If $a in L-k $ satisfies $k(a^n)=L$ (for all $n geq 1$), then $L/k$ is Galois?
Let $k subsetneq L$ be a finite separable field extension, and let $a in L-k$ satisfy: For every $n geq 1$, $k(a^n)=L$.
In other words, all the non-zero powers of the primitive element $a$ are also primitive elements.
Is there something interesting to say about such an extension? Should it be Galois?
Partial answer: According to this question, if the extension is of prime degree, and if there exist infinitely many $m$'s such that $k(b^m) neq L=k(b)$, then $k subsetneq L$ is not Galois.
But I am asking about the opposite direction, namely, if only finitely many (= more precisely, zero) $2 leq m$'s are such that $k(b^m) subsetneq k(b)=L$, then $k subseteq L$ is Galois? Can we reverse the argument in that question?
Special case $k=mathbb{Q}$: Can one find $a in bar{mathbb{Q}}-mathbb{Q}$
satisfying $mathbb{Q}(a)=mathbb{Q}(a^2)=mathbb{Q}(a^3)=ldots$?
The following are two non-examples for $k=mathbb{Q}$:
(1) In the spirit of the comment for this question, notice that here taking $a=p^{frac{1}{m}}$ for some (positive) prime $p geq 2$ and fixed $m geq 2$ will not help, since $a^m=(p^{frac{1}{m}})^m=p$ so $mathbb{Q}(a^m)=mathbb{Q}(p)=mathbb{Q} subsetneq mathbb{Q}(a)$.
(2) $a=frac{-1+sqrt{3}i}{2}$ will not help; indeed, $a^2+a+1=0$,
so $a^2=-a-1$, then $a^3=-a^2-a=-(-a-1)-a=1$, so $mathbb{Q}(a^3)=mathbb{Q}(1)=mathbb{Q} subsetneq mathbb{Q}(a)$.
Any hints are welcome!
field-theory galois-theory extension-field separable-extension
asked 2 days ago
user237522
2,1411617
|
show 7 more comments
Let $k subsetneq L$ be a finite separable field extension, and let $a in L-k$ satisfy: For every $n geq 1$, $k(a^n)=L$.
In other words, all the non-zero powers of the primitive element $a$ are also primitive elements.
Is there something interesting to say about such an extension? Should it be Galois?
Partial answer: According to this question, if the extension is of prime degree, and if there exist infinitely many $m$'s such that $k(b^m) neq L=k(b)$, then $k subsetneq L$ is not Galois.
But I am asking about the opposite direction, namely, if only finitely many (= more precisely, zero) $2 leq m$'s are such that $k(b^m) subsetneq k(b)=L$, then $k subseteq L$ is Galois? Can we reverse the argument in that question?
Special case $k=mathbb{Q}$: Can one find $a in bar{mathbb{Q}}-mathbb{Q}$
satisfying $mathbb{Q}(a)=mathbb{Q}(a^2)=mathbb{Q}(a^3)=ldots$?
The following are two non-examples for $k=mathbb{Q}$:
(1) In the spirit of the comment for this question, notice that here taking $a=p^{frac{1}{m}}$ for some (positive) prime $p geq 2$ and fixed $m geq 2$ will not help, since $a^m=(p^{frac{1}{m}})^m=p$ so $mathbb{Q}(a^m)=mathbb{Q}(p)=mathbb{Q} subsetneq mathbb{Q}(a)$.
(2) $a=frac{-1+sqrt{3}i}{2}$ will not help; indeed, $a^2+a+1=0$,
so $a^2=-a-1$, then $a^3=-a^2-a=-(-a-1)-a=1$, so $mathbb{Q}(a^3)=mathbb{Q}(1)=mathbb{Q} subsetneq mathbb{Q}(a)$.
Any hints are welcome!
field-theory galois-theory extension-field separable-extension
asked 2 days ago
user237522
2,1411617
1
A simple example take $a=1+sqrt{2}$
– mouthetics
2 days ago
Oh, of course.. thank you. Is it true that also any $a=1+p^{frac{1}{m}}$ will work, where $p geq 2$ is prime and $m geq 2$? (Perhaps one has to be careful, and not all such $p$ and $m$ will work).
