What is wrong with my conclusion (compositum of local fields)?
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Let $K = mathbb{Q}_2(zeta_3)$ where $zeta_3$ is a primitive third root of unity, and $F = mathbb{Q}_2(zeta_3,sqrt[3]{2})$. Furthermore, let $L = mathbb{Q}(zeta_3,beta)$ where $beta$ is defined by $beta^3 = zeta_3 sqrt[3]{2}$.
My conclusion: Since $beta^3/zeta_3 = sqrt[3]{2}$, the field $F$ is contained in $L$. Therefore, the compositum $FL$ is $L$ and therefore, $FL/L$ has degree $1$.
Problem: My professor told me briefly that $F$ can not be contained in $L$ because $(frac{beta}{sqrt[3]{2}})^3 = zeta_3$ which means that $FL$ has a primitive ninth root of unity which neither exists in $F$ not in $L$.
Could you please explain me why neither $F$ nor $L$ have a primitive ninth root of unity? And why was my argument wrong? I thought in order to show $Fsubset L$ it enough to show that the elements generating $F$ are in $L$ too, so $zeta_3 in L$ (which is obvious) and $sqrt[3]{2} in L$ (which follows from my computation above).
Any help is really appreciated!
abstract-algebra algebraic-number-theory extension-field local-field
asked Dec 23 '18 at 12:44
DiglettDiglett
1,1081521
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add a comment |
$begingroup$
Let $K = mathbb{Q}_2(zeta_3)$ where $zeta_3$ is a primitive third root of unity, and $F = mathbb{Q}_2(zeta_3,sqrt[3]{2})$. Furthermore, let $L = mathbb{Q}(zeta_3,beta)$ where $beta$ is defined by $beta^3 = zeta_3 sqrt[3]{2}$.
My conclusion: Since $beta^3/zeta_3 = sqrt[3]{2}$, the field $F$ is contained in $L$. Therefore, the compositum $FL$ is $L$ and therefore, $FL/L$ has degree $1$.
Problem: My professor told me briefly that $F$ can not be contained in $L$ because $(frac{beta}{sqrt[3]{2}})^3 = zeta_3$ which means that $FL$ has a primitive ninth root of unity which neither exists in $F$ not in $L$.
Could you please explain me why neither $F$ nor $L$ have a primitive ninth root of unity? And why was my argument wrong? I thought in order to show $Fsubset L$ it enough to show that the elements generating $F$ are in $L$ too, so $zeta_3 in L$ (which is obvious) and $sqrt[3]{2} in L$ (which follows from my computation above).
Any help is really appreciated!
abstract-algebra algebraic-number-theory extension-field local-field
asked Dec 23 '18 at 12:44
DiglettDiglett
1,1081521
$endgroup$
2
$begingroup$
I don't think $(beta/sqrt[3]2)^3$ equals $zeta_3$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 13:00
$begingroup$
@LordSharktheUnknown: Thanks for spotting the issue. It turned out that my professor made a mistake.
$endgroup$
– Diglett
Dec 23 '18 at 14:08
add a comment |
$begingroup$
Let $K = mathbb{Q}_2(zeta_3)$ where $zeta_3$ is a primitive third root of unity, and $F = mathbb{Q}_2(zeta_3,sqrt[3]{2})$. Furthermore, let $L = mathbb{Q}(zeta_3,beta)$ where $beta$ is defined by $beta^3 = zeta_3 sqrt[3]{2}$.
My conclusion: Since $beta^3/zeta_3 = sqrt[3]{2}$, the field $F$ is contained in $L$. Therefore, the compositum $FL$ is $L$ and therefore, $FL/L$ has degree $1$.
Problem: My professor told me briefly that $F$ can not be contained in $L$ because $(frac{beta}{sqrt[3]{2}})^3 = zeta_3$ which means that $FL$ has a primitive ninth root of unity which neither exists in $F$ not in $L$.
Could you please explain me why neither $F$ nor $L$ have a primitive ninth root of unity? And why was my argument wrong? I thought in order to show $Fsubset L$ it enough to show that the elements generating $F$ are in $L$ too, so $zeta_3 in L$ (which is obvious) and $sqrt[3]{2} in L$ (which follows from my computation above).
Any help is really appreciated!
abstract-algebra algebraic-number-theory extension-field local-field
asked Dec 23 '18 at 12:44
DiglettDiglett
1,1081521
$endgroup$
Let $K = mathbb{Q}_2(zeta_3)$ where $zeta_3$ is a primitive third root of unity, and $F = mathbb{Q}_2(zeta_3,sqrt[3]{2})$. Furthermore, let $L = mathbb{Q}(zeta_3,beta)$ where $beta$ is defined by $beta^3 = zeta_3 sqrt[3]{2}$.
My conclusion: Since $beta^3/zeta_3 = sqrt[3]{2}$, the field $F$ is contained in $L$. Therefore, the compositum $FL$ is $L$ and therefore, $FL/L$ has degree $1$.
Problem: My professor told me briefly that $F$ can not be contained in $L$ because $(frac{beta}{sqrt[3]{2}})^3 = zeta_3$ which means that $FL$ has a primitive ninth root of unity which neither exists in $F$ not in $L$.
Could you please explain me why neither $F$ nor $L$ have a primitive ninth root of unity? And why was my argument wrong? I thought in order to show $Fsubset L$ it enough to show that the elements generating $F$ are in $L$ too, so $zeta_3 in L$ (which is obvious) and $sqrt[3]{2} in L$ (which follows from my computation above).
Any help is really appreciated!
abstract-algebra algebraic-number-theory extension-field local-field
abstract-algebra algebraic-number-theory extension-field local-field
asked Dec 23 '18 at 12:44
DiglettDiglett
1,1081521
asked Dec 23 '18 at 12:44
DiglettDiglett
1,1081521
edited Dec 23 '18 at 12:58
Diglett
asked Dec 23 '18 at 12:44
DiglettDiglett
1,1081521
asked Dec 23 '18 at 12:44
DiglettDiglett
1,1081521
asked Dec 23 '18 at 12:44
DiglettDiglett
1,1081521
1,1081521
2
$begingroup$
I don't think $(beta/sqrt[3]2)^3$ equals $zeta_3$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 13:00
$begingroup$
@LordSharktheUnknown: Thanks for spotting the issue. It turned out that my professor made a mistake.
$endgroup$
– Diglett
Dec 23 '18 at 14:08
add a comment |
2
$begingroup$
I don't think $(beta/sqrt[3]2)^3$ equals $zeta_3$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 13:00
$begingroup$
@LordSharktheUnknown: Thanks for spotting the issue. It turned out that my professor made a mistake.
$endgroup$
– Diglett
Dec 23 '18 at 14:08
2
2
$begingroup$
I don't think $(beta/sqrt[3]2)^3$ equals $zeta_3$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 13:00
$begingroup$
I don't think $(beta/sqrt[3]2)^3$ equals $zeta_3$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 13:00
$begingroup$
@LordSharktheUnknown: Thanks for spotting the issue. It turned out that my professor made a mistake.
$endgroup$
– Diglett
Dec 23 '18 at 14:08
$begingroup$
@LordSharktheUnknown: Thanks for spotting the issue. It turned out that my professor made a mistake.
$endgroup$
– Diglett
Dec 23 '18 at 14:08
add a comment |
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$begingroup$
I don't think $(beta/sqrt[3]2)^3$ equals $zeta_3$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 13:00
$begingroup$
@LordSharktheUnknown: Thanks for spotting the issue. It turned out that my professor made a mistake.
$endgroup$
– Diglett
Dec 23 '18 at 14:08