Integral Closure, Galois extension,and Dedekind Domain












3














Let
$A$: Dedekind domain, $K$: $operatorname{Frac}(A)$, $B$: Dedekind domain with $A subset B$, $L$: $operatorname{Frac}(B)$



Let $L/K$: galois extension with galois group: $G$.



$B^G={b in B mid sigma(b)=b text{ for all } sigma in G}=A$
$implies B$ is the integral closure of $A$ in $L$



Is this true?
I already prove the converse, but not sure if this holds.
Thank you in advance.






abstract-algebra number-theory ring-theory field-theory galois-theory







share|cite|improve this question
























  • I think it's not true. Try to take B=A. Then the fixed ring is A, but unless the extension is trivial, this is not the integral closure.
    – Madarb
    Dec 9 at 7:38










  • when A=B, then L=K. Since A is dedekind, A is integrally closed and B(=A) is the integral closure of A in L holds, i think.
    – Kento
    Dec 9 at 7:56










  • I don't think I understood what you say. The claim you want to prove, as I understand it, is that if B is a subring of L, for which the fixed ring under the action of G is A, then B is the integral closure. It is not the true if you take B=A. If I got you wrong, please explain what you wanted to prove :)
    – Madarb
    Dec 9 at 8:00








  • 1




    you understand it right. but i do not get that it does not hold when B=A. when B=A,the extension L/K is trivial, which leads to the conclusion. i think. please tell me if i get something wrong.
    – Kento
    Dec 9 at 8:07








  • 1




    @reuns I think you don't necessarily have $sigma(B) = B$. A possible counterexample is given below.
    – pisco
    Dec 9 at 12:53


















3














Let
$A$: Dedekind domain, $K$: $operatorname{Frac}(A)$, $B$: Dedekind domain with $A subset B$, $L$: $operatorname{Frac}(B)$



Let $L/K$: galois extension with galois group: $G$.



$B^G={b in B mid sigma(b)=b text{ for all } sigma in G}=A$
$implies B$ is the integral closure of $A$ in $L$



Is this true?
I already prove the converse, but not sure if this holds.
Thank you in advance.






abstract-algebra number-theory ring-theory field-theory galois-theory







share|cite|improve this question
























  • I think it's not true. Try to take B=A. Then the fixed ring is A, but unless the extension is trivial, this is not the integral closure.
    – Madarb
    Dec 9 at 7:38










  • when A=B, then L=K. Since A is dedekind, A is integrally closed and B(=A) is the integral closure of A in L holds, i think.
    – Kento
    Dec 9 at 7:56










  • I don't think I understood what you say. The claim you want to prove, as I understand it, is that if B is a subring of L, for which the fixed ring under the action of G is A, then B is the integral closure. It is not the true if you take B=A. If I got you wrong, please explain what you wanted to prove :)
    – Madarb
    Dec 9 at 8:00








  • 1




    you understand it right. but i do not get that it does not hold when B=A. when B=A,the extension L/K is trivial, which leads to the conclusion. i think. please tell me if i get something wrong.
    – Kento
    Dec 9 at 8:07








  • 1




    @reuns I think you don't necessarily have $sigma(B) = B$. A possible counterexample is given below.
    – pisco
    Dec 9 at 12:53
















3












3








3


1





Let
$A$: Dedekind domain, $K$: $operatorname{Frac}(A)$, $B$: Dedekind domain with $A subset B$, $L$: $operatorname{Frac}(B)$



Let $L/K$: galois extension with galois group: $G$.



$B^G={b in B mid sigma(b)=b text{ for all } sigma in G}=A$
$implies B$ is the integral closure of $A$ in $L$



Is this true?
I already prove the converse, but not sure if this holds.
Thank you in advance.






abstract-algebra number-theory ring-theory field-theory galois-theory







share|cite|improve this question















Let
$A$: Dedekind domain, $K$: $operatorname{Frac}(A)$, $B$: Dedekind domain with $A subset B$, $L$: $operatorname{Frac}(B)$



Let $L/K$: galois extension with galois group: $G$.



