Let $E$ be an algebraically closed extension field of a field $F$. Show that the algebraic closure $bar F_E$...
$begingroup$
Question: Let $E$ be an algebraically closed extension field of a field $F$. Show that the algebraic closure $bar F_E$ of $F$ in $E$ is algebraically closed.
Firstly, I conclude a thing, if $bar F_E$ is properly contained in $E$, then it couldn't be algebraically closed, since an algebraically closed field cannot be properly contained. Therefore my idea is to prove $bar F_E=E$.
Then I notice that $E$ is algebraically closed, then for each polynomial $f(x)in E[x]$, it can be split completely into linear factors:
$$f(x)=(x-alpha_1)cdots(x-alpha_n)quadtext{ for $alpha_iin E$, $i=1,2,cdots,n$. }$$
But since $bar F_Eleq E$ obviously, the remaining things to prove is $Eleq bar F_E$, so I want to choose an element $alphain E$, and I want to prove it is algebraic over $F$, but I cannot find any connection between this and the completely factorization of $f(x)$. Any suggestion?
abstract-algebra field-theory extension-field
asked Jan 21 at 0:47
kelvin hong 方kelvin hong 方
65518
$endgroup$
add a comment |
$begingroup$
Question: Let $E$ be an algebraically closed extension field of a field $F$. Show that the algebraic closure $bar F_E$ of $F$ in $E$ is algebraically closed.
Firstly, I conclude a thing, if $bar F_E$ is properly contained in $E$, then it couldn't be algebraically closed, since an algebraically closed field cannot be properly contained. Therefore my idea is to prove $bar F_E=E$.
Then I notice that $E$ is algebraically closed, then for each polynomial $f(x)in E[x]$, it can be split completely into linear factors:
$$f(x)=(x-alpha_1)cdots(x-alpha_n)quadtext{ for $alpha_iin E$, $i=1,2,cdots,n$. }$$
But since $bar F_Eleq E$ obviously, the remaining things to prove is $Eleq bar F_E$, so I want to choose an element $alphain E$, and I want to prove it is algebraic over $F$, but I cannot find any connection between this and the completely factorization of $f(x)$. Any suggestion?
abstract-algebra field-theory extension-field
asked Jan 21 at 0:47
kelvin hong 方kelvin hong 方
65518
$endgroup$
1
$begingroup$
Are you assuming $E$ is an algebraic extension of $F$?
$endgroup$
– Eclipse Sun
Jan 21 at 0:51
$begingroup$
@EclipseSun I think no, it is the assumption of the question to make me think to that direction of proving. Is it $bar F_E$ cannot properly contained in $E$ a wrong idea?
$endgroup$
– kelvin hong 方
Jan 21 at 0:58
$begingroup$
An algebraically closed field, for example $mathbb{C}$ can be properly contained in another algebraically closed field, if you allow that extension to be transcendental.
$endgroup$
– Eclipse Sun
Jan 21 at 1:03
$begingroup$
But isn't that all transcendental numbers are already in $mathbb R$ hence in $mathbb C$? I never heard before $mathbb C$ can be properly contained in other field. Wow!
$endgroup$
– kelvin hong 方
Jan 21 at 1:06
2
$begingroup$
$mathbb{C}$ is contained in $mathbb{C}(x)$, the field of rational functions on $mathbb{C}$.
$endgroup$
– Eclipse Sun
Jan 21 at 1:13
add a comment |
$begingroup$
Question: Let $E$ be an algebraically closed extension field of a field $F$. Show that the algebraic closure $bar F_E$ of $F$ in $E$ is algebraically closed.
Firstly, I conclude a thing, if $bar F_E$ is properly contained in $E$, then it couldn't be algebraically closed, since an algebraically closed field cannot be properly contained. Therefore my idea is to prove $bar F_E=E$.
