Field Extension $k(t^{1/n})/k(t)$ Cyclic
$begingroup$
Let $k$ be a field. We consider the field extension $k(t^{1/n})/k(t)$. My question is why and how to see that it is cyclic, therefore $Gal(k(t^{1/n})/k(t))$ is a cyclic group.
There is one case that is easy:
If we assume that $k(t)$ contains a primitive $n$-th root $zeta_n$ and futhermore $k(t^{1/n})= k(t)[X]/(X^n-t)$
Kummer thery tells that the extension has indeed cyclic Galois group.
By how to argue if $k(t)$ doesn't contain a primitive $n$-th root? Here Kummer can't help. Any ideas?
Futhermore, what about if we play the same game with the extension $$k((t^{1/n}))/k((t))$$ therefore fractions of $k[[t]]$.
I suppose that this extension has the same Galois group as the extension $k(t^{1/n})/k(t)$ above. Does anybody know how to see it formally/ maybe a theorem providing this result?
field-theory galois-theory extension-field class-field-theory
asked Dec 10 '18 at 1:05
KarlPeterKarlPeter
6101315
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field. We consider the field extension $k(t^{1/n})/k(t)$. My question is why and how to see that it is cyclic, therefore $Gal(k(t^{1/n})/k(t))$ is a cyclic group.
There is one case that is easy:
If we assume that $k(t)$ contains a primitive $n$-th root $zeta_n$ and futhermore $k(t^{1/n})= k(t)[X]/(X^n-t)$
Kummer thery tells that the extension has indeed cyclic Galois group.
By how to argue if $k(t)$ doesn't contain a primitive $n$-th root? Here Kummer can't help. Any ideas?
Futhermore, what about if we play the same game with the extension $$k((t^{1/n}))/k((t))$$ therefore fractions of $k[[t]]$.
I suppose that this extension has the same Galois group as the extension $k(t^{1/n})/k(t)$ above. Does anybody know how to see it formally/ maybe a theorem providing this result?
field-theory galois-theory extension-field class-field-theory
asked Dec 10 '18 at 1:05
KarlPeterKarlPeter
6101315
$endgroup$
1
$begingroup$
Well, it isn’t normal if the $n$-th roots aren’t in $k(t^{1/n})$, which in most cases means that those roots aren’t in $k$. And unless normal, you can’t even speak of a Galois group.
$endgroup$
– Lubin
Dec 10 '18 at 1:46
$begingroup$
@Lubin: So we can us restrict to the first case that $k(t)$ contains a primitive $n$-th root. Then I guess we can again use Kummer for $k((t^{1/n}))= k((t))[X]/(X^n-t)$ to show that it also has cyclic Galois group of degree $n$. Hope, I used Kummer in a correct way
$endgroup$
– KarlPeter
Dec 10 '18 at 1:59
$begingroup$
If $zeta_n in k$ then $sigma(f(t^{1/n})) = f(zeta_n t^{1/n})$ is an element of $Aut(k(t^{1/n})/k(t))$ and the subfield fixed by $sigma$ is exactly $k(t)$, whence $k(t^{1/n})/k(t)$ is cyclic with Galois group generated by $sigma$.
$endgroup$
– reuns
Dec 10 '18 at 2:06
$begingroup$
Your use of Kummer seems right.
$endgroup$
– Lubin
Dec 10 '18 at 2:10
add a comment |
$begingroup$
Let $k$ be a field. We consider the field extension $k(t^{1/n})/k(t)$. My question is why and how to see that it is cyclic, therefore $Gal(k(t^{1/n})/k(t))$ is a cyclic group.
There is one case that is easy:
If we assume that $k(t)$ contains a primitive $n$-th root $zeta_n$ and futhermore $k(t^{1/n})= k(t)[X]/(X^n-t)$
Kummer thery tells that the extension has indeed cyclic Galois group.
By how to argue if $k(t)$ doesn't contain a primitive $n$-th root? Here Kummer can't help. Any ideas?
