Mixing
Integrally Closed in Terms of Galois Theory
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Let $K$ be a finite Galois extension of $mathbb{Q}$ and let $A$ be a subring. We say $A$ is integrally closed in $K$ if any element $a in K$ integral over $A$ is contained in $A$.
Does being integrally closed have an interpretation in terms of the automorphisms of $K$?
I have come across a similar observation which goes as follows:
Lemma: Let $L/K$ be a Galois extension of fields, and let $A subset L$ be a subring. If $sigma (A) = A$ for each $sigma in [L, L]_K$, then $A$ is integral over $A cap K$.
Proof: Take $a in A$ and let $f in K[x]$ be the (monic) minimal polynomial for $a$ over $K$. $f(x)$ factors as $prod_{sigma in [L, L]_K }(x - sigma(a))$. Since each $sigma(a)$ belongs to $A$, the coefficients of $f$ belong to $A$, so that they belong to $A cap K$. Therefore $a$ is integral over $A cap K$.
I suspect there should be some way of expressing "integrally closed" with the automorphisms, but I can't quite figure out what it should be.
Of course, this would lead to a description of $mathcal{O}_K subset K$ for a Galois number field $K$ in terms of the Galois group, which is the case I am interested in.
number-theory galois-theory
asked Jan 20 at 3:29
Dean YoungDean Young
1,683721
$endgroup$
add a comment |
$begingroup$
Let $K$ be a finite Galois extension of $mathbb{Q}$ and let $A$ be a subring. We say $A$ is integrally closed in $K$ if any element $a in K$ integral over $A$ is contained in $A$.
Does being integrally closed have an interpretation in terms of the automorphisms of $K$?
I have come across a similar observation which goes as follows:
Lemma: Let $L/K$ be a Galois extension of fields, and let $A subset L$ be a subring. If $sigma (A) = A$ for each $sigma in [L, L]_K$, then $A$ is integral over $A cap K$.
Proof: Take $a in A$ and let $f in K[x]$ be the (monic) minimal polynomial for $a$ over $K$. $f(x)$ factors as $prod_{sigma in [L, L]_K }(x - sigma(a))$. Since each $sigma(a)$ belongs to $A$, the coefficients of $f$ belong to $A$, so that they belong to $A cap K$. Therefore $a$ is integral over $A cap K$.
I suspect there should be some way of expressing "integrally closed" with the automorphisms, but I can't quite figure out what it should be.
Of course, this would lead to a description of $mathcal{O}_K subset K$ for a Galois number field $K$ in terms of the Galois group, which is the case I am interested in.
number-theory galois-theory
asked Jan 20 at 3:29
Dean YoungDean Young
1,683721
$endgroup$
$begingroup$
To clarify, does the notation $[L, L]_{K}$ refer to the field automorphisms of $L$ which fix $K$?
$endgroup$
– Alex Wertheim
Jan 20 at 3:50
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Yes, sorry about that.
$endgroup$
– Dean Young
Jan 20 at 3:50
$begingroup$
$O_K$ is the integral closure of $mathbb{Z}$ means it is the largest finitely generated $mathbb{Z}$-module. You need that characterization to show $(K:I) = {x in K, x I subset O_K} $ is $= a^{-1} J$ for some ideal $J$.
$endgroup$
– reuns
Jan 20 at 4:19
$begingroup$
Do you suspect that can be phrased in terms of the automorphisms?
$endgroup$
– Dean Young
Jan 20 at 4:23
add a comment |
$begingroup$
Let $K$ be a finite Galois extension of $mathbb{Q}$ and let $A$ be a subring. We say $A$ is integrally closed in $K$ if any element $a in K$ integral over $A$ is contained in $A$.
Does being integrally closed have an interpretation in terms of the automorphisms of $K$?
I have come across a similar observation which goes as follows:
Lemma: Let $L/K$ be a Galois extension of fields, and let $A subset L$ be a subring. If $sigma (A) = A$ for each $sigma in [L, L]_K$, then $A$ is integral over $A cap K$.
Proof: Take $a in A$ and let $f in K[x]$ be the (monic) minimal polynomial for $a$ over $K$. $f(x)$ factors as $prod_{sigma in [L, L]_K }(x - sigma(a))$. Since each $sigma(a)$ belongs to $A$, the coefficients of $f$ belong to $A$, so that they belong to $A cap K$. Therefore $a$ is integral over $A cap K$.
I suspect there should be some way of expressing "integrally closed" with the automorphisms, but I can't quite figure out what it should be.
Of course, this would lead to a description of $mathcal{O}_K subset K$ for a Galois number field $K$ in terms of the Galois group, which is the case I am interested in.
number-theory galois-theory
asked Jan 20 at 3:29
Dean YoungDean Young
1,683721
$endgroup$
Let $K$ be a finite Galois extension of $mathbb{Q}$ and let $A$ be a subring. We say $A$ is integrally closed in $K$ if any element $a in K$ integral over $A$ is contained in $A$.
Does being integrally closed have an interpretation in terms of the automorphisms of $K$?
