Let $K/mathbb Q$ be a number field and a Galois extension. Show $K$ contains the field $mathbb Q(sqrt{d_K})$...
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
asked Dec 2 '18 at 22:48
John117
326
add a comment |
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
asked Dec 2 '18 at 22:48
John117
326
1
@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) $.
– hellHound
Dec 3 '18 at 17:31
1
@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 $sqrt{det(B B^top)} in K$.
– reuns
Dec 3 '18 at 20:45
add a comment |
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
asked Dec 2 '18 at 22:48
John117
326
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
asked Dec 2 '18 at 22:48
John117
326
asked Dec 2 '18 at 22:48
John117
326
asked Dec 2 '18 at 22:48
John117
326
asked Dec 2 '18 at 22:48
John117
326
asked Dec 2 '18 at 22:48
John117
326
326
1
@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) $.
– hellHound
Dec 3 '18 at 17:31
1
@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 $sqrt{det(B B^top)} in K$.
– reuns
Dec 3 '18 at 20:45
add a comment |
1
@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) $.
– hellHound
Dec 3 '18 at 17:31
1
@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 $sqrt{det(B B^top)} in K$.
– reuns
Dec 3 '18 at 20:45
1
1
@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) $.
– hellHound
Dec 3 '18 at 17:31
@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) $.
– hellHound
Dec 3 '18 at 17:31
1
1
@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 $sqrt{det(B B^top)} in K$.
– reuns
Dec 3 '18 at 20:45
@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 $sqrt{det(B B^top)} in K$.
– reuns
Dec 3 '18 at 20:45
add a comment |
1 Answer
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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 $.
answered Dec 3 '18 at 17:41
hellHound
48328
add a comment |
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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 $.
answered Dec 3 '18 at 17:41
hellHound
48328
add a comment |
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 $.
answered Dec 3 '18 at 17:41
hellHound
48328
add a comment |
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 $.
answered Dec 3 '18 at 17:41
hellHound
48328
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 $.
answered Dec 3 '18 at 17:41
hellHound
48328
answered Dec 3 '18 at 17:41
hellHound
48328
answered Dec 3 '18 at 17:41
hellHound
48328
answered Dec 3 '18 at 17:41
hellHound
48328
48328
add a comment |
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@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) $.
– hellHound
Dec 3 '18 at 17:31
1
@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 $sqrt{det(B B^top)} in K$.
– reuns
Dec 3 '18 at 20:45