$B^G={b in B mid sigma(b)=b text{ for all } sigma in G}=A$
$implies B$ is the integral closure of $A$ in $L$



Is this true?
I already prove the converse, but not sure if this holds.
Thank you in advance.






abstract-algebra number-theory ring-theory field-theory galois-theory




abstract-algebra number-theory ring-theory field-theory galois-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 at 8:45









Alex Vong

1,294819




1,294819










asked Dec 9 at 7:24









Kento

473




473












  • I think it's not true. Try to take B=A. Then the fixed ring is A, but unless the extension is trivial, this is not the integral closure.
    – Madarb
    Dec 9 at 7:38










  • when A=B, then L=K. Since A is dedekind, A is integrally closed and B(=A) is the integral closure of A in L holds, i think.
    – Kento
    Dec 9 at 7:56










  • I don't think I understood what you say. The claim you want to prove, as I understand it, is that if B is a subring of L, for which the fixed ring under the action of G is A, then B is the integral closure. It is not the true if you take B=A. If I got you wrong, please explain what you wanted to prove :)
    – Madarb
    Dec 9 at 8:00








  • 1




    you understand it right. but i do not get that it does not hold when B=A. when B=A,the extension L/K is trivial, which leads to the conclusion. i think. please tell me if i get something wrong.
    – Kento
    Dec 9 at 8:07








  • 1




    @reuns I think you don't necessarily have $sigma(B) = B$. A possible counterexample is given below.
    – pisco
    Dec 9 at 12:53




















  • I think it's not true. Try to take B=A. Then the fixed ring is A, but unless the extension is trivial, this is not the integral closure.
    – Madarb
    Dec 9 at 7:38










  • when A=B, then L=K. Since A is dedekind, A is integrally closed and B(=A) is the integral closure of A in L holds, i think.
    – Kento
    Dec 9 at 7:56










  • I don't think I understood what you say. The claim you want to prove, as I understand it, is that if B is a subring of L, for which the fixed ring under the action of G is A, then B is the integral closure. It is not the true if you take B=A. If I got you wrong, please explain what you wanted to prove :)
    – Madarb
    Dec 9 at 8:00








  • 1




    you understand it right. but i do not get that it does not hold when B=A. when B=A,the extension L/K is trivial, which leads to the conclusion. i think. please tell me if i get something wrong.
    – Kento
    Dec 9 at 8:07








  • 1




    @reuns I think you don't necessarily have $sigma(B) = B$. A possible counterexample is given below.
    – pisco
    Dec 9 at 12:53


















I think it's not true. Try to take B=A. Then the fixed ring is A, but unless the extension is trivial, this is not the integral closure.
– Madarb
Dec 9 at 7:38




I think it's not true. Try to take B=A. Then the fixed ring is A, but unless the extension is trivial, this is not the integral closure.
– Madarb
Dec 9 at 7:38












when A=B, then L=K. Since A is dedekind, A is integrally closed and B(=A) is the integral closure of A in L holds, i think.
– Kento
Dec 9 at 7:56




when A=B, then L=K. Since A is dedekind, A is integrally closed and B(=A) is the integral closure of A in L holds, i think.
– Kento
Dec 9 at 7:56












I don't think I understood what you say. The claim you want to prove, as I understand it, is that if B is a subring of L, for which the fixed ring under the action of G is A, then B is the integral closure. It is not the true if you take B=A. If I got you wrong, please explain what you wanted to prove :)
– Madarb
Dec 9 at 8:00






I don't think I understood what you say. The claim you want to prove, as I understand it, is that if B is a subring of L, for which the fixed ring under the action of G is A, then B is the integral closure. It is not the true if you take B=A. If I got you wrong, please explain what you wanted to prove :)
– Madarb
Dec 9 at 8:00






1




1




you understand it right. but i do not get that it does not hold when B=A. when B=A,the extension L/K is trivial, which leads to the conclusion. i think. please tell me if i get something wrong.
– Kento
Dec 9 at 8:07






you understand it right. but i do not get that it does not hold when B=A. when B=A,the extension L/K is trivial, which leads to the conclusion. i think. please tell me if i get something wrong.
– Kento
Dec 9 at 8:07






1




1




@reuns I think you don't necessarily have $sigma(B) = B$. A possible counterexample is given below.
– pisco
Dec 9 at 12:53






@reuns I think you don't necessarily have $sigma(B) = B$. A possible counterexample is given below.
– pisco
Dec 9 at 12:53












1 Answer
1






active

oldest

votes


















3














This is not true.



