Mixing
Let $[L:K]=2$ and char $Kneq 2$ show that $L/K$ is a simple 2-radical extension.
$begingroup$
I want to show that
$[L:K]=2$ and char $Kneq 2$ $Rightarrow$ $L/K$ is a simple 2-radical extension.
I know that, since $[L:K]$ is prime, the extension is simple. I also concluded that, since $[L:K]=[K(a):K]=[a:K]$ for all $ain L/K$, the minimal polynomial of $a$ has to have degree 2. But how do I proceed from here ? And how does char$Kneq 2$ help me ?
abstract-algebra field-theory extension-field radicals
asked Feb 3 at 6:55
Christian SingerChristian Singer
429313
$endgroup$
add a comment |
$begingroup$
I want to show that
$[L:K]=2$ and char $Kneq 2$ $Rightarrow$ $L/K$ is a simple 2-radical extension.
I know that, since $[L:K]$ is prime, the extension is simple. I also concluded that, since $[L:K]=[K(a):K]=[a:K]$ for all $ain L/K$, the minimal polynomial of $a$ has to have degree 2. But how do I proceed from here ? And how does char$Kneq 2$ help me ?
abstract-algebra field-theory extension-field radicals
asked Feb 3 at 6:55
Christian SingerChristian Singer
429313
$endgroup$
3
$begingroup$
$a$ is a root of a quadratic equations. You can solve quadratic equations by "completing the square" (except in characteristic two).
$endgroup$
– Lord Shark the Unknown
Feb 3 at 6:59
$begingroup$
So if I'm correct I just have to write down the "abc-formular", square it and hence get a representation of $a^{2}$ with coefficients from $L$. This would then show that $a^{2}in L$. The char$Kneq2$ is needed to gurantee that the denominator of the abc formular is not $0$.
$endgroup$
– Christian Singer
Feb 3 at 7:11
$begingroup$
* I meant to say $K$ instead of $L$
$endgroup$
– Christian Singer
Feb 3 at 7:18
add a comment |
$begingroup$
I want to show that
$[L:K]=2$ and char $Kneq 2$ $Rightarrow$ $L/K$ is a simple 2-radical extension.
I know that, since $[L:K]$ is prime, the extension is simple. I also concluded that, since $[L:K]=[K(a):K]=[a:K]$ for all $ain L/K$, the minimal polynomial of $a$ has to have degree 2. But how do I proceed from here ? And how does char$Kneq 2$ help me ?
abstract-algebra field-theory extension-field radicals
asked Feb 3 at 6:55
Christian SingerChristian Singer
429313
$endgroup$
I want to show that
$[L:K]=2$ and char $Kneq 2$ $Rightarrow$ $L/K$ is a simple 2-radical extension.
I know that, since $[L:K]$ is prime, the extension is simple. I also concluded that, since $[L:K]=[K(a):K]=[a:K]$ for all $ain L/K$, the minimal polynomial of $a$ has to have degree 2. But how do I proceed from here ? And how does char$Kneq 2$ help me ?
abstract-algebra field-theory extension-field radicals
abstract-algebra field-theory extension-field radicals
asked Feb 3 at 6:55
Christian SingerChristian Singer
429313
asked Feb 3 at 6:55
Christian SingerChristian Singer
429313
asked Feb 3 at 6:55
Christian SingerChristian Singer
429313
asked Feb 3 at 6:55
Christian SingerChristian Singer
429313
asked Feb 3 at 6:55
Christian SingerChristian Singer
429313
429313
3
$begingroup$
$a$ is a root of a quadratic equations. You can solve quadratic equations by "completing the square" (except in characteristic two).
$endgroup$
– Lord Shark the Unknown
Feb 3 at 6:59
$begingroup$
So if I'm correct I just have to write down the "abc-formular", square it and hence get a representation of $a^{2}$ with coefficients from $L$. This would then show that $a^{2}in L$. The char$Kneq2$ is needed to gurantee that the denominator of the abc formular is not $0$.