– user237522
2 days ago
No It won't. As you explained in you question.
– mouthetics
2 days ago
@mouthetics, please, is it possible to find $b in mathbb{Q}(sqrt{2})$ such that: $mathbb{Q}(b^n)=mathbb{Q}(sqrt{2})$ (for all $n geq 2$), and also $b-a notin mathbb{Q}$?
– user237522
2 days ago
Well, $b=1+2sqrt2$ =)
– Kenny Lau
2 days ago
|
show 7 more comments
Let $k subsetneq L$ be a finite separable field extension, and let $a in L-k$ satisfy: For every $n geq 1$, $k(a^n)=L$.
In other words, all the non-zero powers of the primitive element $a$ are also primitive elements.
Is there something interesting to say about such an extension? Should it be Galois?
Partial answer: According to this question, if the extension is of prime degree, and if there exist infinitely many $m$'s such that $k(b^m) neq L=k(b)$, then $k subsetneq L$ is not Galois.
But I am asking about the opposite direction, namely, if only finitely many (= more precisely, zero) $2 leq m$'s are such that $k(b^m) subsetneq k(b)=L$, then $k subseteq L$ is Galois? Can we reverse the argument in that question?
Special case $k=mathbb{Q}$: Can one find $a in bar{mathbb{Q}}-mathbb{Q}$
satisfying $mathbb{Q}(a)=mathbb{Q}(a^2)=mathbb{Q}(a^3)=ldots$?
The following are two non-examples for $k=mathbb{Q}$:
(1) In the spirit of the comment for this question, notice that here taking $a=p^{frac{1}{m}}$ for some (positive) prime $p geq 2$ and fixed $m geq 2$ will not help, since $a^m=(p^{frac{1}{m}})^m=p$ so $mathbb{Q}(a^m)=mathbb{Q}(p)=mathbb{Q} subsetneq mathbb{Q}(a)$.
(2) $a=frac{-1+sqrt{3}i}{2}$ will not help; indeed, $a^2+a+1=0$,
so $a^2=-a-1$, then $a^3=-a^2-a=-(-a-1)-a=1$, so $mathbb{Q}(a^3)=mathbb{Q}(1)=mathbb{Q} subsetneq mathbb{Q}(a)$.
Any hints are welcome!
field-theory galois-theory extension-field separable-extension
asked 2 days ago
user237522
2,1411617
Let $k subsetneq L$ be a finite separable field extension, and let $a in L-k$ satisfy: For every $n geq 1$, $k(a^n)=L$.
In other words, all the non-zero powers of the primitive element $a$ are also primitive elements.
Is there something interesting to say about such an extension? Should it be Galois?
Partial answer: According to this question, if the extension is of prime degree, and if there exist infinitely many $m$'s such that $k(b^m) neq L=k(b)$, then $k subsetneq L$ is not Galois.
But I am asking about the opposite direction, namely, if only finitely many (= more precisely, zero) $2 leq m$'s are such that $k(b^m) subsetneq k(b)=L$, then $k subseteq L$ is Galois? Can we reverse the argument in that question?
Special case $k=mathbb{Q}$: Can one find $a in bar{mathbb{Q}}-mathbb{Q}$
satisfying $mathbb{Q}(a)=mathbb{Q}(a^2)=mathbb{Q}(a^3)=ldots$?
The following are two non-examples for $k=mathbb{Q}$:
(1) In the spirit of the comment for this question, notice that here taking $a=p^{frac{1}{m}}$ for some (positive) prime $p geq 2$ and fixed $m geq 2$ will not help, since $a^m=(p^{frac{1}{m}})^m=p$ so $mathbb{Q}(a^m)=mathbb{Q}(p)=mathbb{Q} subsetneq mathbb{Q}(a)$.
(2) $a=frac{-1+sqrt{3}i}{2}$ will not help; indeed, $a^2+a+1=0$,
so $a^2=-a-1$, then $a^3=-a^2-a=-(-a-1)-a=1$, so $mathbb{Q}(a^3)=mathbb{Q}(1)=mathbb{Q} subsetneq mathbb{Q}(a)$.
