Minimal polynomial for normal closure












1












$begingroup$


I came across this problem while studying Normal closures.



Given $K=mathbb{Q}$ and the polynomial $x^3-2in K[x]$. $L/K$ is not normal where $L=mathbb{Q}(2^{frac13})$ since $omeganotin L $ where $omega$ is the cube root of unity. The normal closure of $L/K$ is $M/L$ where $M=L(omega)=mathbb{Q}(2^{frac13},omega)$ and $M/K$ is normal since $M$ is the splitting field for $x^3-2in K[x]$.



$L/K$ is finite extension and $[L:K]=3$ $left(deg(m_{2^{frac13}})=3,; m_{2^{frac13}}=x^3-2 right)$ and $M/L$ is finite extension and $[M:L]=3$ $left(deg(m_omega)=3, ; m_omega=x^3-1 right)$.



$[M:K]=[M:L][L:K]=9$ seems strange. Minimal polynomial associated with $2^{frac13}$ and $omega$ is $x^3-2$ which does not have degree $9$. Where am I wrong?



My question: What is the minimal polynomial of degree 9 associated with $2^{frac13}$ and $omega$?










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$endgroup$








  • 1




    $begingroup$
    $|M:K|=6$, not $9$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 17:47










  • $begingroup$
    @LordSharktheUnknown: So you are saying the minimal polynomial is $(x^3-1)(x^3-2)$. But isn't the product reducible in $mathbb{Q}$? (Minimal=irreducible)
    $endgroup$
    – Yadati Kiran
    Jan 1 at 17:52












  • $begingroup$
    I said no such thing.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 17:57










  • $begingroup$
    @LordSharktheUnknown: Could you give me a hint as to how you said $[M:K]=6$?
    $endgroup$
    – Yadati Kiran
    Jan 1 at 18:00


















1












$begingroup$


I came across this problem while studying Normal closures.



Given $K=mathbb{Q}$ and the polynomial $x^3-2in K[x]$. $L/K$ is not normal where $L=mathbb{Q}(2^{frac13})$ since $omeganotin L $ where $omega$ is the cube root of unity. The normal closure of $L/K$ is $M/L$ where $M=L(omega)=mathbb{Q}(2^{frac13},omega)$ and $M/K$ is normal since $M$ is the splitting field for $x^3-2in K[x]$.



$L/K$ is finite extension and $[L:K]=3$ $left(deg(m_{2^{frac13}})=3,; m_{2^{frac13}}=x^3-2 right)$ and $M/L$ is finite extension and $[M:L]=3$ $left(deg(m_omega)=3, ; m_omega=x^3-1 right)$.



$[M:K]=[M:L][L:K]=9$ seems strange. Minimal polynomial associated with $2^{frac13}$ and $omega$ is $x^3-2$ which does not have degree $9$. Where am I wrong?



My question: What is the minimal polynomial of degree 9 associated with $2^{frac13}$ and $omega$?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $|M:K|=6$, not $9$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 17:47










  • $begingroup$
    @LordSharktheUnknown: So you are saying the minimal polynomial is $(x^3-1)(x^3-2)$. But isn't the product reducible in $mathbb{Q}$? (Minimal=irreducible)
    $endgroup$
    – Yadati Kiran
    Jan 1 at 17:52












  • $begingroup$
    I said no such thing.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 17:57










  • $begingroup$
    @LordSharktheUnknown: Could you give me a hint as to how you said $[M:K]=6$?
    $endgroup$
    – Yadati Kiran
    Jan 1 at 18:00
















1












1








1





$begingroup$


I came across this problem while studying Normal closures.



Given $K=mathbb{Q}$ and the polynomial $x^3-2in K[x]$. $L/K$ is not normal where $L=mathbb{Q}(2^{frac13})$ since $omeganotin L $ where $omega$ is the cube root of unity. The normal closure of $L/K$ is $M/L$ where $M=L(omega)=mathbb{Q}(2^{frac13},omega)$ and $M/K$ is normal since $M$ is the splitting field for $x^3-2in K[x]$.



$L/K$ is finite extension and $[L:K]=3$ $left(deg(m_{2^{frac13}})=3,; m_{2^{frac13}}=x^3-2 right)$ and $M/L$ is finite extension and $[M:L]=3$ $left(deg(m_omega)=3, ; m_omega=x^3-1 right)$.



$[M:K]=[M:L][L:K]=9$ seems strange. Minimal polynomial associated with $2^{frac13}$ and $omega$ is $x^3-2$ which does not have degree $9$. Where am I wrong?



