Solubility of general cubic by radicals
My task is to use the following two theorems to prove that any cubic is soluble by radicals. We are not allowed to use that $Bbb S_3$ is a soluble group.
Theorem 1: Hilbert's Theorem 90
Let $L:K$ be a finite normal extension with cyclic galois group generated by $tau$. Then $a$$in$L has norm 1 iff $a=frac{b}{tau(b)}$, for some $b in L$.
Here, $N(a)=tau_1(a)...tau_n(a)$, where $tau_i$ are the elements of the Galois Group $G$,corresponding to a finite, normal extension $L:K$.
Theorem 2:
Suppose $L:K$ is a finite normal extension with cyclic Galois Group of prime order generated by $tau$. If K has characteristic 0, (or characteristic which is prime to p), and $t^p-1$ splits in K. Then $L=K(a)$, where $a$ is a zero of an irreducible polynomial $t^p-beta$ over K for some $beta in K$.
I am a bit confused, as I am not sure what I am supposed to do.
I start with $Bbb Q$, and then adjoin $omega$, where $omega^3=1$, but is not equal to 1.
Then over $K=Bbb Q(omega)$, we have $N(omega)=1$, so according to Theorem 1, we may write $omega=frac{b}{tau(b)}$.
If I am not mistaken $tau=(123)$, and the Galois Group of the extension $L:K$ is $Bbb A_3$.
According to Ian Stewart's Galois Theory book, $b=omega c+omega tau(omega) tau(c)+ omegatau(omega)tau^2(omega)tau^2(c)=omega c+ omega^2 tau(c) +tau^2(c).$
Then, it follows that $b^3=tau(b^3)$, so $b^3 in <tau>^dagger$.
Now, using theorem 2, since $K$ splits over $t^3-1$, and $L:K$ has cyclic galois group $Bbb A_3$, then $L=K(alpha)$, which is a radical extensions of $Bbb Q$.
I am not sure whether my reasoning is correct. The whole idea is to show that the general cubic, $f(x)$, is soluble by radicals by showing that there exists a field that contains a a splitting field for $f$ such that the extension $L:Bbb Q$ is a radical extension.
polynomials field-theory galois-theory galois-extensions cyclotomic-fields
asked Nov 26 at 14:06
ʎpoqou
1741111
add a comment |
My task is to use the following two theorems to prove that any cubic is soluble by radicals. We are not allowed to use that $Bbb S_3$ is a soluble group.
Theorem 1: Hilbert's Theorem 90
Let $L:K$ be a finite normal extension with cyclic galois group generated by $tau$. Then $a$$in$L has norm 1 iff $a=frac{b}{tau(b)}$, for some $b in L$.
Here, $N(a)=tau_1(a)...tau_n(a)$, where $tau_i$ are the elements of the Galois Group $G$,corresponding to a finite, normal extension $L:K$.
Theorem 2:
Suppose $L:K$ is a finite normal extension with cyclic Galois Group of prime order generated by $tau$. If K has characteristic 0, (or characteristic which is prime to p), and $t^p-1$ splits in K. Then $L=K(a)$, where $a$ is a zero of an irreducible polynomial $t^p-beta$ over K for some $beta in K$.
I am a bit confused, as I am not sure what I am supposed to do.
I start with $Bbb Q$, and then adjoin $omega$, where $omega^3=1$, but is not equal to 1.
Then over $K=Bbb Q(omega)$, we have $N(omega)=1$, so according to Theorem 1, we may write $omega=frac{b}{tau(b)}$.
If I am not mistaken $tau=(123)$, and the Galois Group of the extension $L:K$ is $Bbb A_3$.
According to Ian Stewart's Galois Theory book, $b=omega c+omega tau(omega) tau(c)+ omegatau(omega)tau^2(omega)tau^2(c)=omega c+ omega^2 tau(c) +tau^2(c).$
Then, it follows that $b^3=tau(b^3)$, so $b^3 in <tau>^dagger$.
Now, using theorem 2, since $K$ splits over $t^3-1$, and $L:K$ has cyclic galois group $Bbb A_3$, then $L=K(alpha)$, which is a radical extensions of $Bbb Q$.
I am not sure whether my reasoning is correct. The whole idea is to show that the general cubic, $f(x)$, is soluble by radicals by showing that there exists a field that contains a a splitting field for $f$ such that the extension $L:Bbb Q$ is a radical extension.
polynomials field-theory galois-theory galois-extensions cyclotomic-fields
asked Nov 26 at 14:06
ʎpoqou
1741111
add a comment |
My task is to use the following two theorems to prove that any cubic is soluble by radicals. We are not allowed to use that $Bbb S_3$ is a soluble group.