Let $L/K$ be a Galois extension of number field, let $mathfrak{p}$ be a prime ideal of $mathcal{O}_K$ that splits into more than one primes in $mathcal{O}_L$:
$$mathfrak{p}mathcal{O}_L = mathfrak{P}_1 cdots mathfrak{P}_r$$



Let $A = (mathcal{O}_K)_{mathfrak{p}}$, the localization at $mathfrak{p}$ and $B = (mathcal{O}_L)_{mathfrak{P}_1}$. Both are Dedekind domains. It is easily seen that $B^G = A$, but the integral closure of $A$ in $L$ is the localization of $mathcal{O}_L$ at all $mathfrak{P}_1, cdots, mathfrak{P}_r$, which is a proper subset of $B$.






share|cite|improve this answer



















  • 1




    So $K = mathbf{Q}, L = mathbf{Q}(i),p=5= (2+i)(2-i), A=mathbf{Z}_{(5)} = {frac{u}{v}, (u,v) in mathbf{Z}^2, 5 nmid v }$, $B = mathbf{Z}[i]_{(2+i)} = {frac{u}{v}, (u,v) in mathbf{Z}[i]^2, v not in (2+i)mathbf{Z}[i] }$, $sigma(c+id) = c-id, G= {sigma^2,sigma}$ then $frac1{5^n} not in B$ so $B^G = mathbf{Z}_{(5)}$ but $frac{2+i}{5}=frac{1}{2-i} in B$ is not integral over $mathbf{Z}_{(5)}$. $A,B$ are Dedekind domains since they have only one prime ideal. The problem is that $sigma(B) ne B$.
    – reuns
    Dec 9 at 23:22










  • And replacing $B$ by $bigcap_{sigma in G} sigma(B)$ makes the claim true
    – reuns
    Dec 10 at 18:19











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1 Answer
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3














This is not true.



Let $L/K$ be a Galois extension of number field, let $mathfrak{p}$ be a prime ideal of $mathcal{O}_K$ that splits into more than one primes in $mathcal{O}_L$:
$$mathfrak{p}mathcal{O}_L = mathfrak{P}_1 cdots mathfrak{P}_r$$



Let $A = (mathcal{O}_K)_{mathfrak{p}}$, the localization at $mathfrak{p}$ and $B = (mathcal{O}_L)_{mathfrak{P}_1}$. Both are Dedekind domains. It is easily seen that $B^G = A$, but the integral closure of $A$ in $L$ is the localization of $mathcal{O}_L$ at all $mathfrak{P}_1, cdots, mathfrak{P}_r$, which is a proper subset of $B$.






share|cite|improve this answer



















  • 1




    So $K = mathbf{Q}, L = mathbf{Q}(i),p=5= (2+i)(2-i), A=mathbf{Z}_{(5)} = {frac{u}{v}, (u,v) in mathbf{Z}^2, 5 nmid v }$, $B = mathbf{Z}[i]_{(2+i)} = {frac{u}{v}, (u,v) in mathbf{Z}[i]^2, v not in (2+i)mathbf{Z}[i] }$, $sigma(c+id) = c-id, G= {sigma^2,sigma}$ then $frac1{5^n} not in B$ so $B^G = mathbf{Z}_{(5)}$ but $frac{2+i}{5}=frac{1}{2-i} in B$ is not integral over $mathbf{Z}_{(5)}$. $A,B$ are Dedekind domains since they have only one prime ideal. The problem is that $sigma(B) ne B$.
    – reuns
    Dec 9 at 23:22










  • And replacing $B$ by $bigcap_{sigma in G} sigma(B)$ makes the claim true
    – reuns
    Dec 10 at 18:19
















3














This is not true.



