory algebraic-number-theory
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I have a question concerning the remark after defintion (7.6) in Chapter 2 of Neukirch’s Algebraic number theory. This is the defintion:

($T$ denotes the maximal unramified subextension of $L$)
And this is the remark:
Namely, he writes that we can use this definition for non-henselian fields as well. But the existence of $T$ relies on henselianity (at least, the proof in book uses it). So what does he really mean and what is the „correct“ definition?
number-theory algebraic-number-theory
1
Isn’t it the case that when $v$ has a unique extension to $L$ that there is a maximal unramified of $K$?
– Lubin
Nov 24 at 20:57
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have a question concerning the remark after defintion (7.6) in Chapter 2 of Neukirch’s Algebraic number theory. This is the defintion:

($T$ denotes the maximal unramified subextension of $L$)
And this is the remark:
Namely, he writes that we can use this definition for non-henselian fields as well. But the existence of $T$ relies on henselianity (at least, the proof in book uses it). So what does he really mean and what is the „correct“ definition?
number-theory algebraic-number-theory
I have a question concerning the remark after defintion (7.6) in Chapter 2 of Neukirch’s Algebraic number theory. This is the defintion:

($T$ denotes the maximal unramified subextension of $L$)
And this is the remark:
Namely, he writes that we can use this definition for non-henselian fields as well. But the existence of $T$ relies on henselianity (at least, the proof in book uses it). So what does he really mean and what is the „correct“ definition?
number-theory algebraic-number-theory
number-theory algebraic-number-theory
edited Nov 27 at 1:56
asked Nov 24 at 16:19
Gregg
888
888
1
Isn’t it the case that when $v$ has a unique extension to $L$ that there is a maximal unramified of $K$?
– Lubin
Nov 24 at 20:57
add a comment |
1
Isn’t it the case that when $v$ has a unique extension to $L$ that there is a maximal unramified of $K$?
– Lubin
Nov 24 at 20:57
1
1
Isn’t it the case that when $v$ has a unique extension to $L$ that there is a maximal unramified of $K$?
– Lubin
Nov 24 at 20:57
Isn’t it the case that when $v$ has a unique extension to $L$ that there is a maximal unramified of $K$?
– Lubin
Nov 24 at 20:57
add a comment |
1 Answer
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1
down vote
accepted
Just to give a somewhat complete answer, I will expand on the comment of Prof. Lubin.
As he pointed out, it is indeed true that if the valuation admits a unique extension, then there is the maximal unramified subextension, i.e. the composite of all unramified subextensions. To prove this, one can follow the general strategy of the proof of Proposition (7.2) in Neukirch‘s book and use the following observations:
- If the valuation admits a unique extension to an algebraic extension, then it does so for any subextension.
- In this case, the integral closure of the valuation ring is the valuation ring of the extension.
- Unramified extensions are separable.
- A separable extension generated by $alpha$ admits a unique extension of the valuation iff its minimal polynomial is irreducible over the completion.
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
Just to give a somewhat complete answer, I will expand on the comment of Prof. Lubin.
As he pointed out, it is indeed true that if the valuation admits a unique extension, then there is the maximal unramified subextension, i.e. the composite of all unramified subextensions. To prove this, one can follow the general strategy of the proof of Proposition (7.2) in Neukirch‘s book and use the following observations:
- If the valuation admits a unique extension to an algebraic extension, then it does so for any subextension.
- In this case, the integral closure of the valuation ring is the valuation ring of the extension.
- Unramified extensions are separable.
- A separable extension generated by $alpha$ admits a unique extension of the valuation iff its minimal polynomial is irreducible over the completion.
add a comment |
up vote
1
down vote
accepted
Just to give a somewhat complete answer, I will expand on the comment of Prof. Lubin.
As he pointed out, it is indeed true that if the valuation admits a unique extension, then there is the maximal unramified subextension, i.e. the composite of all unramified subextensions. To prove this, one can follow the general strategy of the proof of Proposition (7.2) in Neukirch‘s book and use the following observations:
- If the valuation admits a unique extension to an algebraic extension, then it does so for any subextension.
- In this case, the integral closure of the valuation ring is the valuation ring of the extension.
- Unramified extensions are separable.
- A separable extension generated by $alpha$ admits a unique extension of the valuation iff its minimal polynomial is irreducible over the completion.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Just to give a somewhat complete answer, I will expand on the comment of Prof. Lubin.
As he pointed out, it is indeed true that if the valuation admits a unique extension, then there is the maximal unramified subextension, i.e. the composite of all unramified subextensions. To prove this, one can follow the general strategy of the proof of Proposition (7.2) in Neukirch‘s book and use the following observations:
- If the valuation admits a unique extension to an algebraic extension, then it does so for any subextension.
- In this case, the integral closure of the valuation ring is the valuation ring of the extension.
- Unramified extensions are separable.
- A separable extension generated by $alpha$ admits a unique extension of the valuation iff its minimal polynomial is irreducible over the completion.
Just to give a somewhat complete answer, I will expand on the comment of Prof. Lubin.
As he pointed out, it is indeed true that if the valuation admits a unique extension, then there is the maximal unramified subextension, i.e. the composite of all unramified subextensions. To prove this, one can follow the general strategy of the proof of Proposition (7.2) in Neukirch‘s book and use the following observations:
- If the valuation admits a unique extension to an algebraic extension, then it does so for any subextension.
- In this case, the integral closure of the valuation ring is the valuation ring of the extension.
- Unramified extensions are separable.
- A separable extension generated by $alpha$ admits a unique extension of the valuation iff its minimal polynomial is irreducible over the completion.
edited Nov 30 at 14:31
answered Nov 27 at 1:53
Gregg
888
888
add a comment |
add a comment |
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1
Isn’t it the case that when $v$ has a unique extension to $L$ that there is a maximal unramified of $K$?
– Lubin
Nov 24 at 20:57