Then I notice that $E$ is algebraically closed, then for each polynomial $f(x)in E[x]$, it can be split completely into linear factors:
$$f(x)=(x-alpha_1)cdots(x-alpha_n)quadtext{ for $alpha_iin E$, $i=1,2,cdots,n$. }$$
But since $bar F_Eleq E$ obviously, the remaining things to prove is $Eleq bar F_E$, so I want to choose an element $alphain E$, and I want to prove it is algebraic over $F$, but I cannot find any connection between this and the completely factorization of $f(x)$. Any suggestion?
abstract-algebra field-theory extension-field
asked Jan 21 at 0:47
kelvin hong 方kelvin hong 方
65518
$endgroup$
Question: Let $E$ be an algebraically closed extension field of a field $F$. Show that the algebraic closure $bar F_E$ of $F$ in $E$ is algebraically closed.
Firstly, I conclude a thing, if $bar F_E$ is properly contained in $E$, then it couldn't be algebraically closed, since an algebraically closed field cannot be properly contained. Therefore my idea is to prove $bar F_E=E$.
Then I notice that $E$ is algebraically closed, then for each polynomial $f(x)in E[x]$, it can be split completely into linear factors:
$$f(x)=(x-alpha_1)cdots(x-alpha_n)quadtext{ for $alpha_iin E$, $i=1,2,cdots,n$. }$$
But since $bar F_Eleq E$ obviously, the remaining things to prove is $Eleq bar F_E$, so I want to choose an element $alphain E$, and I want to prove it is algebraic over $F$, but I cannot find any connection between this and the completely factorization of $f(x)$. Any suggestion?
abstract-algebra field-theory extension-field
abstract-algebra field-theory extension-field
asked Jan 21 at 0:47
kelvin hong 方kelvin hong 方
65518
asked Jan 21 at 0:47
kelvin hong 方kelvin hong 方
65518
asked Jan 21 at 0:47
kelvin hong 方kelvin hong 方
65518
asked Jan 21 at 0:47
kelvin hong 方kelvin hong 方
65518
asked Jan 21 at 0:47
kelvin hong 方kelvin hong 方
65518
65518
1
$begingroup$
Are you assuming $E$ is an algebraic extension of $F$?
$endgroup$
– Eclipse Sun
Jan 21 at 0:51
$begingroup$
@EclipseSun I think no, it is the assumption of the question to make me think to that direction of proving. Is it $bar F_E$ cannot properly contained in $E$ a wrong idea?
$endgroup$
– kelvin hong 方
Jan 21 at 0:58
$begingroup$
An algebraically closed field, for example $mathbb{C}$ can be properly contained in another algebraically closed field, if you allow that extension to be transcendental.
$endgroup$
– Eclipse Sun
Jan 21 at 1:03
$begingroup$
But isn't that all transcendental numbers are already in $mathbb R$ hence in $mathbb C$? I never heard before $mathbb C$ can be properly contained in other field. Wow!
$endgroup$
– kelvin hong 方
Jan 21 at 1:06
2
$begingroup$
$mathbb{C}$ is contained in $mathbb{C}(x)$, the field of rational functions on $mathbb{C}$.
$endgroup$
– Eclipse Sun
Jan 21 at 1:13
add a comment |
1
$begingroup$
Are you assuming $E$ is an algebraic extension of $F$?
$endgroup$
– Eclipse Sun
Jan 21 at 0:51
$begingroup$
@EclipseSun I think no, it is the assumption of the question to make me think to that direction of proving. Is it $bar F_E$ cannot properly contained in $E$ a wrong idea?
$endgroup$
– kelvin hong 方
Jan 21 at 0:58
$begingroup$
An algebraically closed field, for example $mathbb{C}$ can be properly contained in another algebraically closed field, if you allow that extension to be transcendental.
$endgroup$
– Eclipse Sun
Jan 21 at 1:03
$begingroup$
But isn't that all transcendental numbers are already in $mathbb R$ hence in $mathbb C$? I never heard before $mathbb C$ can be properly contained in other field. Wow!
$endgroup$
– kelvin hong 方
Jan 21 at 1:06
2
$begingroup$
$mathbb{C}$ is contained in $mathbb{C}(x)$, the field of rational functions on $mathbb{C}$.
$endgroup$
– Eclipse Sun
Jan 21 at 1:13
1
1
$begingroup$
Are you assuming $E$ is an algebraic extension of $F$?