Futhermore, what about if we play the same game with the extension $$k((t^{1/n}))/k((t))$$ therefore fractions of $k[[t]]$.
I suppose that this extension has the same Galois group as the extension $k(t^{1/n})/k(t)$ above. Does anybody know how to see it formally/ maybe a theorem providing this result?
field-theory galois-theory extension-field class-field-theory
asked Dec 10 '18 at 1:05
KarlPeterKarlPeter
6101315
$endgroup$
Let $k$ be a field. We consider the field extension $k(t^{1/n})/k(t)$. My question is why and how to see that it is cyclic, therefore $Gal(k(t^{1/n})/k(t))$ is a cyclic group.
There is one case that is easy:
If we assume that $k(t)$ contains a primitive $n$-th root $zeta_n$ and futhermore $k(t^{1/n})= k(t)[X]/(X^n-t)$
Kummer thery tells that the extension has indeed cyclic Galois group.
By how to argue if $k(t)$ doesn't contain a primitive $n$-th root? Here Kummer can't help. Any ideas?
Futhermore, what about if we play the same game with the extension $$k((t^{1/n}))/k((t))$$ therefore fractions of $k[[t]]$.
I suppose that this extension has the same Galois group as the extension $k(t^{1/n})/k(t)$ above. Does anybody know how to see it formally/ maybe a theorem providing this result?
field-theory galois-theory extension-field class-field-theory
field-theory galois-theory extension-field class-field-theory
asked Dec 10 '18 at 1:05
KarlPeterKarlPeter
6101315
asked Dec 10 '18 at 1:05
KarlPeterKarlPeter
6101315
asked Dec 10 '18 at 1:05
KarlPeterKarlPeter
6101315
asked Dec 10 '18 at 1:05
KarlPeterKarlPeter
6101315
asked Dec 10 '18 at 1:05
KarlPeterKarlPeter
6101315
6101315
1
$begingroup$
Well, it isn’t normal if the $n$-th roots aren’t in $k(t^{1/n})$, which in most cases means that those roots aren’t in $k$. And unless normal, you can’t even speak of a Galois group.
$endgroup$
– Lubin
Dec 10 '18 at 1:46
$begingroup$
@Lubin: So we can us restrict to the first case that $k(t)$ contains a primitive $n$-th root. Then I guess we can again use Kummer for $k((t^{1/n}))= k((t))[X]/(X^n-t)$ to show that it also has cyclic Galois group of degree $n$. Hope, I used Kummer in a correct way
$endgroup$
– KarlPeter
Dec 10 '18 at 1:59
$begingroup$
If $zeta_n in k$ then $sigma(f(t^{1/n})) = f(zeta_n t^{1/n})$ is an element of $Aut(k(t^{1/n})/k(t))$ and the subfield fixed by $sigma$ is exactly $k(t)$, whence $k(t^{1/n})/k(t)$ is cyclic with Galois group generated by $sigma$.
$endgroup$
– reuns
Dec 10 '18 at 2:06
$begingroup$
Your use of Kummer seems right.
$endgroup$
– Lubin
Dec 10 '18 at 2:10
add a comment |
1
$begingroup$
Well, it isn’t normal if the $n$-th roots aren’t in $k(t^{1/n})$, which in most cases means that those roots aren’t in $k$. And unless normal, you can’t even speak of a Galois group.
$endgroup$
– Lubin
Dec 10 '18 at 1:46
$begingroup$
@Lubin: So we can us restrict to the first case that $k(t)$ contains a primitive $n$-th root. Then I guess we can again use Kummer for $k((t^{1/n}))= k((t))[X]/(X^n-t)$ to show that it also has cyclic Galois group of degree $n$. Hope, I used Kummer in a correct way
$endgroup$
– KarlPeter
Dec 10 '18 at 1:59
$begingroup$
If $zeta_n in k$ then $sigma(f(t^{1/n})) = f(zeta_n t^{1/n})$ is an element of $Aut(k(t^{1/n})/k(t))$ and the subfield fixed by $sigma$ is exactly $k(t)$, whence $k(t^{1/n})/k(t)$ is cyclic with Galois group generated by $sigma$.
