Let $K/mathbb Q$ be a number field and a Galois extension. Show $K$ contains the field $mathbb Q(sqrt{d_K})$...












1












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Let $K/mathbb Q$ be a number field and a Galois extension. Show $K$ contains the field $mathbb Q(sqrt{d_K})$ where $d_K = mathrm{disc}(K)$.



I'm guessing this will somehow use the fact that the $d_K$ in an integer but I'm not sure how this being an Galois extension is used.






field-theory galois-theory algebraic-number-theory







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$endgroup$








  • 1




    $begingroup$
    @reuns: I believe it is much simpler than conductor-discriminant, all you have to do is to write down the formula for $ text{disc}(K) $.
    $endgroup$
    – hellHound
    Dec 3 '18 at 17:31






  • 1




    $begingroup$
    @hellHound right I feel dumb. So it works in any Galois extension $K/F$ : if $b_1,ldots,b_r$ is a $F$-basis of $K$, let $B_{ij} = sigma_j(b_i)$, then $(B B^top)_{il} = Tr_{K/F}(b_i b_l)$, so $det(B)^2 = det(B B^top) in F$ and $sqrt{det(B B^top)} in K$.
    $endgroup$
    – reuns
    Dec 3 '18 at 20:45


















1












$begingroup$


Let $K/mathbb Q$ be a number field and a Galois extension. Show $K$ contains the field $mathbb Q(sqrt{d_K})$ where $d_K = mathrm{disc}(K)$.



I'm guessing this will somehow use the fact that the $d_K$ in an integer but I'm not sure how this being an Galois extension is used.






field-theory galois-theory algebraic-number-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    @reuns: I believe it is much simpler than conductor-discriminant, all you have to do is to write down the formula for $ text{disc}(K) $.
    $endgroup$
    – hellHound
    Dec 3 '18 at 17:31






  • 1




    $begingroup$
    @hellHound right I feel dumb. So it works in any Galois extension $K/F$ : if $b_1,ldots,b_r$ is a $F$-basis of $K$, let $B_{ij} = sigma_j(b_i)$, then $(B B^top)_{il} = Tr_{K/F}(b_i b_l)$, so $det(B)^2 = det(B B^top) in F$ and $sqrt{det(B B^top)} in K$.
    $endgroup$
    – reuns
    Dec 3 '18 at 20:45
















1












1








1





$begingroup$


Let $K/mathbb Q$ be a number field and a Galois extension. Show $K$ contains the field $mathbb Q(sqrt{d_K})$ where $d_K = mathrm{disc}(K)$.



I'm guessing this will somehow use the fact that the $d_K$ in an integer but I'm not sure how this being an Galois extension is used.






field-theory galois-theory algebraic-number-theory







share|cite|improve this question









$endgroup$




Let $K/mathbb Q$ be a number field and a Galois extension. Show $K$ contains the field $mathbb Q(sqrt{d_K})$ where $d_K = mathrm{disc}(K)$.



I'm guessing this will somehow use the fact that the $d_K$ in an integer but I'm not sure how this being an Galois extension is used.






field-theory galois-theory algebraic-number-theory




field-theory galois-theory algebraic-number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Dec 2 '18 at 22:48









John117John117

376




376








  • 1




    $begingroup$
    @reuns: I believe it is much simpler than conductor-discriminant, all you have to do is to write down the formula for $ text{disc}(K) $.
    $endgroup$
    – hellHound
    Dec 3 '18 at 17:31






  • 1




    $begingroup$
    @hellHound right I feel dumb. So it works in any Galois extension $K/F$ : if $b_1,ldots,b_r$ is a $F$-basis of $K$, let $B_{ij} = sigma_j(b_i)$, then $(B B^top)_{il} = Tr_{K/F}(b_i b_l)$, so $det(B)^2 = det(B B^top) in F$ and $sqrt{det(B B^top)} in K$.
    $endgroup$
    – reuns
    Dec 3 '18 at 20:45
















  • 1




    $begingroup$
    @reuns: I believe it is much simpler than conductor-discriminant, all you have to do is to write down the formula for $ text{disc}(K) $.
    $endgroup$
    – hellHound
    Dec 3 '18 at 17:31






  • 1




    $begingroup$
    @hellHound right I feel dumb. So it works in any Galois extension $K/F$ : if $b_1,ldots,b_r$ is a $F$-basis of $K$, let $B_{ij} = sigma_j(b_i)$, then $(B B^top)_{il} = Tr_{K/F}(b_i b_l)$, so $det(B)^2 = det(B B^top) in F$ and $sqrt{det(B B^top)} in K$.
    $endgroup$
    – reuns
    Dec 3 '18 at 20:45










