$a,b in bar{k}$, such that $k(a)=k(b)=k(ab)$
$begingroup$
Let $k$ be a field of characteristic zero, and let $a, b in bar{k}$ ($bar{k}$ is an algebraic closure of $k$) be two distinct elements, such that
$k(a)=k(b)$.
Notice that $k(a)=k(b)$ implies that the degree of the minimal polynomial of $a$ over $k$, $d_a$, equals the degree of the minimal polynomial of $b$ over $k$, $d_b$; denote $d:=d_a=d_b$.
Further assume that $k(ab)=k(a)=k(b)$ (hence the degree of the minimal polynomial of $ab$ over $k$, $d_{ab}$, equals $d$).
(1) Could one find a concrete example to the above situation with, for example, $k=mathbb{Q}$?
(2) Is there something 'interesting' to say about $ab$?
Same questions (1) and (2) for the following special cases:
(i) $b=a-lambda$, for some $lambda in k^{times}$? Should it have 'easier/nicer' answers?
(ii) $d$ is a prime number $geq 3$? Also, should the extension be Galois? See this and this questions.
Notice that this question is not relevant (at least not directly, but maybe something can be obtained from it), since it talks about relatively prime degrees and sum of elements, while here we talk about equal degrees (probably $d>1$) and product of elements.
Perhaps the primitive element theorem can help?
Thank you very much!
In short, I am asking: Assume that the product $ab$ of two distinct primitive elements $a,b in L$ for the field extension $k subseteq L$ (namely, $L=k(a)=k(b)$) is also a primitive element (namely, $L=k(ab)$), does this tell something interesting?
field-theory galois-theory extension-field minimal-polynomials
asked Jan 3 at 0:12
user237522user237522
2,1631617
$endgroup$
|
show 4 more comments
$begingroup$
Let $k$ be a field of characteristic zero, and let $a, b in bar{k}$ ($bar{k}$ is an algebraic closure of $k$) be two distinct elements, such that
$k(a)=k(b)$.
Notice that $k(a)=k(b)$ implies that the degree of the minimal polynomial of $a$ over $k$, $d_a$, equals the degree of the minimal polynomial of $b$ over $k$, $d_b$; denote $d:=d_a=d_b$.
Further assume that $k(ab)=k(a)=k(b)$ (hence the degree of the minimal polynomial of $ab$ over $k$, $d_{ab}$, equals $d$).
(1) Could one find a concrete example to the above situation with, for example, $k=mathbb{Q}$?
(2) Is there something 'interesting' to say about $ab$?
Same questions (1) and (2) for the following special cases:
(i) $b=a-lambda$, for some $lambda in k^{times}$? Should it have 'easier/nicer' answers?
(ii) $d$ is a prime number $geq 3$? Also, should the extension be Galois? See this and this questions.
Notice that this question is not relevant (at least not directly, but maybe something can be obtained from it), since it talks about relatively prime degrees and sum of elements, while here we talk about equal degrees (probably $d>1$) and product of elements.
Perhaps the primitive element theorem can help?
Thank you very much!
In short, I am asking: Assume that the product $ab$ of two distinct primitive elements $a,b in L$ for the field extension $k subseteq L$ (namely, $L=k(a)=k(b)$) is also a primitive element (namely, $L=k(ab)$), does this tell something interesting?
field-theory galois-theory extension-field minimal-polynomials
asked Jan 3 at 0:12
user237522user237522
2,1631617
$endgroup$
1
$begingroup$
An example: over the rationals, $a=sqrt2$, $b=1+sqrt2$.
$endgroup$
– Gerry Myerson
Jan 3 at 2:31
$begingroup$
Nice example, thanks.
$endgroup$
– user237522
Jan 3 at 14:04
1
$begingroup$
Well if $L=Bbb Q(1+sqrt2)$ then $L = Bbb Q((1+sqrt2)^n)$ for all $n ge 1$ =)
$endgroup$
– Kenny Lau
Jan 3 at 18:13
1
$begingroup$
I think these experiences should be enough to show you that you need to think way harder before you ask your questions
$endgroup$
– Kenny Lau
Jan 3 at 18:19
1
$begingroup$
Thanks for the comment. Yes, for a prime degree extension, if we take $a,b in L-k$ such that $ab notin k$, then necessarily $k(ab)=L$ (since $k subseteq L$ does not have intermediate fields, by considerations of degrees).
