Why do minimal polynomial have coefficients in the integral closured ring?
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I'm reading the following theorem:
My question is, why are $a_i$s in $A[alpha_1...alpha_n]$?
abstract-algebra field-theory algebraic-number-theory
add a comment |Â
up vote
2
down vote
favorite
I'm reading the following theorem:
My question is, why are $a_i$s in $A[alpha_1...alpha_n]$?
abstract-algebra field-theory algebraic-number-theory
1
The as are the result of evaluating the elementary symmetric functions on the alphas.
â Mariano Suárez-Ãlvarez
Feb 27 '17 at 4:47
I guess I know the answer now...how stupid I am...
â CYC
Feb 27 '17 at 4:57
3
I suggest you write a complete answer answering your question.
â Mariano Suárez-Ãlvarez
Feb 27 '17 at 5:25
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm reading the following theorem:
My question is, why are $a_i$s in $A[alpha_1...alpha_n]$?
abstract-algebra field-theory algebraic-number-theory
I'm reading the following theorem:
My question is, why are $a_i$s in $A[alpha_1...alpha_n]$?
abstract-algebra field-theory algebraic-number-theory
abstract-algebra field-theory algebraic-number-theory
asked Feb 27 '17 at 4:39
CYC
940711
940711
1
The as are the result of evaluating the elementary symmetric functions on the alphas.
â Mariano Suárez-Ãlvarez
Feb 27 '17 at 4:47
I guess I know the answer now...how stupid I am...
â CYC
Feb 27 '17 at 4:57
3
I suggest you write a complete answer answering your question.
â Mariano Suárez-Ãlvarez
Feb 27 '17 at 5:25
add a comment |Â
1
The as are the result of evaluating the elementary symmetric functions on the alphas.
â Mariano Suárez-Ãlvarez
Feb 27 '17 at 4:47
I guess I know the answer now...how stupid I am...
â CYC
Feb 27 '17 at 4:57
3
I suggest you write a complete answer answering your question.
â Mariano Suárez-Ãlvarez
Feb 27 '17 at 5:25
1
1
The as are the result of evaluating the elementary symmetric functions on the alphas.
â Mariano Suárez-Ãlvarez
Feb 27 '17 at 4:47
The as are the result of evaluating the elementary symmetric functions on the alphas.
â Mariano Suárez-Ãlvarez
Feb 27 '17 at 4:47
I guess I know the answer now...how stupid I am...
â CYC
Feb 27 '17 at 4:57
I guess I know the answer now...how stupid I am...
â CYC
Feb 27 '17 at 4:57
3
3
I suggest you write a complete answer answering your question.
â Mariano Suárez-Ãlvarez
Feb 27 '17 at 5:25
I suggest you write a complete answer answering your question.
â Mariano Suárez-Ãlvarez
Feb 27 '17 at 5:25
add a comment |Â
1 Answer
1
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up vote
0
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accepted
There's an $sigma$ sending $alpha$ to $alpha_i$ fixing $K$, hence $alpha_i$ are integral over $A$ since $alpha$ is, the rest is easy.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
There's an $sigma$ sending $alpha$ to $alpha_i$ fixing $K$, hence $alpha_i$ are integral over $A$ since $alpha$ is, the rest is easy.
add a comment |Â
up vote
0
down vote
accepted
There's an $sigma$ sending $alpha$ to $alpha_i$ fixing $K$, hence $alpha_i$ are integral over $A$ since $alpha$ is, the rest is easy.
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
There's an $sigma$ sending $alpha$ to $alpha_i$ fixing $K$, hence $alpha_i$ are integral over $A$ since $alpha$ is, the rest is easy.
There's an $sigma$ sending $alpha$ to $alpha_i$ fixing $K$, hence $alpha_i$ are integral over $A$ since $alpha$ is, the rest is easy.
answered Sep 9 at 11:02
CYC
940711
940711
add a comment |Â
add a comment |Â
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1
The as are the result of evaluating the elementary symmetric functions on the alphas.
â Mariano Suárez-Ãlvarez
Feb 27 '17 at 4:47
I guess I know the answer now...how stupid I am...
â CYC
Feb 27 '17 at 4:57
3
I suggest you write a complete answer answering your question.
â Mariano Suárez-Ãlvarez
Feb 27 '17 at 5:25