Extension of degree 5 not obtained by adjoining a 5th root and whose normal closure contains a primitive 5th root of unity.Field extension, primitive root of unityExample of field's normal closure that's not Abelian?Subgroups and corresponding fixed fields of Galois group of primitive 24th root of unity.Galois group of finite field extension of p-adic fieldsDistinct primes dividing the order of a solvable Galois extension $E/F$, radical if $F$ contains a primitive $(p_1cdots p_r)$-th root of unityGalois group of $x^5+x-1$Is every simple radical extension containing a primitive nth root of unity cyclic?On a normal closure of a trivial (i.e. degree 1) extensionA field such that no extension has an 11th primitive root of unityCan one write a finite extension of local fields as a compositum of fields whose degrees are prime powers?
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Extension of degree 5 not obtained by adjoining a 5th root and whose normal closure contains a primitive 5th root of unity.
Field extension, primitive root of unityExample of field's normal closure that's not Abelian?Subgroups and corresponding fixed fields of Galois group of primitive 24th root of unity.Galois group of finite field extension of p-adic fieldsDistinct primes dividing the order of a solvable Galois extension $E/F$, radical if $F$ contains a primitive $(p_1cdots p_r)$-th root of unityGalois group of $x^5+x-1$Is every simple radical extension containing a primitive nth root of unity cyclic?On a normal closure of a trivial (i.e. degree 1) extensionA field such that no extension has an 11th primitive root of unityCan one write a finite extension of local fields as a compositum of fields whose degrees are prime powers?
$begingroup$
The question is mainly what the title says, but here is the setup in more details.
Let $K$ be a field not containing a primitive 5th root of unity (for this question, the case $K = mathbbQ$ seems already interesting). Let $L$ be a degree $5$ extension such that $Lsetminus K$ contains no element $x$ with $x^5 in K$. Is it possible for the normal closure of $L$ to contain a primitive 5th root of unity?
EDIT: I am trying to find an example using local fields, and there are some good candidates in $mathbbQ_5$. Indeed, using the table of local fields of Jones and Roberts, I see that $mathbbQ_5$ has $4$ field extensions of degree $5$ with discriminant having $5$-adic valuation $7$, Galois group $F_5cong C_5rtimes C_4$ and which are totally ramified (because the inertia group is the same as the Galois group). Those fields are not of the form $mathbbQ_5[X]/(X^5-a)$ (because the discriminant would be of $5$-adic valuation either $5$ or $9$ depending on the valuation of $a$). The only thing I'm unsure of is how to check whether their normal closure contain a primitive $5$th root of unity. Any idea on how to do that?
I'm actually interested in this question when one replaces all occurrences of $5$ by an arbitrary prime number. Note that for $p = 3$, the answer is no, because the normal closure of L contains a primitive 3rd root of unity if and only if the fundamental discriminant is $-3$, which is equivalent to $L$ being a pure cubic field. But already for $p=5$ I have no idea of how to approach the question, and any suggestion is welcome.
galois-theory local-field galois-extensions
asked 2 days ago
user293657user293657
756
$endgroup$
add a comment |
$begingroup$
The question is mainly what the title says, but here is the setup in more details.
Let $K$ be a field not containing a primitive 5th root of unity (for this question, the case $K = mathbbQ$ seems already interesting). Let $L$ be a degree $5$ extension such that $Lsetminus K$ contains no element $x$ with $x^5 in K$. Is it possible for the normal closure of $L$ to contain a primitive 5th root of unity?
EDIT: I am trying to find an example using local fields, and there are some good candidates in $mathbbQ_5$. Indeed, using the table of local fields of Jones and Roberts, I see that $mathbbQ_5$ has $4$ field extensions of degree $5$ with discriminant having $5$-adic valuation $7$, Galois group $F_5cong C_5rtimes C_4$ and which are totally ramified (because the inertia group is the same as the Galois group). Those fields are not of the form $mathbbQ_5[X]/(X^5-a)$ (because the discriminant would be of $5$-adic valuation either $5$ or $9$ depending on the valuation of $a$). The only thing I'm unsure of is how to check whether their normal closure contain a primitive $5$th root of unity. Any idea on how to do that?
