decomposition group and inertia group, the minimal polynomial,surjectivity of the map $D_{M/P}rightarrow Gal$
$begingroup$
Can anyone explain the underlined sentence?
For notation, A:Dedekind domain, K=Frac(A), L/K:Galois extension, B:The integral closure of A in L, M:A maximal ideal of B, P:The intersection of M and A (hence the maximal ideal of A), $D_{M/P}$ :the decomposition group.
I reckon the way we take $alpha$ is the key, but cannot make it to the conclusion, 'we find that the only non-zero roots of...'.
I read some of the close questions already answered but none of them was using this type of logic.
Thank you in advance.
abstract-algebra number-theory galois-theory algebraic-number-theory arithmetic-geometry
asked Dec 4 '18 at 2:13
KentoKento
473
$endgroup$
add a comment |
$begingroup$
Can anyone explain the underlined sentence?
For notation, A:Dedekind domain, K=Frac(A), L/K:Galois extension, B:The integral closure of A in L, M:A maximal ideal of B, P:The intersection of M and A (hence the maximal ideal of A), $D_{M/P}$ :the decomposition group.
I reckon the way we take $alpha$ is the key, but cannot make it to the conclusion, 'we find that the only non-zero roots of...'.
I read some of the close questions already answered but none of them was using this type of logic.
Thank you in advance.
abstract-algebra number-theory galois-theory algebraic-number-theory arithmetic-geometry
asked Dec 4 '18 at 2:13
KentoKento
473
$endgroup$
add a comment |
$begingroup$
Can anyone explain the underlined sentence?
For notation, A:Dedekind domain, K=Frac(A), L/K:Galois extension, B:The integral closure of A in L, M:A maximal ideal of B, P:The intersection of M and A (hence the maximal ideal of A), $D_{M/P}$ :the decomposition group.
I reckon the way we take $alpha$ is the key, but cannot make it to the conclusion, 'we find that the only non-zero roots of...'.
I read some of the close questions already answered but none of them was using this type of logic.
Thank you in advance.
abstract-algebra number-theory galois-theory algebraic-number-theory arithmetic-geometry
asked Dec 4 '18 at 2:13
KentoKento
473
$endgroup$
Can anyone explain the underlined sentence?
For notation, A:Dedekind domain, K=Frac(A), L/K:Galois extension, B:The integral closure of A in L, M:A maximal ideal of B, P:The intersection of M and A (hence the maximal ideal of A), $D_{M/P}$ :the decomposition group.
I reckon the way we take $alpha$ is the key, but cannot make it to the conclusion, 'we find that the only non-zero roots of...'.
I read some of the close questions already answered but none of them was using this type of logic.
Thank you in advance.
abstract-algebra number-theory galois-theory algebraic-number-theory arithmetic-geometry
abstract-algebra number-theory galois-theory algebraic-number-theory arithmetic-geometry
asked Dec 4 '18 at 2:13
KentoKento
473
asked Dec 4 '18 at 2:13
KentoKento
473
edited Dec 4 '18 at 14:04
Kento
asked Dec 4 '18 at 2:13
KentoKento
473
asked Dec 4 '18 at 2:13
KentoKento
473
asked Dec 4 '18 at 2:13
KentoKento
473
473
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Let me expand on the highlighted part:
$g(y)$ is the min. polynomial of $overline{alpha}$ over $A/P$, so it has to divide the polynomial $overline{f}(y)=f(y) ,mathrm{mod}, P$, since $overline{alpha}$ is a root of $overline{f}(y)$ (and $overline{f}(y)$ is nonzero, take $f(y)$ monic). From this and the expression $f(y)=prod_H(y-sigma(alpha))$ it follows that the roots of $g(y)$ are just some of the roots $sigma(alpha)$ taken modulo $M$, and the goal is to identify which ones.
Now $alpha$ was chosen so that $alpha in sigma(M)$ whenever $sigma notin D_{M/P}$, i.e. $sigma(M)neq M$. Applying $sigma^{-1}$, we have that $sigma^{-1}(alpha) in M$ whenever $sigma(M)neq M$. Changing $sigma^{-1}$ to $sigma$ (note that $sigma^{-1} notin D_{M/P}$ iff $sigma notin D_{M/P}$), we have that $sigma(alpha) in M $ whenever $sigma notin D_{M/P}$. And conversely, we have $alpha notin M$ (because $overline{alpha} neq 0$), so given any $sigma in D_{M/P}$, we have that $sigma(alpha) notin sigma(M)=M$. So altogether: $sigma(alpha) ,mathrm{mod},M$ is nonzero iff $sigma in D_{M/P}$. So the roots of $g(y)$ can come only from these, i.e. in the form $overline{sigma}(overline{alpha})$ (because $g(y)$ cannot have $0$ as a root, it's the min. poly. of $overline{alpha}$). And all of them has to be roots for Galois reasons (all the maps $overline{sigma}$ are elements of the Galois group of the residue field, and $overline{alpha}$ is a root of $g(y)$).