$endgroup$
– user237522
Jan 3 at 20:15
|
show 4 more comments
$begingroup$
Let $k$ be a field of characteristic zero, and let $a, b in bar{k}$ ($bar{k}$ is an algebraic closure of $k$) be two distinct elements, such that
$k(a)=k(b)$.
Notice that $k(a)=k(b)$ implies that the degree of the minimal polynomial of $a$ over $k$, $d_a$, equals the degree of the minimal polynomial of $b$ over $k$, $d_b$; denote $d:=d_a=d_b$.
Further assume that $k(ab)=k(a)=k(b)$ (hence the degree of the minimal polynomial of $ab$ over $k$, $d_{ab}$, equals $d$).
(1) Could one find a concrete example to the above situation with, for example, $k=mathbb{Q}$?
(2) Is there something 'interesting' to say about $ab$?
Same questions (1) and (2) for the following special cases:
(i) $b=a-lambda$, for some $lambda in k^{times}$? Should it have 'easier/nicer' answers?
(ii) $d$ is a prime number $geq 3$? Also, should the extension be Galois? See this and this questions.
Notice that this question is not relevant (at least not directly, but maybe something can be obtained from it), since it talks about relatively prime degrees and sum of elements, while here we talk about equal degrees (probably $d>1$) and product of elements.
Perhaps the primitive element theorem can help?
Thank you very much!
In short, I am asking: Assume that the product $ab$ of two distinct primitive elements $a,b in L$ for the field extension $k subseteq L$ (namely, $L=k(a)=k(b)$) is also a primitive element (namely, $L=k(ab)$), does this tell something interesting?
field-theory galois-theory extension-field minimal-polynomials
asked Jan 3 at 0:12
user237522user237522
2,1631617
$endgroup$
Let $k$ be a field of characteristic zero, and let $a, b in bar{k}$ ($bar{k}$ is an algebraic closure of $k$) be two distinct elements, such that
$k(a)=k(b)$.
Notice that $k(a)=k(b)$ implies that the degree of the minimal polynomial of $a$ over $k$, $d_a$, equals the degree of the minimal polynomial of $b$ over $k$, $d_b$; denote $d:=d_a=d_b$.
Further assume that $k(ab)=k(a)=k(b)$ (hence the degree of the minimal polynomial of $ab$ over $k$, $d_{ab}$, equals $d$).
(1) Could one find a concrete example to the above situation with, for example, $k=mathbb{Q}$?
(2) Is there something 'interesting' to say about $ab$?
Same questions (1) and (2) for the following special cases:
(i) $b=a-lambda$, for some $lambda in k^{times}$? Should it have 'easier/nicer' answers?
(ii) $d$ is a prime number $geq 3$? Also, should the extension be Galois? See this and this questions.
Notice that this question is not relevant (at least not directly, but maybe something can be obtained from it), since it talks about relatively prime degrees and sum of elements, while here we talk about equal degrees (probably $d>1$) and product of elements.
Perhaps the primitive element theorem can help?
Thank you very much!
In short, I am asking: Assume that the product $ab$ of two distinct primitive elements $a,b in L$ for the field extension $k subseteq L$ (namely, $L=k(a)=k(b)$) is also a primitive element (namely, $L=k(ab)$), does this tell something interesting?
field-theory galois-theory extension-field minimal-polynomials
field-theory galois-theory extension-field minimal-polynomials
asked Jan 3 at 0:12
user237522user237522
2,1631617
asked Jan 3 at 0:12
user237522user237522
2,1631617
edited Jan 3 at 1:46
user237522
asked Jan 3 at 0:12
user237522user237522
2,1631617
asked Jan 3 at 0:12
user237522user237522
2,1631617
asked Jan 3 at 0:12
user237522user237522
2,1631617
2,1631617
1
$begingroup$
An example: over the rationals, $a=sqrt2$, $b=1+sqrt2$.
$endgroup$
– Gerry Myerson
Jan 3 at 2:31
$begingroup$
Nice example, thanks.
$endgroup$
– user237522
Jan 3 at 14:04
1
$begingroup$
Well if $L=Bbb Q(1+sqrt2)$ then $L = Bbb Q((1+sqrt2)^n)$ for all $n ge 1$ =)
$endgroup$
– Kenny Lau
Jan 3 at 18:13
1
$begingroup$
I think these experiences should be enough to show you that you need to think way harder before you ask your questions
$endgroup$
– Kenny Lau
Jan 3 at 18:19
1
$begingroup$
Thanks for the comment. Yes, for a prime degree extension, if we take $a,b in L-k$ such that $ab notin k$, then necessarily $k(ab)=L$ (since $k subseteq L$ does not have intermediate fields, by considerations of degrees).