I'm actually interested in this question when one replaces all occurrences of $5$ by an arbitrary prime number. Note that for $p = 3$, the answer is no, because the normal closure of L contains a primitive 3rd root of unity if and only if the fundamental discriminant is $-3$, which is equivalent to $L$ being a pure cubic field. But already for $p=5$ I have no idea of how to approach the question, and any suggestion is welcome.
galois-theory local-field galois-extensions
asked 2 days ago
user293657user293657
756
$endgroup$
add a comment |
$begingroup$
The question is mainly what the title says, but here is the setup in more details.
Let $K$ be a field not containing a primitive 5th root of unity (for this question, the case $K = mathbbQ$ seems already interesting). Let $L$ be a degree $5$ extension such that $Lsetminus K$ contains no element $x$ with $x^5 in K$. Is it possible for the normal closure of $L$ to contain a primitive 5th root of unity?
EDIT: I am trying to find an example using local fields, and there are some good candidates in $mathbbQ_5$. Indeed, using the table of local fields of Jones and Roberts, I see that $mathbbQ_5$ has $4$ field extensions of degree $5$ with discriminant having $5$-adic valuation $7$, Galois group $F_5cong C_5rtimes C_4$ and which are totally ramified (because the inertia group is the same as the Galois group). Those fields are not of the form $mathbbQ_5[X]/(X^5-a)$ (because the discriminant would be of $5$-adic valuation either $5$ or $9$ depending on the valuation of $a$). The only thing I'm unsure of is how to check whether their normal closure contain a primitive $5$th root of unity. Any idea on how to do that?
I'm actually interested in this question when one replaces all occurrences of $5$ by an arbitrary prime number. Note that for $p = 3$, the answer is no, because the normal closure of L contains a primitive 3rd root of unity if and only if the fundamental discriminant is $-3$, which is equivalent to $L$ being a pure cubic field. But already for $p=5$ I have no idea of how to approach the question, and any suggestion is welcome.
galois-theory local-field galois-extensions
asked 2 days ago
user293657user293657
756
$endgroup$
The question is mainly what the title says, but here is the setup in more details.
Let $K$ be a field not containing a primitive 5th root of unity (for this question, the case $K = mathbbQ$ seems already interesting). Let $L$ be a degree $5$ extension such that $Lsetminus K$ contains no element $x$ with $x^5 in K$. Is it possible for the normal closure of $L$ to contain a primitive 5th root of unity?
EDIT: I am trying to find an example using local fields, and there are some good candidates in $mathbbQ_5$. Indeed, using the table of local fields of Jones and Roberts, I see that $mathbbQ_5$ has $4$ field extensions of degree $5$ with discriminant having $5$-adic valuation $7$, Galois group $F_5cong C_5rtimes C_4$ and which are totally ramified (because the inertia group is the same as the Galois group). Those fields are not of the form $mathbbQ_5[X]/(X^5-a)$ (because the discriminant would be of $5$-adic valuation either $5$ or $9$ depending on the valuation of $a$). The only thing I'm unsure of is how to check whether their normal closure contain a primitive $5$th root of unity. Any idea on how to do that?
I'm actually interested in this question when one replaces all occurrences of $5$ by an arbitrary prime number. Note that for $p = 3$, the answer is no, because the normal closure of L contains a primitive 3rd root of unity if and only if the fundamental discriminant is $-3$, which is equivalent to $L$ being a pure cubic field. But already for $p=5$ I have no idea of how to approach the question, and any suggestion is welcome.
galois-theory local-field galois-extensions
galois-theory local-field galois-extensions
asked 2 days ago
user293657user293657
756
asked 2 days ago
user293657user293657
756
edited 2 days ago
user293657
asked 2 days ago
user293657user293657
756
asked 2 days ago
user293657user293657
756
asked 2 days ago
user293657user293657
756
756
add a comment |
add a comment |
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$begingroup$
So it turns out an example can be constructed using $K=mathbbQ_5$. Taking $L = K[X]/(X^5+15X^3+5)$, we get a field extension of discriminant having 5-adic valuation $7$, hence $L$ is not isomorphic to $K[X]/(X^5-a)$ for any $ain mathbbQ_5$ because this would have discriminant $5^5.a^4$ (hopefully the discriminant is an invariant of the extension). Finally, according to what is written in the proof of Proposition 2.3.1 of the paper explaining the table of local fields, the cyclic 4 extension of $mathbbQ_5$ that one has to adjoin to obtain the normal closure of $L$ is $mathbbQ_5[X]/(X^4-45)$, which is indeed isomorphic to $mathbbQ_5[X]/(X^4+X^3+X^2+X+1)$ (again using the table of local fields of Jones and Roberts).