Hope this helps.
answered Dec 5 '18 at 5:01
Pavel ČoupekPavel Čoupek
4,47611126
$endgroup$
$begingroup$
thank you very much. i think i understand .
$endgroup$
– Kento
Dec 5 '18 at 8:31
add a comment |
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$begingroup$
Let me expand on the highlighted part:
$g(y)$ is the min. polynomial of $overline{alpha}$ over $A/P$, so it has to divide the polynomial $overline{f}(y)=f(y) ,mathrm{mod}, P$, since $overline{alpha}$ is a root of $overline{f}(y)$ (and $overline{f}(y)$ is nonzero, take $f(y)$ monic). From this and the expression $f(y)=prod_H(y-sigma(alpha))$ it follows that the roots of $g(y)$ are just some of the roots $sigma(alpha)$ taken modulo $M$, and the goal is to identify which ones.
Now $alpha$ was chosen so that $alpha in sigma(M)$ whenever $sigma notin D_{M/P}$, i.e. $sigma(M)neq M$. Applying $sigma^{-1}$, we have that $sigma^{-1}(alpha) in M$ whenever $sigma(M)neq M$. Changing $sigma^{-1}$ to $sigma$ (note that $sigma^{-1} notin D_{M/P}$ iff $sigma notin D_{M/P}$), we have that $sigma(alpha) in M $ whenever $sigma notin D_{M/P}$. And conversely, we have $alpha notin M$ (because $overline{alpha} neq 0$), so given any $sigma in D_{M/P}$, we have that $sigma(alpha) notin sigma(M)=M$. So altogether: $sigma(alpha) ,mathrm{mod},M$ is nonzero iff $sigma in D_{M/P}$. So the roots of $g(y)$ can come only from these, i.e. in the form $overline{sigma}(overline{alpha})$ (because $g(y)$ cannot have $0$ as a root, it's the min. poly. of $overline{alpha}$). And all of them has to be roots for Galois reasons (all the maps $overline{sigma}$ are elements of the Galois group of the residue field, and $overline{alpha}$ is a root of $g(y)$).
Hope this helps.
answered Dec 5 '18 at 5:01
Pavel ČoupekPavel Čoupek
4,47611126
$endgroup$
$begingroup$
thank you very much. i think i understand .
$endgroup$
– Kento
Dec 5 '18 at 8:31
add a comment |
$begingroup$
Let me expand on the highlighted part:
$g(y)$ is the min. polynomial of $overline{alpha}$ over $A/P$, so it has to divide the polynomial $overline{f}(y)=f(y) ,mathrm{mod}, P$, since $overline{alpha}$ is a root of $overline{f}(y)$ (and $overline{f}(y)$ is nonzero, take $f(y)$ monic). From this and the expression $f(y)=prod_H(y-sigma(alpha))$ it follows that the roots of $g(y)$ are just some of the roots $sigma(alpha)$ taken modulo $M$, and the goal is to identify which ones.
Now $alpha$ was chosen so that $alpha in sigma(M)$ whenever $sigma notin D_{M/P}$, i.e. $sigma(M)neq M$. Applying $sigma^{-1}$, we have that $sigma^{-1}(alpha) in M$ whenever $sigma(M)neq M$. Changing $sigma^{-1}$ to $sigma$ (note that $sigma^{-1} notin D_{M/P}$ iff $sigma notin D_{M/P}$), we have that $sigma(alpha) in M $ whenever $sigma notin D_{M/P}$. And conversely, we have $alpha notin M$ (because $overline{alpha} neq 0$), so given any $sigma in D_{M/P}$, we have that $sigma(alpha) notin sigma(M)=M$. So altogether: $sigma(alpha) ,mathrm{mod},M$ is nonzero iff $sigma in D_{M/P}$. So the roots of $g(y)$ can come only from these, i.e. in the form $overline{sigma}(overline{alpha})$ (because $g(y)$ cannot have $0$ as a root, it's the min. poly. of $overline{alpha}$). And all of them has to be roots for Galois reasons (all the maps $overline{sigma}$ are elements of the Galois group of the residue field, and $overline{alpha}$ is a root of $g(y)$).
Hope this helps.
answered Dec 5 '18 at 5:01
Pavel ČoupekPavel Čoupek
4,47611126
$endgroup$
$begingroup$
thank you very much. i think i understand .