$endgroup$
– Eclipse Sun
Jan 21 at 0:51
$begingroup$
Are you assuming $E$ is an algebraic extension of $F$?
$endgroup$
– Eclipse Sun
Jan 21 at 0:51
$begingroup$
@EclipseSun I think no, it is the assumption of the question to make me think to that direction of proving. Is it $bar F_E$ cannot properly contained in $E$ a wrong idea?
$endgroup$
– kelvin hong 方
Jan 21 at 0:58
$begingroup$
@EclipseSun I think no, it is the assumption of the question to make me think to that direction of proving. Is it $bar F_E$ cannot properly contained in $E$ a wrong idea?
$endgroup$
– kelvin hong 方
Jan 21 at 0:58
$begingroup$
An algebraically closed field, for example $mathbb{C}$ can be properly contained in another algebraically closed field, if you allow that extension to be transcendental.
$endgroup$
– Eclipse Sun
Jan 21 at 1:03
$begingroup$
An algebraically closed field, for example $mathbb{C}$ can be properly contained in another algebraically closed field, if you allow that extension to be transcendental.
$endgroup$
– Eclipse Sun
Jan 21 at 1:03
$begingroup$
But isn't that all transcendental numbers are already in $mathbb R$ hence in $mathbb C$? I never heard before $mathbb C$ can be properly contained in other field. Wow!
$endgroup$
– kelvin hong 方
Jan 21 at 1:06
$begingroup$
But isn't that all transcendental numbers are already in $mathbb R$ hence in $mathbb C$? I never heard before $mathbb C$ can be properly contained in other field. Wow!
$endgroup$
– kelvin hong 方
Jan 21 at 1:06
2
2
$begingroup$
$mathbb{C}$ is contained in $mathbb{C}(x)$, the field of rational functions on $mathbb{C}$.
$endgroup$
– Eclipse Sun
Jan 21 at 1:13
$begingroup$
$mathbb{C}$ is contained in $mathbb{C}(x)$, the field of rational functions on $mathbb{C}$.
$endgroup$
– Eclipse Sun
Jan 21 at 1:13
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Let $F$ be a field. An extension $K/F$ is called an algebraic
closure of $F$ if $K$ is algebraically closed and $K/F$ is an
algebraic extension.
As @EclipseSun mentioned in the comment section, an algebraically closed field can be properly contained in another algebraically closed field.
Define $$overline{F}_E={ain Emid a ;text{is algebraic over $F $}}.$$
Clearly $overline{F}_E$ is algebraic over $F$.
Let $alpha$ be a root of $f(x)in overline{F}_E[x]$. Then $overline{F}_E(alpha) $ is algebraic over $overline{F}_E$ and $overline{F}_E$ is algebraic over $F$. Hence $overline{F}_E(alpha)$ is algebraic over $F$(why?). So $alphain overline{F}_E$, since $alpha $ is algebraic over $F$. Hence $overline{F}_E$ is algebraically closed.
We need to show that $overline{F}_E$ is actually a field. In other words, if $alpha$ and $beta$ are algebraic over $F $, then $alpha+beta,alpha-beta, alphacdotbeta,fracalphabeta(betaneq0)$ are algebraic over $F$. This is easy to prove if you recall that $F(alpha,beta)=F(alpha)(beta)$ and $alpha$ is algebraic over $F$ iff $F(alpha)/F $ is finite. I hope you can take this from here.
answered Jan 21 at 2:45
Thomas ShelbyThomas Shelby
3,4891525
$endgroup$
1
$begingroup$
Thank you, the only thing I missed is the part $bar F_E(alpha)$ is algebraic over $F$ implies $alphainbar F_E$. After filling up this part the solution is completed.
$endgroup$
– kelvin hong 方
Jan 21 at 7:00
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Let $F$ be a field. An extension $K/F$ is called an algebraic
closure of $F$ if $K$ is algebraically closed and $K/F$ is an
algebraic extension.
As @EclipseSun mentioned in the comment section, an algebraically closed field can be properly contained in another algebraically closed field.
Define $$overline{F}_E={ain Emid a ;text{is algebraic over $F $}}.$$
Clearly $overline{F}_E$ is algebraic over $F$.
