Mixing
Extension of valuation in iterated Laurent series fields
$begingroup$
according to well known theorems, every valuation of a field $K$ can be extended to a valuation of a field extension $L$ of $K$ and this can be explicit in the case of finite extensions.
My question is how explicit this can be made in the case of an iterated Laurent series field: If we have $K=F((t_1))$ for a field $F$ and $L=K((t_2))=F((t_1))((t_2))$, how can the extensions of the $t_1$-adic valuation on $L$ look like? Is there something that can be said about the valuation on $L$ (archimedean? complete? uniformizer...)?
Thanks!
valuation-theory
edited Jan 23 at 14:52
Bernard
122k741116
asked Jan 23 at 14:14
nicosiekirkanicosiekirka
213
$endgroup$
add a comment |
$begingroup$
according to well known theorems, every valuation of a field $K$ can be extended to a valuation of a field extension $L$ of $K$ and this can be explicit in the case of finite extensions.
My question is how explicit this can be made in the case of an iterated Laurent series field: If we have $K=F((t_1))$ for a field $F$ and $L=K((t_2))=F((t_1))((t_2))$, how can the extensions of the $t_1$-adic valuation on $L$ look like? Is there something that can be said about the valuation on $L$ (archimedean? complete? uniformizer...)?
Thanks!
valuation-theory
edited Jan 23 at 14:52
Bernard
122k741116
asked Jan 23 at 14:14
nicosiekirkanicosiekirka
213
$endgroup$
$begingroup$
The $t_1$ valuation on $F((t_1))$ extends in an obvious way to $E((t_1))$ for any field extension $E/F$ such that every $a in F((t_1)), a not in F$ is transcendental over $E$, so the goal is to find (using a transcendental basis of $F((t_1))((t_2))/F$ ?) such a $E$ such that $F((t_1))((t_2))cong E((t_1))$, and the difficulty is that $E$ is not $F((t_2))$
$endgroup$
– reuns
Jan 23 at 16:28
add a comment |
$begingroup$
according to well known theorems, every valuation of a field $K$ can be extended to a valuation of a field extension $L$ of $K$ and this can be explicit in the case of finite extensions.
My question is how explicit this can be made in the case of an iterated Laurent series field: If we have $K=F((t_1))$ for a field $F$ and $L=K((t_2))=F((t_1))((t_2))$, how can the extensions of the $t_1$-adic valuation on $L$ look like? Is there something that can be said about the valuation on $L$ (archimedean? complete? uniformizer...)?
Thanks!
valuation-theory
edited Jan 23 at 14:52
Bernard
122k741116
asked Jan 23 at 14:14
nicosiekirkanicosiekirka
213
$endgroup$
according to well known theorems, every valuation of a field $K$ can be extended to a valuation of a field extension $L$ of $K$ and this can be explicit in the case of finite extensions.
My question is how explicit this can be made in the case of an iterated Laurent series field: If we have $K=F((t_1))$ for a field $F$ and $L=K((t_2))=F((t_1))((t_2))$, how can the extensions of the $t_1$-adic valuation on $L$ look like? Is there something that can be said about the valuation on $L$ (archimedean? complete? uniformizer...)?
Thanks!
valuation-theory
valuation-theory
edited Jan 23 at 14:52
Bernard
122k741116
asked Jan 23 at 14:14
nicosiekirkanicosiekirka
213
edited Jan 23 at 14:52
Bernard
122k741116
asked Jan 23 at 14:14
nicosiekirkanicosiekirka
213
edited Jan 23 at 14:52
Bernard
122k741116
edited Jan 23 at 14:52
Bernard
122k741116
edited Jan 23 at 14:52
Bernard
122k741116
122k741116
asked Jan 23 at 14:14
nicosiekirkanicosiekirka
213
asked Jan 23 at 14:14
nicosiekirkanicosiekirka
213
asked Jan 23 at 14:14
nicosiekirkanicosiekirka
213
213
$begingroup$
The $t_1$ valuation on $F((t_1))$ extends in an obvious way to $E((t_1))$ for any field extension $E/F$ such that every $a in F((t_1)), a not in F$ is transcendental over $E$, so the goal is to find (using a transcendental basis of $F((t_1))((t_2))/F$ ?) such a $E$ such that $F((t_1))((t_2))cong E((t_1))$, and the difficulty is that $E$ is not $F((t_2))$
$endgroup$
– reuns
Jan 23 at 16:28
add a comment |
$begingroup$
The $t_1$ valuation on $F((t_1))$ extends in an obvious way to $E((t_1))$ for any field extension $E/F$ such that every $a in F((t_1)), a not in F$ is transcendental over $E$, so the goal is to find (using a transcendental basis of $F((t_1))((t_2))/F$ ?) such a $E$ such that $F((t_1))((t_2))cong E((t_1))$, and the difficulty is that $E$ is not $F((t_2))$
$endgroup$
– reuns
Jan 23 at 16:28
$begingroup$
The $t_1$ valuation on $F((t_1))$ extends in an obvious way to $E((t_1))$ for any field extension $E/F$ such that every $a in F((t_1)), a not in F$ is transcendental over $E$, so the goal is to find (using a transcendental basis of $F((t_1))((t_2))/F$ ?) such a $E$ such that $F((t_1))((t_2))cong E((t_1))$, and the difficulty is that $E$ is not $F((t_2))$
$endgroup$
– reuns
Jan 23 at 16:28
$begingroup$
The $t_1$ valuation on $F((t_1))$ extends in an obvious way to $E((t_1))$ for any field extension $E/F$ such that every $a in F((t_1)), a not in F$ is transcendental over $E$, so the goal is to find (using a transcendental basis of $F((t_1))((t_2))/F$ ?) such a $E$ such that $F((t_1))((t_2))cong E((t_1))$, and the difficulty is that $E$ is not $F((t_2))$
$endgroup$
– reuns
Jan 23 at 16:28
add a comment |
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$begingroup$
The $t_1$ valuation on $F((t_1))$ extends in an obvious way to $E((t_1))$ for any field extension $E/F$ such that every $a in F((t_1)), a not in F$ is transcendental over $E$, so the goal is to find (using a transcendental basis of $F((t_1))((t_2))/F$ ?) such a $E$ such that $F((t_1))((t_2))cong E((t_1))$, and the difficulty is that $E$ is not $F((t_2))$
$endgroup$
– reuns
Jan 23 at 16:28