$endgroup$
– Christian Singer
Feb 3 at 7:11
$begingroup$
* I meant to say $K$ instead of $L$
$endgroup$
– Christian Singer
Feb 3 at 7:18
add a comment |
3
$begingroup$
$a$ is a root of a quadratic equations. You can solve quadratic equations by "completing the square" (except in characteristic two).
$endgroup$
– Lord Shark the Unknown
Feb 3 at 6:59
$begingroup$
So if I'm correct I just have to write down the "abc-formular", square it and hence get a representation of $a^{2}$ with coefficients from $L$. This would then show that $a^{2}in L$. The char$Kneq2$ is needed to gurantee that the denominator of the abc formular is not $0$.
$endgroup$
– Christian Singer
Feb 3 at 7:11
$begingroup$
* I meant to say $K$ instead of $L$
$endgroup$
– Christian Singer
Feb 3 at 7:18
3
3
$begingroup$
$a$ is a root of a quadratic equations. You can solve quadratic equations by "completing the square" (except in characteristic two).
$endgroup$
– Lord Shark the Unknown
Feb 3 at 6:59
$begingroup$
$a$ is a root of a quadratic equations. You can solve quadratic equations by "completing the square" (except in characteristic two).
$endgroup$
– Lord Shark the Unknown
Feb 3 at 6:59
$begingroup$
So if I'm correct I just have to write down the "abc-formular", square it and hence get a representation of $a^{2}$ with coefficients from $L$. This would then show that $a^{2}in L$. The char$Kneq2$ is needed to gurantee that the denominator of the abc formular is not $0$.
$endgroup$
– Christian Singer
Feb 3 at 7:11
$begingroup$
So if I'm correct I just have to write down the "abc-formular", square it and hence get a representation of $a^{2}$ with coefficients from $L$. This would then show that $a^{2}in L$. The char$Kneq2$ is needed to gurantee that the denominator of the abc formular is not $0$.
$endgroup$
– Christian Singer
Feb 3 at 7:11
$begingroup$
* I meant to say $K$ instead of $L$
$endgroup$
– Christian Singer
Feb 3 at 7:18
$begingroup$
* I meant to say $K$ instead of $L$
$endgroup$
– Christian Singer
Feb 3 at 7:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Choose any
$a in L setminus K; tag 1$
then since
$[L:K] = 2, tag 2$
it follows that
${1, a } subset L tag 3$
forms a basis for $L$ over $K$; thus the set
${1, a, a^2 } tag 4$
is linearly dependent over $K$; hence we have
$sigma, tau, rho in K, tag 5$
not all zero, such that
$sigma a^2 + tau a + rho = sigma a^2 + tau a + rho 1 = 0; tag 6$
we note that
$sigma ne 0, tag 7$
lest
$tau a + rho = 0; tag 8$
but then
$tau ne 0 Longrightarrow a = -dfrac{rho}{tau} in K Rightarrow Leftarrow a in L setminus K, tag 9$
whereas
$tau = 0 Longrightarrow rho = 0 Longrightarrow sigma, tau, rho = 0, tag{10}$
contradicting our hypothesis that there is a non-zero element amongst $sigma$, $tau$, $rho$; thus (7) binds and setting
$alpha = dfrac{tau}{sigma}, ; beta = dfrac{rho}{sigma}, tag{11}$
we write (6) in the form
$a^2 + alpha a + beta = 0, tag{12}$
that is, $a$ satisfies the monic quadratic polynomial
$x^2 + alpha x+ beta in K[x]; tag{13}$
(12) yields
$a^2 + alpha a = -beta, tag{14}$
whence we have, since $text{char}(K) ne 2$,
$left (a + dfrac{alpha}{2} right )^2 = a^2 + alpha a + dfrac{alpha}{4} = dfrac{alpha}{4} - beta in K; tag{15}$
we thus see that
$b = a + dfrac{alpha}{2} in L setminus K tag{16}$
with
$b^2 in K. tag{17}$
It is now manifestly evident that
$L = K(a) = K left (a + dfrac{alpha}{2} right ) = K(b) tag{18}$
is a simple extension of $K$ (since it is generated by adjoining either of the single elements $a$ or $b$ to $K$), and by virtue of (17), is what our OP Christian Singer refers to a a "2-radical extension", since we may write
$b = sqrt { dfrac{alpha}{4} - beta }. tag{19}$
answered Feb 3 at 19:48
Robert LewisRobert Lewis
49k23168
$endgroup$
$begingroup$
Thanks alot for the in depth answer!