Let $alpha$ be a root of $f(x)in overline{F}_E[x]$. Then $overline{F}_E(alpha) $ is algebraic over $overline{F}_E$ and $overline{F}_E$ is algebraic over $F$. Hence $overline{F}_E(alpha)$ is algebraic over $F$(why?). So $alphain overline{F}_E$, since $alpha $ is algebraic over $F$. Hence $overline{F}_E$ is algebraically closed.
We need to show that $overline{F}_E$ is actually a field. In other words, if $alpha$ and $beta$ are algebraic over $F $, then $alpha+beta,alpha-beta, alphacdotbeta,fracalphabeta(betaneq0)$ are algebraic over $F$. This is easy to prove if you recall that $F(alpha,beta)=F(alpha)(beta)$ and $alpha$ is algebraic over $F$ iff $F(alpha)/F $ is finite. I hope you can take this from here.
answered Jan 21 at 2:45
Thomas ShelbyThomas Shelby
3,4891525
$endgroup$
1
$begingroup$
Thank you, the only thing I missed is the part $bar F_E(alpha)$ is algebraic over $F$ implies $alphainbar F_E$. After filling up this part the solution is completed.
$endgroup$
– kelvin hong 方
Jan 21 at 7:00
add a comment |
$begingroup$
Let $F$ be a field. An extension $K/F$ is called an algebraic
closure of $F$ if $K$ is algebraically closed and $K/F$ is an
algebraic extension.
As @EclipseSun mentioned in the comment section, an algebraically closed field can be properly contained in another algebraically closed field.
Define $$overline{F}_E={ain Emid a ;text{is algebraic over $F $}}.$$
Clearly $overline{F}_E$ is algebraic over $F$.
Let $alpha$ be a root of $f(x)in overline{F}_E[x]$. Then $overline{F}_E(alpha) $ is algebraic over $overline{F}_E$ and $overline{F}_E$ is algebraic over $F$. Hence $overline{F}_E(alpha)$ is algebraic over $F$(why?). So $alphain overline{F}_E$, since $alpha $ is algebraic over $F$. Hence $overline{F}_E$ is algebraically closed.
We need to show that $overline{F}_E$ is actually a field. In other words, if $alpha$ and $beta$ are algebraic over $F $, then $alpha+beta,alpha-beta, alphacdotbeta,fracalphabeta(betaneq0)$ are algebraic over $F$. This is easy to prove if you recall that $F(alpha,beta)=F(alpha)(beta)$ and $alpha$ is algebraic over $F$ iff $F(alpha)/F $ is finite. I hope you can take this from here.
answered Jan 21 at 2:45
Thomas ShelbyThomas Shelby
3,4891525
$endgroup$
1
$begingroup$
Thank you, the only thing I missed is the part $bar F_E(alpha)$ is algebraic over $F$ implies $alphainbar F_E$. After filling up this part the solution is completed.
$endgroup$
– kelvin hong 方
Jan 21 at 7:00
add a comment |
$begingroup$
Let $F$ be a field. An extension $K/F$ is called an algebraic
closure of $F$ if $K$ is algebraically closed and $K/F$ is an
algebraic extension.
As @EclipseSun mentioned in the comment section, an algebraically closed field can be properly contained in another algebraically closed field.
Define $$overline{F}_E={ain Emid a ;text{is algebraic over $F $}}.$$
Clearly $overline{F}_E$ is algebraic over $F$.
Let $alpha$ be a root of $f(x)in overline{F}_E[x]$. Then $overline{F}_E(alpha) $ is algebraic over $overline{F}_E$ and $overline{F}_E$ is algebraic over $F$. Hence $overline{F}_E(alpha)$ is algebraic over $F$(why?). So $alphain overline{F}_E$, since $alpha $ is algebraic over $F$. Hence $overline{F}_E$ is algebraically closed.