Any hints are welcome!
field-theory galois-theory extension-field separable-extension
field-theory galois-theory extension-field separable-extension
asked 2 days ago
user237522
2,1411617
asked 2 days ago
user237522
2,1411617
edited 2 days ago
asked 2 days ago
user237522
2,1411617
asked 2 days ago
user237522
2,1411617
asked 2 days ago
user237522
2,1411617
2,1411617
1
A simple example take $a=1+sqrt{2}$
– mouthetics
2 days ago
Oh, of course.. thank you. Is it true that also any $a=1+p^{frac{1}{m}}$ will work, where $p geq 2$ is prime and $m geq 2$? (Perhaps one has to be careful, and not all such $p$ and $m$ will work).
– user237522
2 days ago
No It won't. As you explained in you question.
– mouthetics
2 days ago
@mouthetics, please, is it possible to find $b in mathbb{Q}(sqrt{2})$ such that: $mathbb{Q}(b^n)=mathbb{Q}(sqrt{2})$ (for all $n geq 2$), and also $b-a notin mathbb{Q}$?
– user237522
2 days ago
Well, $b=1+2sqrt2$ =)
– Kenny Lau
2 days ago
|
show 7 more comments
1
A simple example take $a=1+sqrt{2}$
– mouthetics
2 days ago
Oh, of course.. thank you. Is it true that also any $a=1+p^{frac{1}{m}}$ will work, where $p geq 2$ is prime and $m geq 2$? (Perhaps one has to be careful, and not all such $p$ and $m$ will work).
– user237522
2 days ago
No It won't. As you explained in you question.
– mouthetics
2 days ago
@mouthetics, please, is it possible to find $b in mathbb{Q}(sqrt{2})$ such that: $mathbb{Q}(b^n)=mathbb{Q}(sqrt{2})$ (for all $n geq 2$), and also $b-a notin mathbb{Q}$?
– user237522
2 days ago
Well, $b=1+2sqrt2$ =)
– Kenny Lau
2 days ago
1
1
A simple example take $a=1+sqrt{2}$
– mouthetics
2 days ago
A simple example take $a=1+sqrt{2}$
– mouthetics
2 days ago
Oh, of course.. thank you. Is it true that also any $a=1+p^{frac{1}{m}}$ will work, where $p geq 2$ is prime and $m geq 2$? (Perhaps one has to be careful, and not all such $p$ and $m$ will work).
– user237522
2 days ago
Oh, of course.. thank you. Is it true that also any $a=1+p^{frac{1}{m}}$ will work, where $p geq 2$ is prime and $m geq 2$? (Perhaps one has to be careful, and not all such $p$ and $m$ will work).
– user237522
2 days ago
No It won't. As you explained in you question.
– mouthetics
2 days ago
No It won't. As you explained in you question.
– mouthetics
2 days ago
@mouthetics, please, is it possible to find $b in mathbb{Q}(sqrt{2})$ such that: $mathbb{Q}(b^n)=mathbb{Q}(sqrt{2})$ (for all $n geq 2$), and also $b-a notin mathbb{Q}$?
– user237522
2 days ago
@mouthetics, please, is it possible to find $b in mathbb{Q}(sqrt{2})$ such that: $mathbb{Q}(b^n)=mathbb{Q}(sqrt{2})$ (for all $n geq 2$), and also $b-a notin mathbb{Q}$?
– user237522
2 days ago
Well, $b=1+2sqrt2$ =)
– Kenny Lau
2 days ago
Well, $b=1+2sqrt2$ =)
– Kenny Lau
2 days ago
|
show 7 more comments
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1
A simple example take $a=1+sqrt{2}$
– mouthetics
2 days ago
Oh, of course.. thank you. Is it true that also any $a=1+p^{frac{1}{m}}$ will work, where $p geq 2$ is prime and $m geq 2$? (Perhaps one has to be careful, and not all such $p$ and $m$ will work).
– user237522
2 days ago
No It won't. As you explained in you question.
– mouthetics
2 days ago
@mouthetics, please, is it possible to find $b in mathbb{Q}(sqrt{2})$ such that: $mathbb{Q}(b^n)=mathbb{Q}(sqrt{2})$ (for all $n geq 2$), and also $b-a notin mathbb{Q}$?
– user237522
2 days ago
Well, $b=1+2sqrt2$ =)
– Kenny Lau
2 days ago