Double cosets (Neukirch's Algebraic Number Theory)
$begingroup$
This is a question from Neukirch's Algebraic Number Theory, Ch.1 $S$9.
Let $A$ be a Dedekind domain with quotient field $K$, $L$ a finite separable field extension of $K$ and $B$ the integral closure of $A$ in $L$. Let $N/K$ be a Galois closure of $L/K$ and set $G=operatorname{Gal}(N/K)$, $H=operatorname{Gal}(N/L)$.
Let $mathfrak p$ be a prime ideal of $A$ and $P_mathfrak p$ the set of prime ideals of $L$ above $mathfrak p$. If $mathfrak P$ is a prime ideal of $N$ above $mathfrak p$ let $G_mathfrak P$ be the decomposition subgroup of $mathfrak P$ over $K$.
Consider the map $Hbackslash G/G_mathfrak Prightarrow P_mathfrak p$ defined by $Hsigma G_mathfrak P mapsto sigma mathfrak Pcap L$.
Question: how do I show that it is bijective?
I can show that it is well-defined and injective but surjection seems more tricky. I know that the number of distinct primes in $N$ above $mathfrak p$ is $[G:G_mathfrak P]$. Does this imply that the two sets are of equal size?
Many thanks.
abstract-algebra ring-theory algebraic-number-theory galois-theory
edited Jan 12 at 13:05
darij grinberg
11.2k33167
asked Apr 14 '16 at 13:08
eddieeddie
447210
$endgroup$
add a comment |
$begingroup$
This is a question from Neukirch's Algebraic Number Theory, Ch.1 $S$9.
Let $A$ be a Dedekind domain with quotient field $K$, $L$ a finite separable field extension of $K$ and $B$ the integral closure of $A$ in $L$. Let $N/K$ be a Galois closure of $L/K$ and set $G=operatorname{Gal}(N/K)$, $H=operatorname{Gal}(N/L)$.
Let $mathfrak p$ be a prime ideal of $A$ and $P_mathfrak p$ the set of prime ideals of $L$ above $mathfrak p$. If $mathfrak P$ is a prime ideal of $N$ above $mathfrak p$ let $G_mathfrak P$ be the decomposition subgroup of $mathfrak P$ over $K$.
Consider the map $Hbackslash G/G_mathfrak Prightarrow P_mathfrak p$ defined by $Hsigma G_mathfrak P mapsto sigma mathfrak Pcap L$.
Question: how do I show that it is bijective?
I can show that it is well-defined and injective but surjection seems more tricky. I know that the number of distinct primes in $N$ above $mathfrak p$ is $[G:G_mathfrak P]$. Does this imply that the two sets are of equal size?
Many thanks.
abstract-algebra ring-theory algebraic-number-theory galois-theory
edited Jan 12 at 13:05
darij grinberg
11.2k33167
asked Apr 14 '16 at 13:08
eddieeddie
447210
$endgroup$
add a comment |
$begingroup$
This is a question from Neukirch's Algebraic Number Theory, Ch.1 $S$9.
Let $A$ be a Dedekind domain with quotient field $K$, $L$ a finite separable field extension of $K$ and $B$ the integral closure of $A$ in $L$. Let $N/K$ be a Galois closure of $L/K$ and set $G=operatorname{Gal}(N/K)$, $H=operatorname{Gal}(N/L)$.
Let $mathfrak p$ be a prime ideal of $A$ and $P_mathfrak p$ the set of prime ideals of $L$ above $mathfrak p$. If $mathfrak P$ is a prime ideal of $N$ above $mathfrak p$ let $G_mathfrak P$ be the decomposition subgroup of $mathfrak P$ over $K$.
Consider the map $Hbackslash G/G_mathfrak Prightarrow P_mathfrak p$ defined by $Hsigma G_mathfrak P mapsto sigma mathfrak Pcap L$.
Question: how do I show that it is bijective?
I can show that it is well-defined and injective but surjection seems more tricky. I know that the number of distinct primes in $N$ above $mathfrak p$ is $[G:G_mathfrak P]$. Does this imply that the two sets are of equal size?