$endgroup$
– Kento
Dec 5 '18 at 8:31
add a comment |
$begingroup$
Let me expand on the highlighted part:
$g(y)$ is the min. polynomial of $overline{alpha}$ over $A/P$, so it has to divide the polynomial $overline{f}(y)=f(y) ,mathrm{mod}, P$, since $overline{alpha}$ is a root of $overline{f}(y)$ (and $overline{f}(y)$ is nonzero, take $f(y)$ monic). From this and the expression $f(y)=prod_H(y-sigma(alpha))$ it follows that the roots of $g(y)$ are just some of the roots $sigma(alpha)$ taken modulo $M$, and the goal is to identify which ones.
Now $alpha$ was chosen so that $alpha in sigma(M)$ whenever $sigma notin D_{M/P}$, i.e. $sigma(M)neq M$. Applying $sigma^{-1}$, we have that $sigma^{-1}(alpha) in M$ whenever $sigma(M)neq M$. Changing $sigma^{-1}$ to $sigma$ (note that $sigma^{-1} notin D_{M/P}$ iff $sigma notin D_{M/P}$), we have that $sigma(alpha) in M $ whenever $sigma notin D_{M/P}$. And conversely, we have $alpha notin M$ (because $overline{alpha} neq 0$), so given any $sigma in D_{M/P}$, we have that $sigma(alpha) notin sigma(M)=M$. So altogether: $sigma(alpha) ,mathrm{mod},M$ is nonzero iff $sigma in D_{M/P}$. So the roots of $g(y)$ can come only from these, i.e. in the form $overline{sigma}(overline{alpha})$ (because $g(y)$ cannot have $0$ as a root, it's the min. poly. of $overline{alpha}$). And all of them has to be roots for Galois reasons (all the maps $overline{sigma}$ are elements of the Galois group of the residue field, and $overline{alpha}$ is a root of $g(y)$).
Hope this helps.
answered Dec 5 '18 at 5:01
Pavel ČoupekPavel Čoupek
4,47611126
$endgroup$
Let me expand on the highlighted part:
$g(y)$ is the min. polynomial of $overline{alpha}$ over $A/P$, so it has to divide the polynomial $overline{f}(y)=f(y) ,mathrm{mod}, P$, since $overline{alpha}$ is a root of $overline{f}(y)$ (and $overline{f}(y)$ is nonzero, take $f(y)$ monic). From this and the expression $f(y)=prod_H(y-sigma(alpha))$ it follows that the roots of $g(y)$ are just some of the roots $sigma(alpha)$ taken modulo $M$, and the goal is to identify which ones.
Now $alpha$ was chosen so that $alpha in sigma(M)$ whenever $sigma notin D_{M/P}$, i.e. $sigma(M)neq M$. Applying $sigma^{-1}$, we have that $sigma^{-1}(alpha) in M$ whenever $sigma(M)neq M$. Changing $sigma^{-1}$ to $sigma$ (note that $sigma^{-1} notin D_{M/P}$ iff $sigma notin D_{M/P}$), we have that $sigma(alpha) in M $ whenever $sigma notin D_{M/P}$. And conversely, we have $alpha notin M$ (because $overline{alpha} neq 0$), so given any $sigma in D_{M/P}$, we have that $sigma(alpha) notin sigma(M)=M$. So altogether: $sigma(alpha) ,mathrm{mod},M$ is nonzero iff $sigma in D_{M/P}$. So the roots of $g(y)$ can come only from these, i.e. in the form $overline{sigma}(overline{alpha})$ (because $g(y)$ cannot have $0$ as a root, it's the min. poly. of $overline{alpha}$). And all of them has to be roots for Galois reasons (all the maps $overline{sigma}$ are elements of the Galois group of the residue field, and $overline{alpha}$ is a root of $g(y)$).
Hope this helps.
answered Dec 5 '18 at 5:01
Pavel ČoupekPavel Čoupek
4,47611126
edited Dec 10 '18 at 21:44
answered Dec 5 '18 at 5:01
Pavel ČoupekPavel Čoupek
4,47611126
answered Dec 5 '18 at 5:01
Pavel ČoupekPavel Čoupek
4,47611126
answered Dec 5 '18 at 5:01
Pavel ČoupekPavel Čoupek
4,47611126
4,47611126
$begingroup$
thank you very much. i think i understand .
$endgroup$
– Kento
Dec 5 '18 at 8:31
add a comment |
$begingroup$
thank you very much. i think i understand .
$endgroup$
– Kento
Dec 5 '18 at 8:31
$begingroup$
thank you very much. i think i understand .
$endgroup$
– Kento
Dec 5 '18 at 8:31
$begingroup$
thank you very much. i think i understand .
$endgroup$
– Kento
Dec 5 '18 at 8:31
add a comment |
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