My question: What is the minimal polynomial of degree 9 associated with $2^{frac13}$ and $omega$?










share|cite|improve this question











$endgroup$




I came across this problem while studying Normal closures.



Given $K=mathbb{Q}$ and the polynomial $x^3-2in K[x]$. $L/K$ is not normal where $L=mathbb{Q}(2^{frac13})$ since $omeganotin L $ where $omega$ is the cube root of unity. The normal closure of $L/K$ is $M/L$ where $M=L(omega)=mathbb{Q}(2^{frac13},omega)$ and $M/K$ is normal since $M$ is the splitting field for $x^3-2in K[x]$.



$L/K$ is finite extension and $[L:K]=3$ $left(deg(m_{2^{frac13}})=3,; m_{2^{frac13}}=x^3-2 right)$ and $M/L$ is finite extension and $[M:L]=3$ $left(deg(m_omega)=3, ; m_omega=x^3-1 right)$.



$[M:K]=[M:L][L:K]=9$ seems strange. Minimal polynomial associated with $2^{frac13}$ and $omega$ is $x^3-2$ which does not have degree $9$. Where am I wrong?



My question: What is the minimal polynomial of degree 9 associated with $2^{frac13}$ and $omega$?







field-theory splitting-field






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 18:02







Yadati Kiran

















asked Jan 1 at 17:41









Yadati KiranYadati Kiran

1,7951619




1,7951619








  • 1




    $begingroup$
    $|M:K|=6$, not $9$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 17:47










  • $begingroup$
    @LordSharktheUnknown: So you are saying the minimal polynomial is $(x^3-1)(x^3-2)$. But isn't the product reducible in $mathbb{Q}$? (Minimal=irreducible)
    $endgroup$
    – Yadati Kiran
    Jan 1 at 17:52












  • $begingroup$
    I said no such thing.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 17:57










  • $begingroup$
    @LordSharktheUnknown: Could you give me a hint as to how you said $[M:K]=6$?
    $endgroup$
    – Yadati Kiran
    Jan 1 at 18:00
















  • 1




    $begingroup$
    $|M:K|=6$, not $9$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 17:47










  • $begingroup$
    @LordSharktheUnknown: So you are saying the minimal polynomial is $(x^3-1)(x^3-2)$. But isn't the product reducible in $mathbb{Q}$? (Minimal=irreducible)
    $endgroup$
    – Yadati Kiran
    Jan 1 at 17:52












  • $begingroup$
    I said no such thing.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 17:57










  • $begingroup$
    @LordSharktheUnknown: Could you give me a hint as to how you said $[M:K]=6$?
    $endgroup$
    – Yadati Kiran
    Jan 1 at 18:00










1




1




$begingroup$
$|M:K|=6$, not $9$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 17:47




$begingroup$
$|M:K|=6$, not $9$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 17:47












$begingroup$
@LordSharktheUnknown: So you are saying the minimal polynomial is $(x^3-1)(x^3-2)$. But isn't the product reducible in $mathbb{Q}$? (Minimal=irreducible)
$endgroup$
– Yadati Kiran
Jan 1 at 17:52






$begingroup$
@LordSharktheUnknown: So you are saying the minimal polynomial is $(x^3-1)(x^3-2)$. But isn't the product reducible in $mathbb{Q}$? (Minimal=irreducible)
$endgroup$
– Yadati Kiran
Jan 1 at 17:52














$begingroup$
I said no such thing.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 17:57




$begingroup$
I said no such thing.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 17:57












$begingroup$
@LordSharktheUnknown: Could you give me a hint as to how you said $[M:K]=6$?
$endgroup$
– Yadati Kiran
Jan 1 at 18:00






$begingroup$
@LordSharktheUnknown: Could you give me a hint as to how you said $[M:K]=6$?
$endgroup$
– Yadati Kiran
Jan 1 at 18:00












1 Answer
1






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1












$begingroup$

The minimal polynomial of $omega$ over $mathbb Q(sqrt[3]{2})$ is actually
$$ m(X) = 1 + X + X^2 in mathbb Q(sqrt[3]{2})[X].$$ Indeed, $m(X)$ is irreducible over $mathbb Q(sqrt[3]{2})$, and has $omega$ as one of its roots. $m(X)$ has degree two, which implies that
$$[mathbb Q(sqrt[3]{2}, omega) : mathbb Q(sqrt[3]{2})] = 2,$$ and so, by the tower law,$$[mathbb Q(sqrt[3]{2}, omega) : mathbb Q] = [mathbb Q(sqrt[3]{2}, omega) : mathbb Q(sqrt[3]{2})]times[ mathbb Q(sqrt[3]{2}) : mathbb Q] = 2 times 3 = 6.$$



As for your second remark, the splitting field of a degree $d$ polynomial $f(X) in mathbb Q[X]$ does not need to be a degree $d$ extension of $mathbb Q$. It can be anything from a degree $1$ extension to a degree $d!$ extension.