Many thanks.
abstract-algebra ring-theory algebraic-number-theory galois-theory
edited Jan 12 at 13:05
darij grinberg
11.2k33167
asked Apr 14 '16 at 13:08
eddieeddie
447210
$endgroup$
This is a question from Neukirch's Algebraic Number Theory, Ch.1 $S$9.
Let $A$ be a Dedekind domain with quotient field $K$, $L$ a finite separable field extension of $K$ and $B$ the integral closure of $A$ in $L$. Let $N/K$ be a Galois closure of $L/K$ and set $G=operatorname{Gal}(N/K)$, $H=operatorname{Gal}(N/L)$.
Let $mathfrak p$ be a prime ideal of $A$ and $P_mathfrak p$ the set of prime ideals of $L$ above $mathfrak p$. If $mathfrak P$ is a prime ideal of $N$ above $mathfrak p$ let $G_mathfrak P$ be the decomposition subgroup of $mathfrak P$ over $K$.
Consider the map $Hbackslash G/G_mathfrak Prightarrow P_mathfrak p$ defined by $Hsigma G_mathfrak P mapsto sigma mathfrak Pcap L$.
Question: how do I show that it is bijective?
I can show that it is well-defined and injective but surjection seems more tricky. I know that the number of distinct primes in $N$ above $mathfrak p$ is $[G:G_mathfrak P]$. Does this imply that the two sets are of equal size?
Many thanks.
abstract-algebra ring-theory algebraic-number-theory galois-theory
abstract-algebra ring-theory algebraic-number-theory galois-theory
edited Jan 12 at 13:05
darij grinberg
11.2k33167
asked Apr 14 '16 at 13:08
eddieeddie
447210
edited Jan 12 at 13:05
darij grinberg
11.2k33167
asked Apr 14 '16 at 13:08
eddieeddie
447210
edited Jan 12 at 13:05
darij grinberg
11.2k33167
edited Jan 12 at 13:05
darij grinberg
11.2k33167
edited Jan 12 at 13:05
darij grinberg
11.2k33167
11.2k33167
asked Apr 14 '16 at 13:08
eddieeddie
447210
asked Apr 14 '16 at 13:08
eddieeddie
447210
asked Apr 14 '16 at 13:08
eddieeddie
447210
447210
add a comment |
add a comment |
1 Answer
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$begingroup$
Surjectivity is easy, as clearly when $sigma$ traverses all the cosets of the decomposition group of $G_P$, $G_Pcap L$ traverses all prime ideals above $p$ in $L$, as each such prime has some prime $P$ in $N$ above it.
Injectivity follows from the fact that if $P$ and $Q$ are two different primes in $N$ above $p$ having the same intersection with $L$, then if $sigma(P) = Q$, we can show that $sigma$ is in the same doubled coset as the identity. Proof is as follows: Denote by $q$ the intersection of $P$ with $L$ ($Pcap L = Qcap L = q$), then since $N|L$ is Galois, and $P,Q$ are above $q$, there is an element of $H$, which stabilizes $L$ and takes $P$ to $Q$, let us denote it $tau$. Then $tau^{-1}sigma(P) = P$, implying that $tau^{-1}sigmain G_P$ (apologies for not knowing how to write Gothic $P$). We therefore have: $sigmain tau G_P subseteq HG_P$, which is the doubled coset of the identity.
answered Jan 12 at 12:47
kindasortakindasorta
9810
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add a comment |
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$begingroup$
Surjectivity is easy, as clearly when $sigma$ traverses all the cosets of the decomposition group of $G_P$, $G_Pcap L$ traverses all prime ideals above $p$ in $L$, as each such prime has some prime $P$ in $N$ above it.