$endgroup$
– Christian Singer
Feb 3 at 20:04
$begingroup$
My pleasure sir, and thanks for the "acceptance"!
$endgroup$
– Robert Lewis
Feb 3 at 20:05
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Choose any
$a in L setminus K; tag 1$
then since
$[L:K] = 2, tag 2$
it follows that
${1, a } subset L tag 3$
forms a basis for $L$ over $K$; thus the set
${1, a, a^2 } tag 4$
is linearly dependent over $K$; hence we have
$sigma, tau, rho in K, tag 5$
not all zero, such that
$sigma a^2 + tau a + rho = sigma a^2 + tau a + rho 1 = 0; tag 6$
we note that
$sigma ne 0, tag 7$
lest
$tau a + rho = 0; tag 8$
but then
$tau ne 0 Longrightarrow a = -dfrac{rho}{tau} in K Rightarrow Leftarrow a in L setminus K, tag 9$
whereas
$tau = 0 Longrightarrow rho = 0 Longrightarrow sigma, tau, rho = 0, tag{10}$
contradicting our hypothesis that there is a non-zero element amongst $sigma$, $tau$, $rho$; thus (7) binds and setting
$alpha = dfrac{tau}{sigma}, ; beta = dfrac{rho}{sigma}, tag{11}$
we write (6) in the form
$a^2 + alpha a + beta = 0, tag{12}$
that is, $a$ satisfies the monic quadratic polynomial
$x^2 + alpha x+ beta in K[x]; tag{13}$
(12) yields
$a^2 + alpha a = -beta, tag{14}$
whence we have, since $text{char}(K) ne 2$,
$left (a + dfrac{alpha}{2} right )^2 = a^2 + alpha a + dfrac{alpha}{4} = dfrac{alpha}{4} - beta in K; tag{15}$
we thus see that
$b = a + dfrac{alpha}{2} in L setminus K tag{16}$
with
$b^2 in K. tag{17}$
It is now manifestly evident that
$L = K(a) = K left (a + dfrac{alpha}{2} right ) = K(b) tag{18}$
is a simple extension of $K$ (since it is generated by adjoining either of the single elements $a$ or $b$ to $K$), and by virtue of (17), is what our OP Christian Singer refers to a a "2-radical extension", since we may write
$b = sqrt { dfrac{alpha}{4} - beta }. tag{19}$
answered Feb 3 at 19:48
Robert LewisRobert Lewis
49k23168
$endgroup$
$begingroup$
Thanks alot for the in depth answer!
$endgroup$
– Christian Singer
Feb 3 at 20:04
$begingroup$
My pleasure sir, and thanks for the "acceptance"!