Let $L/K$ be a Galois extension of number field, let $mathfrak{p}$ be a prime ideal of $mathcal{O}_K$ that splits into more than one primes in $mathcal{O}_L$:
$$mathfrak{p}mathcal{O}_L = mathfrak{P}_1 cdots mathfrak{P}_r$$



Let $A = (mathcal{O}_K)_{mathfrak{p}}$, the localization at $mathfrak{p}$ and $B = (mathcal{O}_L)_{mathfrak{P}_1}$. Both are Dedekind domains. It is easily seen that $B^G = A$, but the integral closure of $A$ in $L$ is the localization of $mathcal{O}_L$ at all $mathfrak{P}_1, cdots, mathfrak{P}_r$, which is a proper subset of $B$.






share|cite|improve this answer



















  • 1




    So $K = mathbf{Q}, L = mathbf{Q}(i),p=5= (2+i)(2-i), A=mathbf{Z}_{(5)} = {frac{u}{v}, (u,v) in mathbf{Z}^2, 5 nmid v }$, $B = mathbf{Z}[i]_{(2+i)} = {frac{u}{v}, (u,v) in mathbf{Z}[i]^2, v not in (2+i)mathbf{Z}[i] }$, $sigma(c+id) = c-id, G= {sigma^2,sigma}$ then $frac1{5^n} not in B$ so $B^G = mathbf{Z}_{(5)}$ but $frac{2+i}{5}=frac{1}{2-i} in B$ is not integral over $mathbf{Z}_{(5)}$. $A,B$ are Dedekind domains since they have only one prime ideal. The problem is that $sigma(B) ne B$.
    – reuns
    Dec 9 at 23:22










  • And replacing $B$ by $bigcap_{sigma in G} sigma(B)$ makes the claim true
    – reuns
    Dec 10 at 18:19














3












3








3






This is not true.



Let $L/K$ be a Galois extension of number field, let $mathfrak{p}$ be a prime ideal of $mathcal{O}_K$ that splits into more than one primes in $mathcal{O}_L$:
$$mathfrak{p}mathcal{O}_L = mathfrak{P}_1 cdots mathfrak{P}_r$$



Let $A = (mathcal{O}_K)_{mathfrak{p}}$, the localization at $mathfrak{p}$ and $B = (mathcal{O}_L)_{mathfrak{P}_1}$. Both are Dedekind domains. It is easily seen that $B^G = A$, but the integral closure of $A$ in $L$ is the localization of $mathcal{O}_L$ at all $mathfrak{P}_1, cdots, mathfrak{P}_r$, which is a proper subset of $B$.






share|cite|improve this answer














This is not true.



Let $L/K$ be a Galois extension of number field, let $mathfrak{p}$ be a prime ideal of $mathcal{O}_K$ that splits into more than one primes in $mathcal{O}_L$:
$$mathfrak{p}mathcal{O}_L = mathfrak{P}_1 cdots mathfrak{P}_r$$



Let $A = (mathcal{O}_K)_{mathfrak{p}}$, the localization at $mathfrak{p}$ and $B = (mathcal{O}_L)_{mathfrak{P}_1}$. Both are Dedekind domains. It is easily seen that $B^G = A$, but the integral closure of $A$ in $L$ is the localization of $mathcal{O}_L$ at all $mathfrak{P}_1, cdots, mathfrak{P}_r$, which is a proper subset of $B$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 at 16:53

























answered Dec 9 at 12:20









pisco

11.5k21742




11.5k21742








  • 1




    So $K = mathbf{Q}, L = mathbf{Q}(i),p=5= (2+i)(2-i), A=mathbf{Z}_{(5)} = {frac{u}{v}, (u,v) in mathbf{Z}^2, 5 nmid v }$, $B = mathbf{Z}[i]_{(2+i)} = {frac{u}{v}, (u,v) in mathbf{Z}[i]^2, v not in (2+i)mathbf{Z}[i] }$, $sigma(c+id) = c-id, G= {sigma^2,sigma}$ then $frac1{5^n} not in B$ so $B^G = mathbf{Z}_{(5)}$ but $frac{2+i}{5}=frac{1}{2-i} in B$ is not integral over $mathbf{Z}_{(5)}$. $A,B$ are Dedekind domains since they have only one prime ideal. The problem is that $sigma(B) ne B$.
    – reuns
    Dec 9 at 23:22










  • And replacing $B$ by $bigcap_{sigma in G} sigma(B)$ makes the claim true
    – reuns
    Dec 10 at 18:19