Take a look at these examples of degree three polynomials in $mathbb Q[X]$:




  • $ f(X) = X^3$ splits completely over $mathbb Q$, so its splitting field is simply $mathbb Q$, which is (trivially) a degree one extension of $mathbb Q$.


  • $f(X) = X^3 - 1$ has splitting field $mathbb Q(omega)$, which is a degree two extension of $mathbb Q$.


  • $f(X) = X^3 - 3X + 1$ is discussed here, where it is shown that its splitting field is a degree three extension of $mathbb Q$.


  • $f(X) = X^3 - 2$ is the example you're interested in. We showed that its splitting field is $mathbb Q(sqrt[3]{2}, omega) $, a degree six extension of $mathbb Q$.



So there is nothing to worry about.



Finally, a point about terminology: there is no such thing as "the minimal polynomial of $sqrt[3]{2}$ and $omega$". You can talk about the minimal polynomial of $sqrt[3]{2}$ over $mathbb Q$ (which is $X^3 - 2$), and you can talk about the minimal polynomial of $omega$ over $mathbb Q$ (which is $1 + X + X^2$). You can also talk about the minimal polyomial of $omega$ over $mathbb Q(sqrt[3]{2})$ (which also happens to be $1 + X + X^2$). But it doesn't make sense to talk about "the minimal polynomial of
$sqrt[3]{2}$ and $omega$". Nor is such a concept required in order to find the degree of the extension $[mathbb Q(sqrt[3]{2}, omega) : mathbb Q]$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the detailed answer. I shall take note of your point regarding the terminology.
    $endgroup$
    – Yadati Kiran
    Jan 2 at 12:13











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1 Answer
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1 Answer
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1












$begingroup$

The minimal polynomial of $omega$ over $mathbb Q(sqrt[3]{2})$ is actually
$$ m(X) = 1 + X + X^2 in mathbb Q(sqrt[3]{2})[X].$$ Indeed, $m(X)$ is irreducible over $mathbb Q(sqrt[3]{2})$, and has $omega$ as one of its roots. $m(X)$ has degree two, which implies that
$$[mathbb Q(sqrt[3]{2}, omega) : mathbb Q(sqrt[3]{2})] = 2,$$ and so, by the tower law,$$[mathbb Q(sqrt[3]{2}, omega) : mathbb Q] = [mathbb Q(sqrt[3]{2}, omega) : mathbb Q(sqrt[3]{2})]times[ mathbb Q(sqrt[3]{2}) : mathbb Q] = 2 times 3 = 6.$$



As for your second remark, the splitting field of a degree $d$ polynomial $f(X) in mathbb Q[X]$ does not need to be a degree $d$ extension of $mathbb Q$. It can be anything from a degree $1$ extension to a degree $d!$ extension.



Take a look at these examples of degree three polynomials in $mathbb Q[X]$:




  • $ f(X) = X^3$ splits completely over $mathbb Q$, so its splitting field is simply $mathbb Q$, which is (trivially) a degree one extension of $mathbb Q$.


  • $f(X) = X^3 - 1$ has splitting field $mathbb Q(omega)$, which is a degree two extension of $mathbb Q$.


  • $f(X) = X^3 - 3X + 1$ is discussed here, where it is shown that its splitting field is a degree three extension of $mathbb Q$.


  • $f(X) = X^3 - 2$ is the example you're interested in. We showed that its splitting field is $mathbb Q(sqrt[3]{2}, omega) $, a degree six extension of $mathbb Q$.



So there is nothing to worry about.