$endgroup$
– Robert Lewis
Feb 3 at 20:05
add a comment |
$begingroup$
Choose any
$a in L setminus K; tag 1$
then since
$[L:K] = 2, tag 2$
it follows that
${1, a } subset L tag 3$
forms a basis for $L$ over $K$; thus the set
${1, a, a^2 } tag 4$
is linearly dependent over $K$; hence we have
$sigma, tau, rho in K, tag 5$
not all zero, such that
$sigma a^2 + tau a + rho = sigma a^2 + tau a + rho 1 = 0; tag 6$
we note that
$sigma ne 0, tag 7$
lest
$tau a + rho = 0; tag 8$
but then
$tau ne 0 Longrightarrow a = -dfrac{rho}{tau} in K Rightarrow Leftarrow a in L setminus K, tag 9$
whereas
$tau = 0 Longrightarrow rho = 0 Longrightarrow sigma, tau, rho = 0, tag{10}$
contradicting our hypothesis that there is a non-zero element amongst $sigma$, $tau$, $rho$; thus (7) binds and setting
$alpha = dfrac{tau}{sigma}, ; beta = dfrac{rho}{sigma}, tag{11}$
we write (6) in the form
$a^2 + alpha a + beta = 0, tag{12}$
that is, $a$ satisfies the monic quadratic polynomial
$x^2 + alpha x+ beta in K[x]; tag{13}$
(12) yields
$a^2 + alpha a = -beta, tag{14}$
whence we have, since $text{char}(K) ne 2$,
$left (a + dfrac{alpha}{2} right )^2 = a^2 + alpha a + dfrac{alpha}{4} = dfrac{alpha}{4} - beta in K; tag{15}$
we thus see that
$b = a + dfrac{alpha}{2} in L setminus K tag{16}$
with
$b^2 in K. tag{17}$
It is now manifestly evident that
$L = K(a) = K left (a + dfrac{alpha}{2} right ) = K(b) tag{18}$
is a simple extension of $K$ (since it is generated by adjoining either of the single elements $a$ or $b$ to $K$), and by virtue of (17), is what our OP Christian Singer refers to a a "2-radical extension", since we may write
$b = sqrt { dfrac{alpha}{4} - beta }. tag{19}$
answered Feb 3 at 19:48
Robert LewisRobert Lewis
49k23168
$endgroup$
$begingroup$
Thanks alot for the in depth answer!
$endgroup$
– Christian Singer
Feb 3 at 20:04
$begingroup$
My pleasure sir, and thanks for the "acceptance"!
$endgroup$
– Robert Lewis
Feb 3 at 20:05
add a comment |
$begingroup$
Choose any
$a in L setminus K; tag 1$
then since
$[L:K] = 2, tag 2$
it follows that
${1, a } subset L tag 3$
forms a basis for $L$ over $K$; thus the set
${1, a, a^2 } tag 4$
is linearly dependent over $K$; hence we have
$sigma, tau, rho in K, tag 5$
not all zero, such that
$sigma a^2 + tau a + rho = sigma a^2 + tau a + rho 1 = 0; tag 6$
we note that
$sigma ne 0, tag 7$
lest
$tau a + rho = 0; tag 8$
but then
$tau ne 0 Longrightarrow a = -dfrac{rho}{tau} in K Rightarrow Leftarrow a in L setminus K, tag 9$
whereas
$tau = 0 Longrightarrow rho = 0 Longrightarrow sigma, tau, rho = 0, tag{10}$
contradicting our hypothesis that there is a non-zero element amongst $sigma$, $tau$, $rho$; thus (7) binds and setting
$alpha = dfrac{tau}{sigma}, ; beta = dfrac{rho}{sigma}, tag{11}$
we write (6) in the form
$a^2 + alpha a + beta = 0, tag{12}$
that is, $a$ satisfies the monic quadratic polynomial
$x^2 + alpha x+ beta in K[x]; tag{13}$
(12) yields
$a^2 + alpha a = -beta, tag{14}$
whence we have, since $text{char}(K) ne 2$,
$left (a + dfrac{alpha}{2} right )^2 = a^2 + alpha a + dfrac{alpha}{4} = dfrac{alpha}{4} - beta in K; tag{15}$
we thus see that
$b = a + dfrac{alpha}{2} in L setminus K tag{16}$
with
$b^2 in K. tag{17}$
It is now manifestly evident that
$L = K(a) = K left (a + dfrac{alpha}{2} right ) = K(b) tag{18}$
is a simple extension of $K$ (since it is generated by adjoining either of the single elements $a$ or $b$ to $K$), and by virtue of (17), is what our OP Christian Singer refers to a a "2-radical extension", since we may write