answered 2 days ago
user293657user293657
756
$endgroup$
add a comment |
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$begingroup$
So it turns out an example can be constructed using $K=mathbbQ_5$. Taking $L = K[X]/(X^5+15X^3+5)$, we get a field extension of discriminant having 5-adic valuation $7$, hence $L$ is not isomorphic to $K[X]/(X^5-a)$ for any $ain mathbbQ_5$ because this would have discriminant $5^5.a^4$ (hopefully the discriminant is an invariant of the extension). Finally, according to what is written in the proof of Proposition 2.3.1 of the paper explaining the table of local fields, the cyclic 4 extension of $mathbbQ_5$ that one has to adjoin to obtain the normal closure of $L$ is $mathbbQ_5[X]/(X^4-45)$, which is indeed isomorphic to $mathbbQ_5[X]/(X^4+X^3+X^2+X+1)$ (again using the table of local fields of Jones and Roberts).
answered 2 days ago
user293657user293657
756
$endgroup$
add a comment |
$begingroup$
So it turns out an example can be constructed using $K=mathbbQ_5$. Taking $L = K[X]/(X^5+15X^3+5)$, we get a field extension of discriminant having 5-adic valuation $7$, hence $L$ is not isomorphic to $K[X]/(X^5-a)$ for any $ain mathbbQ_5$ because this would have discriminant $5^5.a^4$ (hopefully the discriminant is an invariant of the extension). Finally, according to what is written in the proof of Proposition 2.3.1 of the paper explaining the table of local fields, the cyclic 4 extension of $mathbbQ_5$ that one has to adjoin to obtain the normal closure of $L$ is $mathbbQ_5[X]/(X^4-45)$, which is indeed isomorphic to $mathbbQ_5[X]/(X^4+X^3+X^2+X+1)$ (again using the table of local fields of Jones and Roberts).
answered 2 days ago
user293657user293657
756
$endgroup$
add a comment |
$begingroup$
So it turns out an example can be constructed using $K=mathbbQ_5$. Taking $L = K[X]/(X^5+15X^3+5)$, we get a field extension of discriminant having 5-adic valuation $7$, hence $L$ is not isomorphic to $K[X]/(X^5-a)$ for any $ain mathbbQ_5$ because this would have discriminant $5^5.a^4$ (hopefully the discriminant is an invariant of the extension). Finally, according to what is written in the proof of Proposition 2.3.1 of the paper explaining the table of local fields, the cyclic 4 extension of $mathbbQ_5$ that one has to adjoin to obtain the normal closure of $L$ is $mathbbQ_5[X]/(X^4-45)$, which is indeed isomorphic to $mathbbQ_5[X]/(X^4+X^3+X^2+X+1)$ (again using the table of local fields of Jones and Roberts).
answered 2 days ago
user293657user293657
756
$endgroup$
So it turns out an example can be constructed using $K=mathbbQ_5$. Taking $L = K[X]/(X^5+15X^3+5)$, we get a field extension of discriminant having 5-adic valuation $7$, hence $L$ is not isomorphic to $K[X]/(X^5-a)$ for any $ain mathbbQ_5$ because this would have discriminant $5^5.a^4$ (hopefully the discriminant is an invariant of the extension). Finally, according to what is written in the proof of Proposition 2.3.1 of the paper explaining the table of local fields, the cyclic 4 extension of $mathbbQ_5$ that one has to adjoin to obtain the normal closure of $L$ is $mathbbQ_5[X]/(X^4-45)$, which is indeed isomorphic to $mathbbQ_5[X]/(X^4+X^3+X^2+X+1)$ (again using the table of local fields of Jones and Roberts).
answered 2 days ago
user293657user293657
756
answered 2 days ago
user293657user293657
756
answered 2 days ago
user293657user293657
756
answered 2 days ago
user293657user293657
756
756
add a comment |
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