Finally, a point about terminology: there is no such thing as "the minimal polynomial of $sqrt[3]{2}$ and $omega$". You can talk about the minimal polynomial of $sqrt[3]{2}$ over $mathbb Q$ (which is $X^3 - 2$), and you can talk about the minimal polynomial of $omega$ over $mathbb Q$ (which is $1 + X + X^2$). You can also talk about the minimal polyomial of $omega$ over $mathbb Q(sqrt[3]{2})$ (which also happens to be $1 + X + X^2$). But it doesn't make sense to talk about "the minimal polynomial of
$sqrt[3]{2}$ and $omega$". Nor is such a concept required in order to find the degree of the extension $[mathbb Q(sqrt[3]{2}, omega) : mathbb Q]$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the detailed answer. I shall take note of your point regarding the terminology.
    $endgroup$
    – Yadati Kiran
    Jan 2 at 12:13
















1












$begingroup$

The minimal polynomial of $omega$ over $mathbb Q(sqrt[3]{2})$ is actually
$$ m(X) = 1 + X + X^2 in mathbb Q(sqrt[3]{2})[X].$$ Indeed, $m(X)$ is irreducible over $mathbb Q(sqrt[3]{2})$, and has $omega$ as one of its roots. $m(X)$ has degree two, which implies that
$$[mathbb Q(sqrt[3]{2}, omega) : mathbb Q(sqrt[3]{2})] = 2,$$ and so, by the tower law,$$[mathbb Q(sqrt[3]{2}, omega) : mathbb Q] = [mathbb Q(sqrt[3]{2}, omega) : mathbb Q(sqrt[3]{2})]times[ mathbb Q(sqrt[3]{2}) : mathbb Q] = 2 times 3 = 6.$$



As for your second remark, the splitting field of a degree $d$ polynomial $f(X) in mathbb Q[X]$ does not need to be a degree $d$ extension of $mathbb Q$. It can be anything from a degree $1$ extension to a degree $d!$ extension.



Take a look at these examples of degree three polynomials in $mathbb Q[X]$:




  • $ f(X) = X^3$ splits completely over $mathbb Q$, so its splitting field is simply $mathbb Q$, which is (trivially) a degree one extension of $mathbb Q$.


  • $f(X) = X^3 - 1$ has splitting field $mathbb Q(omega)$, which is a degree two extension of $mathbb Q$.


  • $f(X) = X^3 - 3X + 1$ is discussed here, where it is shown that its splitting field is a degree three extension of $mathbb Q$.


  • $f(X) = X^3 - 2$ is the example you're interested in. We showed that its splitting field is $mathbb Q(sqrt[3]{2}, omega) $, a degree six extension of $mathbb Q$.



So there is nothing to worry about.



Finally, a point about terminology: there is no such thing as "the minimal polynomial of $sqrt[3]{2}$ and $omega$". You can talk about the minimal polynomial of $sqrt[3]{2}$ over $mathbb Q$ (which is $X^3 - 2$), and you can talk about the minimal polynomial of $omega$ over $mathbb Q$ (which is $1 + X + X^2$). You can also talk about the minimal polyomial of $omega$ over $mathbb Q(sqrt[3]{2})$ (which also happens to be $1 + X + X^2$). But it doesn't make sense to talk about "the minimal polynomial of
$sqrt[3]{2}$ and $omega$". Nor is such a concept required in order to find the degree of the extension $[mathbb Q(sqrt[3]{2}, omega) : mathbb Q]$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the detailed answer. I shall take note of your point regarding the terminology.
    $endgroup$
    – Yadati Kiran
    Jan 2 at 12:13














1












1








1





$begingroup$

The minimal polynomial of $omega$ over $mathbb Q(sqrt[3]{2})$ is actually
$$ m(X) = 1 + X + X^2 in mathbb Q(sqrt[3]{2})[X].$$ Indeed, $m(X)$ is irreducible over $mathbb Q(sqrt[3]{2})$, and has $omega$ as one of its roots. $m(X)$ has degree two, which implies that
$$[mathbb Q(sqrt[3]{2}, omega) : mathbb Q(sqrt[3]{2})] = 2,$$ and so, by the tower law,$$[mathbb Q(sqrt[3]{2}, omega) : mathbb Q] = [mathbb Q(sqrt[3]{2}, omega) : mathbb Q(sqrt[3]{2})]times[ mathbb Q(sqrt[3]{2}) : mathbb Q] = 2 times 3 = 6.$$



As for your second remark, the splitting field of a degree $d$ polynomial $f(X) in mathbb Q[X]$ does not need to be a degree $d$ extension of $mathbb Q$. It can be anything from a degree $1$ extension to a degree $d!$ extension.



Take a look at these examples of degree three polynomials in $mathbb Q[X]$:




  • $ f(X) = X^3$ splits completely over $mathbb Q$, so its splitting field is simply $mathbb Q$, which is (trivially) a degree one extension of $mathbb Q$.


  • $f(X) = X^3 - 1$ has splitting field $mathbb Q(omega)$, which is a degree two extension of $mathbb Q$.