$b = sqrt { dfrac{alpha}{4} - beta }. tag{19}$
answered Feb 3 at 19:48
Robert LewisRobert Lewis
49k23168
$endgroup$
Choose any
$a in L setminus K; tag 1$
then since
$[L:K] = 2, tag 2$
it follows that
${1, a } subset L tag 3$
forms a basis for $L$ over $K$; thus the set
${1, a, a^2 } tag 4$
is linearly dependent over $K$; hence we have
$sigma, tau, rho in K, tag 5$
not all zero, such that
$sigma a^2 + tau a + rho = sigma a^2 + tau a + rho 1 = 0; tag 6$
we note that
$sigma ne 0, tag 7$
lest
$tau a + rho = 0; tag 8$
but then
$tau ne 0 Longrightarrow a = -dfrac{rho}{tau} in K Rightarrow Leftarrow a in L setminus K, tag 9$
whereas
$tau = 0 Longrightarrow rho = 0 Longrightarrow sigma, tau, rho = 0, tag{10}$
contradicting our hypothesis that there is a non-zero element amongst $sigma$, $tau$, $rho$; thus (7) binds and setting
$alpha = dfrac{tau}{sigma}, ; beta = dfrac{rho}{sigma}, tag{11}$
we write (6) in the form
$a^2 + alpha a + beta = 0, tag{12}$
that is, $a$ satisfies the monic quadratic polynomial
$x^2 + alpha x+ beta in K[x]; tag{13}$
(12) yields
$a^2 + alpha a = -beta, tag{14}$
whence we have, since $text{char}(K) ne 2$,
$left (a + dfrac{alpha}{2} right )^2 = a^2 + alpha a + dfrac{alpha}{4} = dfrac{alpha}{4} - beta in K; tag{15}$
we thus see that
$b = a + dfrac{alpha}{2} in L setminus K tag{16}$
with
$b^2 in K. tag{17}$
It is now manifestly evident that
$L = K(a) = K left (a + dfrac{alpha}{2} right ) = K(b) tag{18}$
is a simple extension of $K$ (since it is generated by adjoining either of the single elements $a$ or $b$ to $K$), and by virtue of (17), is what our OP Christian Singer refers to a a "2-radical extension", since we may write
$b = sqrt { dfrac{alpha}{4} - beta }. tag{19}$
answered Feb 3 at 19:48
Robert LewisRobert Lewis
49k23168
answered Feb 3 at 19:48
Robert LewisRobert Lewis
49k23168
answered Feb 3 at 19:48
Robert LewisRobert Lewis
49k23168
answered Feb 3 at 19:48
Robert LewisRobert Lewis
49k23168
49k23168
$begingroup$
Thanks alot for the in depth answer!
$endgroup$
– Christian Singer
Feb 3 at 20:04
$begingroup$
My pleasure sir, and thanks for the "acceptance"!
$endgroup$
– Robert Lewis
Feb 3 at 20:05
add a comment |
$begingroup$
Thanks alot for the in depth answer!
$endgroup$
– Christian Singer
Feb 3 at 20:04
$begingroup$
My pleasure sir, and thanks for the "acceptance"!
$endgroup$
– Robert Lewis
Feb 3 at 20:05
$begingroup$
Thanks alot for the in depth answer!
$endgroup$
– Christian Singer
Feb 3 at 20:04
$begingroup$
Thanks alot for the in depth answer!
$endgroup$
– Christian Singer
Feb 3 at 20:04
$begingroup$
My pleasure sir, and thanks for the "acceptance"!
$endgroup$
– Robert Lewis
Feb 3 at 20:05
$begingroup$
My pleasure sir, and thanks for the "acceptance"!
$endgroup$
– Robert Lewis
Feb 3 at 20:05
add a comment |
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3
$begingroup$
$a$ is a root of a quadratic equations. You can solve quadratic equations by "completing the square" (except in characteristic two).
$endgroup$
– Lord Shark the Unknown
Feb 3 at 6:59
$begingroup$
So if I'm correct I just have to write down the "abc-formular", square it and hence get a representation of $a^{2}$ with coefficients from $L$. This would then show that $a^{2}in L$. The char$Kneq2$ is needed to gurantee that the denominator of the abc formular is not $0$.
$endgroup$
– Christian Singer
Feb 3 at 7:11
$begingroup$
* I meant to say $K$ instead of $L$
$endgroup$
– Christian Singer
Feb 3 at 7:18