$endgroup$
– reuns
Dec 10 '18 at 2:06
$begingroup$
Your use of Kummer seems right.
$endgroup$
– Lubin
Dec 10 '18 at 2:10
1
1
$begingroup$
Well, it isn’t normal if the $n$-th roots aren’t in $k(t^{1/n})$, which in most cases means that those roots aren’t in $k$. And unless normal, you can’t even speak of a Galois group.
$endgroup$
– Lubin
Dec 10 '18 at 1:46
$begingroup$
Well, it isn’t normal if the $n$-th roots aren’t in $k(t^{1/n})$, which in most cases means that those roots aren’t in $k$. And unless normal, you can’t even speak of a Galois group.
$endgroup$
– Lubin
Dec 10 '18 at 1:46
$begingroup$
@Lubin: So we can us restrict to the first case that $k(t)$ contains a primitive $n$-th root. Then I guess we can again use Kummer for $k((t^{1/n}))= k((t))[X]/(X^n-t)$ to show that it also has cyclic Galois group of degree $n$. Hope, I used Kummer in a correct way
$endgroup$
– KarlPeter
Dec 10 '18 at 1:59
$begingroup$
@Lubin: So we can us restrict to the first case that $k(t)$ contains a primitive $n$-th root. Then I guess we can again use Kummer for $k((t^{1/n}))= k((t))[X]/(X^n-t)$ to show that it also has cyclic Galois group of degree $n$. Hope, I used Kummer in a correct way
$endgroup$
– KarlPeter
Dec 10 '18 at 1:59
$begingroup$
If $zeta_n in k$ then $sigma(f(t^{1/n})) = f(zeta_n t^{1/n})$ is an element of $Aut(k(t^{1/n})/k(t))$ and the subfield fixed by $sigma$ is exactly $k(t)$, whence $k(t^{1/n})/k(t)$ is cyclic with Galois group generated by $sigma$.
$endgroup$
– reuns
Dec 10 '18 at 2:06
$begingroup$
If $zeta_n in k$ then $sigma(f(t^{1/n})) = f(zeta_n t^{1/n})$ is an element of $Aut(k(t^{1/n})/k(t))$ and the subfield fixed by $sigma$ is exactly $k(t)$, whence $k(t^{1/n})/k(t)$ is cyclic with Galois group generated by $sigma$.
$endgroup$
– reuns
Dec 10 '18 at 2:06
$begingroup$
Your use of Kummer seems right.
$endgroup$
– Lubin
Dec 10 '18 at 2:10
$begingroup$
Your use of Kummer seems right.
$endgroup$
– Lubin
Dec 10 '18 at 2:10
add a comment |
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$begingroup$
Well, it isn’t normal if the $n$-th roots aren’t in $k(t^{1/n})$, which in most cases means that those roots aren’t in $k$. And unless normal, you can’t even speak of a Galois group.
$endgroup$
– Lubin
Dec 10 '18 at 1:46
$begingroup$
@Lubin: So we can us restrict to the first case that $k(t)$ contains a primitive $n$-th root. Then I guess we can again use Kummer for $k((t^{1/n}))= k((t))[X]/(X^n-t)$ to show that it also has cyclic Galois group of degree $n$. Hope, I used Kummer in a correct way
$endgroup$
– KarlPeter
Dec 10 '18 at 1:59
$begingroup$
If $zeta_n in k$ then $sigma(f(t^{1/n})) = f(zeta_n t^{1/n})$ is an element of $Aut(k(t^{1/n})/k(t))$ and the subfield fixed by $sigma$ is exactly $k(t)$, whence $k(t^{1/n})/k(t)$ is cyclic with Galois group generated by $sigma$.
$endgroup$
– reuns
Dec 10 '18 at 2:06
$begingroup$
Your use of Kummer seems right.
$endgroup$
– Lubin
Dec 10 '18 at 2:10