  • $f(X) = X^3 - 3X + 1$ is discussed here, where it is shown that its splitting field is a degree three extension of $mathbb Q$.


  • $f(X) = X^3 - 2$ is the example you're interested in. We showed that its splitting field is $mathbb Q(sqrt[3]{2}, omega) $, a degree six extension of $mathbb Q$.



So there is nothing to worry about.



Finally, a point about terminology: there is no such thing as "the minimal polynomial of $sqrt[3]{2}$ and $omega$". You can talk about the minimal polynomial of $sqrt[3]{2}$ over $mathbb Q$ (which is $X^3 - 2$), and you can talk about the minimal polynomial of $omega$ over $mathbb Q$ (which is $1 + X + X^2$). You can also talk about the minimal polyomial of $omega$ over $mathbb Q(sqrt[3]{2})$ (which also happens to be $1 + X + X^2$). But it doesn't make sense to talk about "the minimal polynomial of
$sqrt[3]{2}$ and $omega$". Nor is such a concept required in order to find the degree of the extension $[mathbb Q(sqrt[3]{2}, omega) : mathbb Q]$.






share|cite|improve this answer











$endgroup$



The minimal polynomial of $omega$ over $mathbb Q(sqrt[3]{2})$ is actually
$$ m(X) = 1 + X + X^2 in mathbb Q(sqrt[3]{2})[X].$$ Indeed, $m(X)$ is irreducible over $mathbb Q(sqrt[3]{2})$, and has $omega$ as one of its roots. $m(X)$ has degree two, which implies that
$$[mathbb Q(sqrt[3]{2}, omega) : mathbb Q(sqrt[3]{2})] = 2,$$ and so, by the tower law,$$[mathbb Q(sqrt[3]{2}, omega) : mathbb Q] = [mathbb Q(sqrt[3]{2}, omega) : mathbb Q(sqrt[3]{2})]times[ mathbb Q(sqrt[3]{2}) : mathbb Q] = 2 times 3 = 6.$$



As for your second remark, the splitting field of a degree $d$ polynomial $f(X) in mathbb Q[X]$ does not need to be a degree $d$ extension of $mathbb Q$. It can be anything from a degree $1$ extension to a degree $d!$ extension.



Take a look at these examples of degree three polynomials in $mathbb Q[X]$:




  • $ f(X) = X^3$ splits completely over $mathbb Q$, so its splitting field is simply $mathbb Q$, which is (trivially) a degree one extension of $mathbb Q$.


  • $f(X) = X^3 - 1$ has splitting field $mathbb Q(omega)$, which is a degree two extension of $mathbb Q$.


  • $f(X) = X^3 - 3X + 1$ is discussed here, where it is shown that its splitting field is a degree three extension of $mathbb Q$.


  • $f(X) = X^3 - 2$ is the example you're interested in. We showed that its splitting field is $mathbb Q(sqrt[3]{2}, omega) $, a degree six extension of $mathbb Q$.



So there is nothing to worry about.



Finally, a point about terminology: there is no such thing as "the minimal polynomial of $sqrt[3]{2}$ and $omega$". You can talk about the minimal polynomial of $sqrt[3]{2}$ over $mathbb Q$ (which is $X^3 - 2$), and you can talk about the minimal polynomial of $omega$ over $mathbb Q$ (which is $1 + X + X^2$). You can also talk about the minimal polyomial of $omega$ over $mathbb Q(sqrt[3]{2})$ (which also happens to be $1 + X + X^2$). But it doesn't make sense to talk about "the minimal polynomial of
$sqrt[3]{2}$ and $omega$". Nor is such a concept required in order to find the degree of the extension $[mathbb Q(sqrt[3]{2}, omega) : mathbb Q]$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 1 at 18:31

























answered Jan 1 at 18:25









Kenny WongKenny Wong

19k21440




19k21440












  • $begingroup$
    Thank you for the detailed answer. I shall take note of your point regarding the terminology.
    $endgroup$
    – Yadati Kiran
    Jan 2 at 12:13


















  • $begingroup$
    Thank you for the detailed answer. I shall take note of your point regarding the terminology.
    $endgroup$
    – Yadati Kiran
    Jan 2 at 12:13
















$begingroup$
Thank you for the detailed answer. I shall take note of your point regarding the terminology.
$endgroup$
– Yadati Kiran
Jan 2 at 12:13




$begingroup$
Thank you for the detailed answer. I shall take note of your point regarding the terminology.
$endgroup$
– Yadati Kiran
Jan 2 at 12:13


















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