Parametrization of cubic fields












5












$begingroup$


A parametrization of quadratic fields is $mathbb{Q}(sqrt{m})$, where $mne1$ is a squarefree integer. That is, $mathbb{Q}(sqrt{m})$ is a quadratic field as $m$ varies, and all quadratic fields are included. There is another parametrization of quadratic fields, given by the 1-1 correspondence between $m$ and fundamental discriminants $d$. In particular, for each fundamental discriminant $d$ there is exactly one quadratic field of that discriminant.



Is there a similar parametrization for cubic fields?






algebraic-number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Kummer theory generalizes your parametrization of the quadratic fields. A special case of Kummer theory says that if $K$ is a field with characteristic different from a prime $p$, and if $K$ contains all the $p^{text{th}}$ roots of unity, then the extensions of $K$ with Galois group $C_p$ (cyclic of order $p$) are parametrized by non-trivial elements of $K^times/K^{times p}$, where $K(sqrt[p]a)$ corresponds to $a in K^times/K^{times p}$.
    $endgroup$
    – Kenny Lau
    Jan 3 at 18:09












  • $begingroup$
    I believe if $K/Bbb Q$ is a cubic extension and $K=Bbb Q(alpha)$ then $Bbb Q(omega,alpha)/Bbb Q(omega)$ is still a cubic extension... in that case $Bbb Q(omega,alpha) = Bbb Q(omega,(p+qomega)^{1/3})$ for some $p,q in Bbb Z$ for a choice of the cube root by Kummer theory... but I also suspect that it is possible to have $Bbb Q(omega,alpha) = Bbb Q(omega,beta)$ with $Bbb Q(alpha) ne Bbb Q(beta)$
    $endgroup$
    – Kenny Lau
    Jan 3 at 18:47










  • $begingroup$
    If $K=Bbb Q(alpha)/Bbb Q$ is cubic, i.e. $f(alpha) = 0$ for cubic $f$, then we have two situations depending on the discriminant of $f$: if the discriminant is a perfect square, then $K$ has only one conjugate (itself); otherwise, $K$ has three conjugates. Using this, we can shift the question to classifying splitting fields of (irreducible) cubic polynomials.
    $endgroup$
    – Kenny Lau
    Jan 3 at 18:54












  • $begingroup$
    An elementary observation is that if $K/Bbb Q$ is cubic then $K=Bbb Q(alpha)$ for every $alpha in K setminus Bbb Q$.
    $endgroup$
    – Kenny Lau
    Jan 3 at 19:05










  • $begingroup$
    @Kenny Lau: Thanks for the answers. Are there isomorphic unequal cubic fields?
    $endgroup$
    – sai
    Jan 5 at 2:41
















5












$begingroup$


A parametrization of quadratic fields is $mathbb{Q}(sqrt{m})$, where $mne1$ is a squarefree integer. That is, $mathbb{Q}(sqrt{m})$ is a quadratic field as $m$ varies, and all quadratic fields are included. There is another parametrization of quadratic fields, given by the 1-1 correspondence between $m$ and fundamental discriminants $d$. In particular, for each fundamental discriminant $d$ there is exactly one quadratic field of that discriminant.



Is there a similar parametrization for cubic fields?






algebraic-number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Kummer theory generalizes your parametrization of the quadratic fields. A special case of Kummer theory says that if $K$ is a field with characteristic different from a prime $p$, and if $K$ contains all the $p^{text{th}}$ roots of unity, then the extensions of $K$ with Galois group $C_p$ (cyclic of order $p$) are parametrized by non-trivial elements of $K^times/K^{times p}$, where $K(sqrt[p]a)$ corresponds to $a in K^times/K^{times p}$.
    $endgroup$
    – Kenny Lau
    Jan 3 at 18:09












  • $begingroup$
    I believe if $K/Bbb Q$ is a cubic extension and $K=Bbb Q(alpha)$ then $Bbb Q(omega,alpha)/Bbb Q(omega)$ is still a cubic extension... in that case $Bbb Q(omega,alpha) = Bbb Q(omega,(p+qomega)^{1/3})$ for some $p,q in Bbb Z$ for a choice of the cube root by Kummer theory... but I also suspect that it is possible to have $Bbb Q(omega,alpha) = Bbb Q(omega,beta)$ with $Bbb Q(alpha) ne Bbb Q(beta)$
    $endgroup$
    – Kenny Lau
    Jan 3 at 18:47










  • $begingroup$
    If $K=Bbb Q(alpha)/Bbb Q$ is cubic, i.e. $f(alpha) = 0$ for cubic $f$, then we have two situations depending on the discriminant of $f$: if the discriminant is a perfect square, then $K$ has only one conjugate (itself); otherwise, $K$ has three conjugates. Using this, we can shift the question to classifying splitting fields of (irreducible) cubic polynomials.
    $endgroup$
    – Kenny Lau
    Jan 3 at 18:54