  • 1




    So $K = mathbf{Q}, L = mathbf{Q}(i),p=5= (2+i)(2-i), A=mathbf{Z}_{(5)} = {frac{u}{v}, (u,v) in mathbf{Z}^2, 5 nmid v }$, $B = mathbf{Z}[i]_{(2+i)} = {frac{u}{v}, (u,v) in mathbf{Z}[i]^2, v not in (2+i)mathbf{Z}[i] }$, $sigma(c+id) = c-id, G= {sigma^2,sigma}$ then $frac1{5^n} not in B$ so $B^G = mathbf{Z}_{(5)}$ but $frac{2+i}{5}=frac{1}{2-i} in B$ is not integral over $mathbf{Z}_{(5)}$. $A,B$ are Dedekind domains since they have only one prime ideal. The problem is that $sigma(B) ne B$.
    – reuns
    Dec 9 at 23:22










  • And replacing $B$ by $bigcap_{sigma in G} sigma(B)$ makes the claim true
    – reuns
    Dec 10 at 18:19








1




1




So $K = mathbf{Q}, L = mathbf{Q}(i),p=5= (2+i)(2-i), A=mathbf{Z}_{(5)} = {frac{u}{v}, (u,v) in mathbf{Z}^2, 5 nmid v }$, $B = mathbf{Z}[i]_{(2+i)} = {frac{u}{v}, (u,v) in mathbf{Z}[i]^2, v not in (2+i)mathbf{Z}[i] }$, $sigma(c+id) = c-id, G= {sigma^2,sigma}$ then $frac1{5^n} not in B$ so $B^G = mathbf{Z}_{(5)}$ but $frac{2+i}{5}=frac{1}{2-i} in B$ is not integral over $mathbf{Z}_{(5)}$. $A,B$ are Dedekind domains since they have only one prime ideal. The problem is that $sigma(B) ne B$.
– reuns
Dec 9 at 23:22




So $K = mathbf{Q}, L = mathbf{Q}(i),p=5= (2+i)(2-i), A=mathbf{Z}_{(5)} = {frac{u}{v}, (u,v) in mathbf{Z}^2, 5 nmid v }$, $B = mathbf{Z}[i]_{(2+i)} = {frac{u}{v}, (u,v) in mathbf{Z}[i]^2, v not in (2+i)mathbf{Z}[i] }$, $sigma(c+id) = c-id, G= {sigma^2,sigma}$ then $frac1{5^n} not in B$ so $B^G = mathbf{Z}_{(5)}$ but $frac{2+i}{5}=frac{1}{2-i} in B$ is not integral over $mathbf{Z}_{(5)}$. $A,B$ are Dedekind domains since they have only one prime ideal. The problem is that $sigma(B) ne B$.
– reuns
Dec 9 at 23:22












And replacing $B$ by $bigcap_{sigma in G} sigma(B)$ makes the claim true
– reuns
Dec 10 at 18:19




And replacing $B$ by $bigcap_{sigma in G} sigma(B)$ makes the claim true
– reuns
Dec 10 at 18:19


















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VLT Carioca

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VLT Carioca Informações Proprietário Prefeitura do Rio de Janeiro Local [[Rio de Janeiro, RJ]] País {{country data Brasil| country flag2| name = Brasil| variant = | size = }} Tipo de transporte Bonde (Veículo leve sobre trilhos) Número de linhas 2 (+1 em construção) Número de estações 26 (+3 em construção) Tráfego 80 mil passageiros/dia [ 1 ] (dezembro/2018) Website www.vltrio.com.br Funcionamento Início de funcionamento 5 de junho de 2016 (2 anos) Operadora(s) Concessionária do VLT Carioca S.A. Número de veículos 32 Alstom Citadis Comprimento dos veículos 44 m (144 ft) Dados técnicos Extensão do sistema 28 km (17,4 mi) [ 2 ] Frequência 5–10 minutos Bitola Bitola padrão/internacional 1 435 mm (4,71 ft) Eletrificação APS, 750V DC Velocidade média 15 km/h (9,32 mph) [ 2 ] Velocidade máxima 50 km/h (31,1 mph) ...
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