Injectivity follows from the fact that if $P$ and $Q$ are two different primes in $N$ above $p$ having the same intersection with $L$, then if $sigma(P) = Q$, we can show that $sigma$ is in the same doubled coset as the identity. Proof is as follows: Denote by $q$ the intersection of $P$ with $L$ ($Pcap L = Qcap L = q$), then since $N|L$ is Galois, and $P,Q$ are above $q$, there is an element of $H$, which stabilizes $L$ and takes $P$ to $Q$, let us denote it $tau$. Then $tau^{-1}sigma(P) = P$, implying that $tau^{-1}sigmain G_P$ (apologies for not knowing how to write Gothic $P$). We therefore have: $sigmain tau G_P subseteq HG_P$, which is the doubled coset of the identity.
answered Jan 12 at 12:47
kindasortakindasorta
9810
$endgroup$
add a comment |
$begingroup$
Surjectivity is easy, as clearly when $sigma$ traverses all the cosets of the decomposition group of $G_P$, $G_Pcap L$ traverses all prime ideals above $p$ in $L$, as each such prime has some prime $P$ in $N$ above it.
Injectivity follows from the fact that if $P$ and $Q$ are two different primes in $N$ above $p$ having the same intersection with $L$, then if $sigma(P) = Q$, we can show that $sigma$ is in the same doubled coset as the identity. Proof is as follows: Denote by $q$ the intersection of $P$ with $L$ ($Pcap L = Qcap L = q$), then since $N|L$ is Galois, and $P,Q$ are above $q$, there is an element of $H$, which stabilizes $L$ and takes $P$ to $Q$, let us denote it $tau$. Then $tau^{-1}sigma(P) = P$, implying that $tau^{-1}sigmain G_P$ (apologies for not knowing how to write Gothic $P$). We therefore have: $sigmain tau G_P subseteq HG_P$, which is the doubled coset of the identity.
answered Jan 12 at 12:47
kindasortakindasorta
9810
$endgroup$
add a comment |
$begingroup$
Surjectivity is easy, as clearly when $sigma$ traverses all the cosets of the decomposition group of $G_P$, $G_Pcap L$ traverses all prime ideals above $p$ in $L$, as each such prime has some prime $P$ in $N$ above it.
Injectivity follows from the fact that if $P$ and $Q$ are two different primes in $N$ above $p$ having the same intersection with $L$, then if $sigma(P) = Q$, we can show that $sigma$ is in the same doubled coset as the identity. Proof is as follows: Denote by $q$ the intersection of $P$ with $L$ ($Pcap L = Qcap L = q$), then since $N|L$ is Galois, and $P,Q$ are above $q$, there is an element of $H$, which stabilizes $L$ and takes $P$ to $Q$, let us denote it $tau$. Then $tau^{-1}sigma(P) = P$, implying that $tau^{-1}sigmain G_P$ (apologies for not knowing how to write Gothic $P$). We therefore have: $sigmain tau G_P subseteq HG_P$, which is the doubled coset of the identity.
answered Jan 12 at 12:47
kindasortakindasorta
9810
$endgroup$
Surjectivity is easy, as clearly when $sigma$ traverses all the cosets of the decomposition group of $G_P$, $G_Pcap L$ traverses all prime ideals above $p$ in $L$, as each such prime has some prime $P$ in $N$ above it.
Injectivity follows from the fact that if $P$ and $Q$ are two different primes in $N$ above $p$ having the same intersection with $L$, then if $sigma(P) = Q$, we can show that $sigma$ is in the same doubled coset as the identity. Proof is as follows: Denote by $q$ the intersection of $P$ with $L$ ($Pcap L = Qcap L = q$), then since $N|L$ is Galois, and $P,Q$ are above $q$, there is an element of $H$, which stabilizes $L$ and takes $P$ to $Q$, let us denote it $tau$. Then $tau^{-1}sigma(P) = P$, implying that $tau^{-1}sigmain G_P$ (apologies for not knowing how to write Gothic $P$). We therefore have: $sigmain tau G_P subseteq HG_P$, which is the doubled coset of the identity.
answered Jan 12 at 12:47
kindasortakindasorta
9810
answered Jan 12 at 12:47
kindasortakindasorta
9810
answered Jan 12 at 12:47
kindasortakindasorta
9810
answered Jan 12 at 12:47
kindasortakindasorta
9810
9810
add a comment |
add a comment |
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