$endgroup$
– user237522
Jan 3 at 20:15
|
show 4 more comments
1
$begingroup$
An example: over the rationals, $a=sqrt2$, $b=1+sqrt2$.
$endgroup$
– Gerry Myerson
Jan 3 at 2:31
$begingroup$
Nice example, thanks.
$endgroup$
– user237522
Jan 3 at 14:04
1
$begingroup$
Well if $L=Bbb Q(1+sqrt2)$ then $L = Bbb Q((1+sqrt2)^n)$ for all $n ge 1$ =)
$endgroup$
– Kenny Lau
Jan 3 at 18:13
1
$begingroup$
I think these experiences should be enough to show you that you need to think way harder before you ask your questions
$endgroup$
– Kenny Lau
Jan 3 at 18:19
1
$begingroup$
Thanks for the comment. Yes, for a prime degree extension, if we take $a,b in L-k$ such that $ab notin k$, then necessarily $k(ab)=L$ (since $k subseteq L$ does not have intermediate fields, by considerations of degrees).
$endgroup$
– user237522
Jan 3 at 20:15
1
1
$begingroup$
An example: over the rationals, $a=sqrt2$, $b=1+sqrt2$.
$endgroup$
– Gerry Myerson
Jan 3 at 2:31
$begingroup$
An example: over the rationals, $a=sqrt2$, $b=1+sqrt2$.
$endgroup$
– Gerry Myerson
Jan 3 at 2:31
$begingroup$
Nice example, thanks.
$endgroup$
– user237522
Jan 3 at 14:04
$begingroup$
Nice example, thanks.
$endgroup$
– user237522
Jan 3 at 14:04
1
1
$begingroup$
Well if $L=Bbb Q(1+sqrt2)$ then $L = Bbb Q((1+sqrt2)^n)$ for all $n ge 1$ =)
$endgroup$
– Kenny Lau
Jan 3 at 18:13
$begingroup$
Well if $L=Bbb Q(1+sqrt2)$ then $L = Bbb Q((1+sqrt2)^n)$ for all $n ge 1$ =)
$endgroup$
– Kenny Lau
Jan 3 at 18:13
1
1
$begingroup$
I think these experiences should be enough to show you that you need to think way harder before you ask your questions
$endgroup$
– Kenny Lau
Jan 3 at 18:19
$begingroup$
I think these experiences should be enough to show you that you need to think way harder before you ask your questions
$endgroup$
– Kenny Lau
Jan 3 at 18:19
1
1
$begingroup$
Thanks for the comment. Yes, for a prime degree extension, if we take $a,b in L-k$ such that $ab notin k$, then necessarily $k(ab)=L$ (since $k subseteq L$ does not have intermediate fields, by considerations of degrees).
$endgroup$
– user237522
Jan 3 at 20:15
$begingroup$
Thanks for the comment. Yes, for a prime degree extension, if we take $a,b in L-k$ such that $ab notin k$, then necessarily $k(ab)=L$ (since $k subseteq L$ does not have intermediate fields, by considerations of degrees).
$endgroup$
– user237522
Jan 3 at 20:15
|
show 4 more comments
0
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1
$begingroup$
An example: over the rationals, $a=sqrt2$, $b=1+sqrt2$.
$endgroup$
– Gerry Myerson
Jan 3 at 2:31
$begingroup$
Nice example, thanks.
$endgroup$
– user237522
Jan 3 at 14:04
1
$begingroup$
Well if $L=Bbb Q(1+sqrt2)$ then $L = Bbb Q((1+sqrt2)^n)$ for all $n ge 1$ =)
$endgroup$
– Kenny Lau
Jan 3 at 18:13
1
$begingroup$
I think these experiences should be enough to show you that you need to think way harder before you ask your questions
$endgroup$
– Kenny Lau
Jan 3 at 18:19
1
$begingroup$
Thanks for the comment. Yes, for a prime degree extension, if we take $a,b in L-k$ such that $ab notin k$, then necessarily $k(ab)=L$ (since $k subseteq L$ does not have intermediate fields, by considerations of degrees).
$endgroup$
– user237522
Jan 3 at 20:15