We need to show that $overline{F}_E$ is actually a field. In other words, if $alpha$ and $beta$ are algebraic over $F $, then $alpha+beta,alpha-beta, alphacdotbeta,fracalphabeta(betaneq0)$ are algebraic over $F$. This is easy to prove if you recall that $F(alpha,beta)=F(alpha)(beta)$ and $alpha$ is algebraic over $F$ iff $F(alpha)/F $ is finite. I hope you can take this from here.
answered Jan 21 at 2:45
Thomas ShelbyThomas Shelby
3,4891525
$endgroup$
Let $F$ be a field. An extension $K/F$ is called an algebraic
closure of $F$ if $K$ is algebraically closed and $K/F$ is an
algebraic extension.
As @EclipseSun mentioned in the comment section, an algebraically closed field can be properly contained in another algebraically closed field.
Define $$overline{F}_E={ain Emid a ;text{is algebraic over $F $}}.$$
Clearly $overline{F}_E$ is algebraic over $F$.
Let $alpha$ be a root of $f(x)in overline{F}_E[x]$. Then $overline{F}_E(alpha) $ is algebraic over $overline{F}_E$ and $overline{F}_E$ is algebraic over $F$. Hence $overline{F}_E(alpha)$ is algebraic over $F$(why?). So $alphain overline{F}_E$, since $alpha $ is algebraic over $F$. Hence $overline{F}_E$ is algebraically closed.
We need to show that $overline{F}_E$ is actually a field. In other words, if $alpha$ and $beta$ are algebraic over $F $, then $alpha+beta,alpha-beta, alphacdotbeta,fracalphabeta(betaneq0)$ are algebraic over $F$. This is easy to prove if you recall that $F(alpha,beta)=F(alpha)(beta)$ and $alpha$ is algebraic over $F$ iff $F(alpha)/F $ is finite. I hope you can take this from here.
answered Jan 21 at 2:45
Thomas ShelbyThomas Shelby
3,4891525
answered Jan 21 at 2:45
Thomas ShelbyThomas Shelby
3,4891525
answered Jan 21 at 2:45
Thomas ShelbyThomas Shelby
3,4891525
answered Jan 21 at 2:45
Thomas ShelbyThomas Shelby
3,4891525
3,4891525
1
$begingroup$
Thank you, the only thing I missed is the part $bar F_E(alpha)$ is algebraic over $F$ implies $alphainbar F_E$. After filling up this part the solution is completed.
$endgroup$
– kelvin hong 方
Jan 21 at 7:00
add a comment |
1
$begingroup$
Thank you, the only thing I missed is the part $bar F_E(alpha)$ is algebraic over $F$ implies $alphainbar F_E$. After filling up this part the solution is completed.
$endgroup$
– kelvin hong 方
Jan 21 at 7:00
1
1
$begingroup$
Thank you, the only thing I missed is the part $bar F_E(alpha)$ is algebraic over $F$ implies $alphainbar F_E$. After filling up this part the solution is completed.
$endgroup$
– kelvin hong 方
Jan 21 at 7:00
$begingroup$
Thank you, the only thing I missed is the part $bar F_E(alpha)$ is algebraic over $F$ implies $alphainbar F_E$. After filling up this part the solution is completed.
$endgroup$
– kelvin hong 方
Jan 21 at 7:00
add a comment |
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$begingroup$
Are you assuming $E$ is an algebraic extension of $F$?
$endgroup$
– Eclipse Sun
Jan 21 at 0:51
$begingroup$
@EclipseSun I think no, it is the assumption of the question to make me think to that direction of proving. Is it $bar F_E$ cannot properly contained in $E$ a wrong idea?
$endgroup$
– kelvin hong 方
Jan 21 at 0:58
$begingroup$
An algebraically closed field, for example $mathbb{C}$ can be properly contained in another algebraically closed field, if you allow that extension to be transcendental.
$endgroup$
– Eclipse Sun
Jan 21 at 1:03
$begingroup$
But isn't that all transcendental numbers are already in $mathbb R$ hence in $mathbb C$? I never heard before $mathbb C$ can be properly contained in other field. Wow!
$endgroup$
– kelvin hong 方
Jan 21 at 1:06
2
$begingroup$
$mathbb{C}$ is contained in $mathbb{C}(x)$, the field of rational functions on $mathbb{C}$.
$endgroup$
– Eclipse Sun
Jan 21 at 1:13