  • $begingroup$
    An elementary observation is that if $K/Bbb Q$ is cubic then $K=Bbb Q(alpha)$ for every $alpha in K setminus Bbb Q$.
    $endgroup$
    – Kenny Lau
    Jan 3 at 19:05










  • $begingroup$
    @Kenny Lau: Thanks for the answers. Are there isomorphic unequal cubic fields?
    $endgroup$
    – sai
    Jan 5 at 2:41














5












5








5


1



$begingroup$


A parametrization of quadratic fields is $mathbb{Q}(sqrt{m})$, where $mne1$ is a squarefree integer. That is, $mathbb{Q}(sqrt{m})$ is a quadratic field as $m$ varies, and all quadratic fields are included. There is another parametrization of quadratic fields, given by the 1-1 correspondence between $m$ and fundamental discriminants $d$. In particular, for each fundamental discriminant $d$ there is exactly one quadratic field of that discriminant.



Is there a similar parametrization for cubic fields?






algebraic-number-theory







share|cite|improve this question









$endgroup$




A parametrization of quadratic fields is $mathbb{Q}(sqrt{m})$, where $mne1$ is a squarefree integer. That is, $mathbb{Q}(sqrt{m})$ is a quadratic field as $m$ varies, and all quadratic fields are included. There is another parametrization of quadratic fields, given by the 1-1 correspondence between $m$ and fundamental discriminants $d$. In particular, for each fundamental discriminant $d$ there is exactly one quadratic field of that discriminant.



Is there a similar parametrization for cubic fields?






algebraic-number-theory




algebraic-number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 3 at 11:55









saisai

1376




1376












  • $begingroup$
    Kummer theory generalizes your parametrization of the quadratic fields. A special case of Kummer theory says that if $K$ is a field with characteristic different from a prime $p$, and if $K$ contains all the $p^{text{th}}$ roots of unity, then the extensions of $K$ with Galois group $C_p$ (cyclic of order $p$) are parametrized by non-trivial elements of $K^times/K^{times p}$, where $K(sqrt[p]a)$ corresponds to $a in K^times/K^{times p}$.
    $endgroup$
    – Kenny Lau
    Jan 3 at 18:09












  • $begingroup$
    I believe if $K/Bbb Q$ is a cubic extension and $K=Bbb Q(alpha)$ then $Bbb Q(omega,alpha)/Bbb Q(omega)$ is still a cubic extension... in that case $Bbb Q(omega,alpha) = Bbb Q(omega,(p+qomega)^{1/3})$ for some $p,q in Bbb Z$ for a choice of the cube root by Kummer theory... but I also suspect that it is possible to have $Bbb Q(omega,alpha) = Bbb Q(omega,beta)$ with $Bbb Q(alpha) ne Bbb Q(beta)$
    $endgroup$
    – Kenny Lau
    Jan 3 at 18:47










  • $begingroup$
    If $K=Bbb Q(alpha)/Bbb Q$ is cubic, i.e. $f(alpha) = 0$ for cubic $f$, then we have two situations depending on the discriminant of $f$: if the discriminant is a perfect square, then $K$ has only one conjugate (itself); otherwise, $K$ has three conjugates. Using this, we can shift the question to classifying splitting fields of (irreducible) cubic polynomials.
    $endgroup$
    – Kenny Lau
    Jan 3 at 18:54












  • $begingroup$
    An elementary observation is that if $K/Bbb Q$ is cubic then $K=Bbb Q(alpha)$ for every $alpha in K setminus Bbb Q$.
    $endgroup$
    – Kenny Lau
    Jan 3 at 19:05










  • $begingroup$
    @Kenny Lau: Thanks for the answers. Are there isomorphic unequal cubic fields?
    $endgroup$
    – sai
    Jan 5 at 2:41


















  • $begingroup$
    Kummer theory generalizes your parametrization of the quadratic fields. A special case of Kummer theory says that if $K$ is a field with characteristic different from a prime $p$, and if $K$ contains all the $p^{text{th}}$ roots of unity, then the extensions of $K$ with Galois group $C_p$ (cyclic of order $p$) are parametrized by non-trivial elements of $K^times/K^{times p}$, where $K(sqrt[p]a)$ corresponds to $a in K^times/K^{times p}$.
    $endgroup$
    – Kenny Lau
    Jan 3 at 18:09












  • $begingroup$
    I believe if $K/Bbb Q$ is a cubic extension and $K=Bbb Q(alpha)$ then $Bbb Q(omega,alpha)/Bbb Q(omega)$ is still a cubic extension... in that case $Bbb Q(omega,alpha) = Bbb Q(omega,(p+qomega)^{1/3})$ for some $p,q in Bbb Z$ for a choice of the cube root by Kummer theory... but I also suspect that it is possible to have $Bbb Q(omega,alpha) = Bbb Q(omega,beta)$ with $Bbb Q(alpha) ne Bbb Q(beta)$
    $endgroup$
    – Kenny Lau
    Jan 3 at 18:47