Theorem 1: Hilbert's Theorem 90
Let $L:K$ be a finite normal extension with cyclic galois group generated by $tau$. Then $a$$in$L has norm 1 iff $a=frac{b}{tau(b)}$, for some $b in L$.
Here, $N(a)=tau_1(a)...tau_n(a)$, where $tau_i$ are the elements of the Galois Group $G$,corresponding to a finite, normal extension $L:K$.
Theorem 2:
Suppose $L:K$ is a finite normal extension with cyclic Galois Group of prime order generated by $tau$. If K has characteristic 0, (or characteristic which is prime to p), and $t^p-1$ splits in K. Then $L=K(a)$, where $a$ is a zero of an irreducible polynomial $t^p-beta$ over K for some $beta in K$.
I am a bit confused, as I am not sure what I am supposed to do.
I start with $Bbb Q$, and then adjoin $omega$, where $omega^3=1$, but is not equal to 1.
Then over $K=Bbb Q(omega)$, we have $N(omega)=1$, so according to Theorem 1, we may write $omega=frac{b}{tau(b)}$.
If I am not mistaken $tau=(123)$, and the Galois Group of the extension $L:K$ is $Bbb A_3$.
According to Ian Stewart's Galois Theory book, $b=omega c+omega tau(omega) tau(c)+ omegatau(omega)tau^2(omega)tau^2(c)=omega c+ omega^2 tau(c) +tau^2(c).$
Then, it follows that $b^3=tau(b^3)$, so $b^3 in <tau>^dagger$.
Now, using theorem 2, since $K$ splits over $t^3-1$, and $L:K$ has cyclic galois group $Bbb A_3$, then $L=K(alpha)$, which is a radical extensions of $Bbb Q$.
I am not sure whether my reasoning is correct. The whole idea is to show that the general cubic, $f(x)$, is soluble by radicals by showing that there exists a field that contains a a splitting field for $f$ such that the extension $L:Bbb Q$ is a radical extension.
polynomials field-theory galois-theory galois-extensions cyclotomic-fields
asked Nov 26 at 14:06
ʎpoqou
1741111
My task is to use the following two theorems to prove that any cubic is soluble by radicals. We are not allowed to use that $Bbb S_3$ is a soluble group.
Theorem 1: Hilbert's Theorem 90
Let $L:K$ be a finite normal extension with cyclic galois group generated by $tau$. Then $a$$in$L has norm 1 iff $a=frac{b}{tau(b)}$, for some $b in L$.
Here, $N(a)=tau_1(a)...tau_n(a)$, where $tau_i$ are the elements of the Galois Group $G$,corresponding to a finite, normal extension $L:K$.
Theorem 2:
Suppose $L:K$ is a finite normal extension with cyclic Galois Group of prime order generated by $tau$. If K has characteristic 0, (or characteristic which is prime to p), and $t^p-1$ splits in K. Then $L=K(a)$, where $a$ is a zero of an irreducible polynomial $t^p-beta$ over K for some $beta in K$.
I am a bit confused, as I am not sure what I am supposed to do.
I start with $Bbb Q$, and then adjoin $omega$, where $omega^3=1$, but is not equal to 1.
Then over $K=Bbb Q(omega)$, we have $N(omega)=1$, so according to Theorem 1, we may write $omega=frac{b}{tau(b)}$.
If I am not mistaken $tau=(123)$, and the Galois Group of the extension $L:K$ is $Bbb A_3$.
According to Ian Stewart's Galois Theory book, $b=omega c+omega tau(omega) tau(c)+ omegatau(omega)tau^2(omega)tau^2(c)=omega c+ omega^2 tau(c) +tau^2(c).$
Then, it follows that $b^3=tau(b^3)$, so $b^3 in <tau>^dagger$.
Now, using theorem 2, since $K$ splits over $t^3-1$, and $L:K$ has cyclic galois group $Bbb A_3$, then $L=K(alpha)$, which is a radical extensions of $Bbb Q$.
I am not sure whether my reasoning is correct. The whole idea is to show that the general cubic, $f(x)$, is soluble by radicals by showing that there exists a field that contains a a splitting field for $f$ such that the extension $L:Bbb Q$ is a radical extension.
polynomials field-theory galois-theory galois-extensions cyclotomic-fields
polynomials field-theory galois-theory galois-extensions cyclotomic-fields
asked Nov 26 at 14:06
ʎpoqou
1741111
asked Nov 26 at 14:06
ʎpoqou
1741111
asked Nov 26 at 14:06
ʎpoqou
1741111
asked Nov 26 at 14:06
ʎpoqou
1741111
asked Nov 26 at 14:06
ʎpoqou
1741111
1741111
add a comment |
add a comment |
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