1




1




$begingroup$
@reuns: I believe it is much simpler than conductor-discriminant, all you have to do is to write down the formula for $ text{disc}(K) $.
$endgroup$
– hellHound
Dec 3 '18 at 17:31




$begingroup$
@reuns: I believe it is much simpler than conductor-discriminant, all you have to do is to write down the formula for $ text{disc}(K) $.
$endgroup$
– hellHound
Dec 3 '18 at 17:31




1




1




$begingroup$
@hellHound right I feel dumb. So it works in any Galois extension $K/F$ : if $b_1,ldots,b_r$ is a $F$-basis of $K$, let $B_{ij} = sigma_j(b_i)$, then $(B B^top)_{il} = Tr_{K/F}(b_i b_l)$, so $det(B)^2 = det(B B^top) in F$ and $sqrt{det(B B^top)} in K$.
$endgroup$
– reuns
Dec 3 '18 at 20:45






$begingroup$
@hellHound right I feel dumb. So it works in any Galois extension $K/F$ : if $b_1,ldots,b_r$ is a $F$-basis of $K$, let $B_{ij} = sigma_j(b_i)$, then $(B B^top)_{il} = Tr_{K/F}(b_i b_l)$, so $det(B)^2 = det(B B^top) in F$ and $sqrt{det(B B^top)} in K$.
$endgroup$
– reuns
Dec 3 '18 at 20:45












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Let $ [K:mathbb{Q}] = r $ and $ text{Gal}(K/mathbb{Q} )= { sigma_1, cdots, sigma_r } $. If $ alpha_1, cdots, alpha_r $ is a $ mathbb{Z} $-basis for the ring of integers $ O_K $ of $ K $, then one definition of the discriminant is $ d_K = [det (sigma_i(alpha_j))_{i,j}]^2 $. As $ K/ mathbb{Q} $ is Galois, the entries $ sigma_i(alpha_j) $ of the matrix are in $ K $. So just expanding this determinant out shows you that $ K $ contains a square root of $ d_K $.






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    $begingroup$

    Let $ [K:mathbb{Q}] = r $ and $ text{Gal}(K/mathbb{Q} )= { sigma_1, cdots, sigma_r } $. If $ alpha_1, cdots, alpha_r $ is a $ mathbb{Z} $-basis for the ring of integers $ O_K $ of $ K $, then one definition of the discriminant is $ d_K = [det (sigma_i(alpha_j))_{i,j}]^2 $. As $ K/ mathbb{Q} $ is Galois, the entries $ sigma_i(alpha_j) $ of the matrix are in $ K $. So just expanding this determinant out shows you that $ K $ contains a square root of $ d_K $.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Let $ [K:mathbb{Q}] = r $ and $ text{Gal}(K/mathbb{Q} )= { sigma_1, cdots, sigma_r } $. If $ alpha_1, cdots, alpha_r $ is a $ mathbb{Z} $-basis for the ring of integers $ O_K $ of $ K $, then one definition of the discriminant is $ d_K = [det (sigma_i(alpha_j))_{i,j}]^2 $. As $ K/ mathbb{Q} $ is Galois, the entries $ sigma_i(alpha_j) $ of the matrix are in $ K $. So just expanding this determinant out shows you that $ K $ contains a square root of $ d_K $.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Let $ [K:mathbb{Q}] = r $ and $ text{Gal}(K/mathbb{Q} )= { sigma_1, cdots, sigma_r } $. If $ alpha_1, cdots, alpha_r $ is a $ mathbb{Z} $-basis for the ring of integers $ O_K $ of $ K $, then one definition of the discriminant is $ d_K = [det (sigma_i(alpha_j))_{i,j}]^2 $. As $ K/ mathbb{Q} $ is Galois, the entries $ sigma_i(alpha_j) $ of the matrix are in $ K $. So just expanding this determinant out shows you that $ K $ contains a square root of $ d_K $.






        share|cite|improve this answer









        $endgroup$



        Let $ [K:mathbb{Q}] = r $ and $ text{Gal}(K/mathbb{Q} )= { sigma_1, cdots, sigma_r } $. If $ alpha_1, cdots, alpha_r $ is a $ mathbb{Z} $-basis for the ring of integers $ O_K $ of $ K $, then one definition of the discriminant is $ d_K = [det (sigma_i(alpha_j))_{i,j}]^2 $. As $ K/ mathbb{Q} $ is Galois, the entries $ sigma_i(alpha_j) $ of the matrix are in $ K $. So just expanding this determinant out shows you that $ K $ contains a square root of $ d_K $.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 17:41









        hellHoundhellHound

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