  • $begingroup$
    If $K=Bbb Q(alpha)/Bbb Q$ is cubic, i.e. $f(alpha) = 0$ for cubic $f$, then we have two situations depending on the discriminant of $f$: if the discriminant is a perfect square, then $K$ has only one conjugate (itself); otherwise, $K$ has three conjugates. Using this, we can shift the question to classifying splitting fields of (irreducible) cubic polynomials.
    $endgroup$
    – Kenny Lau
    Jan 3 at 18:54












  • $begingroup$
    An elementary observation is that if $K/Bbb Q$ is cubic then $K=Bbb Q(alpha)$ for every $alpha in K setminus Bbb Q$.
    $endgroup$
    – Kenny Lau
    Jan 3 at 19:05










  • $begingroup$
    @Kenny Lau: Thanks for the answers. Are there isomorphic unequal cubic fields?
    $endgroup$
    – sai
    Jan 5 at 2:41
















$begingroup$
Kummer theory generalizes your parametrization of the quadratic fields. A special case of Kummer theory says that if $K$ is a field with characteristic different from a prime $p$, and if $K$ contains all the $p^{text{th}}$ roots of unity, then the extensions of $K$ with Galois group $C_p$ (cyclic of order $p$) are parametrized by non-trivial elements of $K^times/K^{times p}$, where $K(sqrt[p]a)$ corresponds to $a in K^times/K^{times p}$.
$endgroup$
– Kenny Lau
Jan 3 at 18:09






$begingroup$
Kummer theory generalizes your parametrization of the quadratic fields. A special case of Kummer theory says that if $K$ is a field with characteristic different from a prime $p$, and if $K$ contains all the $p^{text{th}}$ roots of unity, then the extensions of $K$ with Galois group $C_p$ (cyclic of order $p$) are parametrized by non-trivial elements of $K^times/K^{times p}$, where $K(sqrt[p]a)$ corresponds to $a in K^times/K^{times p}$.
$endgroup$
– Kenny Lau
Jan 3 at 18:09














$begingroup$
I believe if $K/Bbb Q$ is a cubic extension and $K=Bbb Q(alpha)$ then $Bbb Q(omega,alpha)/Bbb Q(omega)$ is still a cubic extension... in that case $Bbb Q(omega,alpha) = Bbb Q(omega,(p+qomega)^{1/3})$ for some $p,q in Bbb Z$ for a choice of the cube root by Kummer theory... but I also suspect that it is possible to have $Bbb Q(omega,alpha) = Bbb Q(omega,beta)$ with $Bbb Q(alpha) ne Bbb Q(beta)$
$endgroup$
– Kenny Lau
Jan 3 at 18:47




$begingroup$
I believe if $K/Bbb Q$ is a cubic extension and $K=Bbb Q(alpha)$ then $Bbb Q(omega,alpha)/Bbb Q(omega)$ is still a cubic extension... in that case $Bbb Q(omega,alpha) = Bbb Q(omega,(p+qomega)^{1/3})$ for some $p,q in Bbb Z$ for a choice of the cube root by Kummer theory... but I also suspect that it is possible to have $Bbb Q(omega,alpha) = Bbb Q(omega,beta)$ with $Bbb Q(alpha) ne Bbb Q(beta)$
$endgroup$
– Kenny Lau
Jan 3 at 18:47












$begingroup$
If $K=Bbb Q(alpha)/Bbb Q$ is cubic, i.e. $f(alpha) = 0$ for cubic $f$, then we have two situations depending on the discriminant of $f$: if the discriminant is a perfect square, then $K$ has only one conjugate (itself); otherwise, $K$ has three conjugates. Using this, we can shift the question to classifying splitting fields of (irreducible) cubic polynomials.
$endgroup$
– Kenny Lau
Jan 3 at 18:54






$begingroup$
If $K=Bbb Q(alpha)/Bbb Q$ is cubic, i.e. $f(alpha) = 0$ for cubic $f$, then we have two situations depending on the discriminant of $f$: if the discriminant is a perfect square, then $K$ has only one conjugate (itself); otherwise, $K$ has three conjugates. Using this, we can shift the question to classifying splitting fields of (irreducible) cubic polynomials.
$endgroup$
– Kenny Lau
Jan 3 at 18:54














$begingroup$
An elementary observation is that if $K/Bbb Q$ is cubic then $K=Bbb Q(alpha)$ for every $alpha in K setminus Bbb Q$.
$endgroup$
– Kenny Lau
Jan 3 at 19:05




$begingroup$
An elementary observation is that if $K/Bbb Q$ is cubic then $K=Bbb Q(alpha)$ for every $alpha in K setminus Bbb Q$.
$endgroup$
– Kenny Lau
Jan 3 at 19:05












$begingroup$
@Kenny Lau: Thanks for the answers. Are there isomorphic unequal cubic fields?
$endgroup$
– sai
Jan 5 at 2:41




$begingroup$
@Kenny Lau: Thanks for the answers. Are there isomorphic unequal cubic fields?
$endgroup$
– sai
Jan 5 at 2:41










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