I have come across a similar observation which goes as follows:
Lemma: Let $L/K$ be a Galois extension of fields, and let $A subset L$ be a subring. If $sigma (A) = A$ for each $sigma in [L, L]_K$, then $A$ is integral over $A cap K$.
Proof: Take $a in A$ and let $f in K[x]$ be the (monic) minimal polynomial for $a$ over $K$. $f(x)$ factors as $prod_{sigma in [L, L]_K }(x - sigma(a))$. Since each $sigma(a)$ belongs to $A$, the coefficients of $f$ belong to $A$, so that they belong to $A cap K$. Therefore $a$ is integral over $A cap K$.
I suspect there should be some way of expressing "integrally closed" with the automorphisms, but I can't quite figure out what it should be.
Of course, this would lead to a description of $mathcal{O}_K subset K$ for a Galois number field $K$ in terms of the Galois group, which is the case I am interested in.
number-theory galois-theory
number-theory galois-theory
asked Jan 20 at 3:29
Dean YoungDean Young
1,683721
asked Jan 20 at 3:29
Dean YoungDean Young
1,683721
asked Jan 20 at 3:29
Dean YoungDean Young
1,683721
asked Jan 20 at 3:29
Dean YoungDean Young
1,683721
asked Jan 20 at 3:29
Dean YoungDean Young
1,683721
1,683721
$begingroup$
To clarify, does the notation $[L, L]_{K}$ refer to the field automorphisms of $L$ which fix $K$?
$endgroup$
– Alex Wertheim
Jan 20 at 3:50
$begingroup$
Yes, sorry about that.
$endgroup$
– Dean Young
Jan 20 at 3:50
$begingroup$
$O_K$ is the integral closure of $mathbb{Z}$ means it is the largest finitely generated $mathbb{Z}$-module. You need that characterization to show $(K:I) = {x in K, x I subset O_K} $ is $= a^{-1} J$ for some ideal $J$.
$endgroup$
– reuns
Jan 20 at 4:19
$begingroup$
Do you suspect that can be phrased in terms of the automorphisms?
$endgroup$
– Dean Young
Jan 20 at 4:23
add a comment |
$begingroup$
To clarify, does the notation $[L, L]_{K}$ refer to the field automorphisms of $L$ which fix $K$?
$endgroup$
– Alex Wertheim
Jan 20 at 3:50
$begingroup$
Yes, sorry about that.
$endgroup$
– Dean Young
Jan 20 at 3:50
$begingroup$
$O_K$ is the integral closure of $mathbb{Z}$ means it is the largest finitely generated $mathbb{Z}$-module. You need that characterization to show $(K:I) = {x in K, x I subset O_K} $ is $= a^{-1} J$ for some ideal $J$.
$endgroup$
– reuns
Jan 20 at 4:19
$begingroup$
Do you suspect that can be phrased in terms of the automorphisms?
$endgroup$
– Dean Young
Jan 20 at 4:23
$begingroup$
To clarify, does the notation $[L, L]_{K}$ refer to the field automorphisms of $L$ which fix $K$?
$endgroup$
– Alex Wertheim
Jan 20 at 3:50
$begingroup$
To clarify, does the notation $[L, L]_{K}$ refer to the field automorphisms of $L$ which fix $K$?
$endgroup$
– Alex Wertheim
Jan 20 at 3:50
$begingroup$
Yes, sorry about that.
$endgroup$
– Dean Young
Jan 20 at 3:50
$begingroup$
Yes, sorry about that.
$endgroup$
– Dean Young
Jan 20 at 3:50
$begingroup$
$O_K$ is the integral closure of $mathbb{Z}$ means it is the largest finitely generated $mathbb{Z}$-module. You need that characterization to show $(K:I) = {x in K, x I subset O_K} $ is $= a^{-1} J$ for some ideal $J$.
$endgroup$
– reuns
Jan 20 at 4:19
$begingroup$
$O_K$ is the integral closure of $mathbb{Z}$ means it is the largest finitely generated $mathbb{Z}$-module. You need that characterization to show $(K:I) = {x in K, x I subset O_K} $ is $= a^{-1} J$ for some ideal $J$.
$endgroup$
– reuns
Jan 20 at 4:19
$begingroup$
Do you suspect that can be phrased in terms of the automorphisms?
$endgroup$
– Dean Young
Jan 20 at 4:23
$begingroup$
Do you suspect that can be phrased in terms of the automorphisms?
$endgroup$
– Dean Young
Jan 20 at 4:23
add a comment |
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$begingroup$
To clarify, does the notation $[L, L]_{K}$ refer to the field automorphisms of $L$ which fix $K$?
$endgroup$
– Alex Wertheim
Jan 20 at 3:50
$begingroup$
Yes, sorry about that.
$endgroup$
– Dean Young
Jan 20 at 3:50
$begingroup$
$O_K$ is the integral closure of $mathbb{Z}$ means it is the largest finitely generated $mathbb{Z}$-module. You need that characterization to show $(K:I) = {x in K, x I subset O_K} $ is $= a^{-1} J$ for some ideal $J$.
$endgroup$
– reuns
Jan 20 at 4:19
$begingroup$
Do you suspect that can be phrased in terms of the automorphisms?
$endgroup$
– Dean Young
Jan 20 at 4:23