$K subset L$ a field extension and $L$ algebraically closed implies that $A$ is the algebraic closure of $K$...

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  Given a field, the field of algebraic elements over it in an algebraically closed field extension is again algebraically closed? [duplicate]
  
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Let $K subset L$ be a field extension and $A:={l in L,:,atext{ algebraic over }K}$. I already have shown that $A$ again is a field and $K subset A$ is algebraic. Also: If $l in L$ is algebraic over $A$ it's also algebraic over $K$.

I now have to prove the following statement:

Suppose $L$ is algebraically closed. Show that $A$ is algebraically closed and that $K subset A$ is algebraic.

As mentioned the latter I already have proven. Any hints how to proceed? I thought about considering a polynomial $a(x) in A[x] setminus A$ which decomposes into linear factors of $L[x]$ since $L$ is closed. From there on it would've been quite nice to show that these linear factors indeed are contained in $A[x]$ but I unfortunately do not know how to proceed.

Thanks for checking in! :)

field-theory extension-field

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edited Jan 24 at 5:54

Jyrki Lahtonen

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asked Jan 23 at 21:22

C. BrendelC. Brendel

538

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marked as duplicate by Dietrich Burde, Paul Frost, Cesareo, darij grinberg, clathratus Jan 24 at 3:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Given a field, the field of algebraic elements over it in an algebraically closed field extension is again algebraically closed? [duplicate]
  
  2 answers
  
  

Let $K subset L$ be a field extension and $A:={l in L,:,atext{ algebraic over }K}$. I already have shown that $A$ again is a field and $K subset A$ is algebraic. Also: If $l in L$ is algebraic over $A$ it's also algebraic over $K$.

I now have to prove the following statement:

Suppose $L$ is algebraically closed. Show that $A$ is algebraically closed and that $K subset A$ is algebraic.

As mentioned the latter I already have proven. Any hints how to proceed? I thought about considering a polynomial $a(x) in A[x] setminus A$ which decomposes into linear factors of $L[x]$ since $L$ is closed. From there on it would've been quite nice to show that these linear factors indeed are contained in $A[x]$ but I unfortunately do not know how to proceed.

Thanks for checking in! :)

field-theory extension-field

share|cite|improve this question

edited Jan 24 at 5:54

Jyrki Lahtonen

110k13171385

asked Jan 23 at 21:22

C. BrendelC. Brendel

538

$endgroup$

marked as duplicate by Dietrich Burde, Paul Frost, Cesareo, darij grinberg, clathratus Jan 24 at 3:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

0

0

0

$begingroup$

This question already has an answer here:

• 
  Given a field, the field of algebraic elements over it in an algebraically closed field extension is again algebraically closed? [duplicate]
  
  2 answers
  
  

Let $K subset L$ be a field extension and $A:={l in L,:,atext{ algebraic over }K}$. I already have shown that $A$ again is a field and $K subset A$ is algebraic. Also: If $l in L$ is algebraic over $A$ it's also algebraic over $K$.

I now have to prove the following statement:

Suppose $L$ is algebraically closed. Show that $A$ is algebraically closed and that $K subset A$ is algebraic.

As mentioned the latter I already have proven. Any hints how to proceed? I thought about considering a polynomial $a(x) in A[x] setminus A$ which decomposes into linear factors of $L[x]$ since $L$ is closed. From there on it would've been quite nice to show that these linear factors indeed are contained in $A[x]$ but I unfortunately do not know how to proceed.

Thanks for checking in! :)

field-theory extension-field

share|cite|improve this question

edited Jan 24 at 5:54

Jyrki Lahtonen

110k13171385

asked Jan 23 at 21:22

C. BrendelC. Brendel

538

$endgroup$

This question already has an answer here:

• 
  Given a field, the field of algebraic elements over it in an algebraically closed field extension is again algebraically closed? [duplicate]
  
  2 answers
  
  

Let $K subset L$ be a field extension and $A:={l in L,:,atext{ algebraic over }K}$. I already have shown that $A$ again is a field and $K subset A$ is algebraic. Also: If $l in L$ is algebraic over $A$ it's also algebraic over $K$.

I now have to prove the following statement:

Suppose $L$ is algebraically closed. Show that $A$ is algebraically closed and that $K subset A$ is algebraic.

As mentioned the latter I already have proven. Any hints how to proceed? I thought about considering a polynomial $a(x) in A[x] setminus A$ which decomposes into linear factors of $L[x]$ since $L$ is closed. From there on it would've been quite nice to show that these linear factors indeed are contained in $A[x]$ but I unfortunately do not know how to proceed.

Thanks for checking in! :)

This question already has an answer here:

• 
  Given a field, the field of algebraic elements over it in an algebraically closed field extension is again algebraically closed? [duplicate]
  
  2 answers
  
  

field-theory extension-field

field-theory extension-field

share|cite|improve this question

edited Jan 24 at 5:54

Jyrki Lahtonen

110k13171385

asked Jan 23 at 21:22

C. BrendelC. Brendel

538

share|cite|improve this question

edited Jan 24 at 5:54

Jyrki Lahtonen

110k13171385

asked Jan 23 at 21:22

C. BrendelC. Brendel

538

share|cite|improve this question

share|cite|improve this question

edited Jan 24 at 5:54

Jyrki Lahtonen

110k13171385

edited Jan 24 at 5:54

Jyrki Lahtonen

110k13171385

edited Jan 24 at 5:54

Jyrki Lahtonen

110k13171385

110k13171385

asked Jan 23 at 21:22

C. BrendelC. Brendel

538

asked Jan 23 at 21:22

C. BrendelC. Brendel

538

asked Jan 23 at 21:22

C. BrendelC. Brendel

538

538

marked as duplicate by Dietrich Burde, Paul Frost, Cesareo, darij grinberg, clathratus Jan 24 at 3:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Dietrich Burde, Paul Frost, Cesareo, darij grinberg, clathratus Jan 24 at 3:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

add a comment | 

2 Answers
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Suppose $L$ is algebraically closed. Then if $f(x) in A[x]$ is a monic polynomial, we may view it as a polynomial $f(x) in L[x]$. Since $L$ is algebraically closed, this has a root $alpha in L$. We want to show that actually $alpha in A subset L$. But since $alpha$ is the root of a polynomial with coefficients in $A$, we immediately have that $alpha$ is algebraic over $A$, and you said you've already shown that this implies that $alpha$ is algebraic over $K$. But then by definition of $A$, we have that $alpha in A$. Thus $A$ is algebraically closed.

The reason that $K subset A$ is algebraic is simply by definition: any element $alpha in A$ is by definition algebraic over $K$.

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answered Jan 23 at 21:37

Ashwin IyengarAshwin Iyengar

1,181615

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah yes! That's very nice and elegantly simple! Thanks a lot! :)
  $endgroup$
  – C. Brendel
  Jan 23 at 22:17
  

  
  
  
  

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1

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Let $f(x)=a_0+a_1x+dots+a_nx^nin A[x]$ have positive degree. Then $f(x)$ has a root $b$ in $L$.

Now note that $b$ has finite degree over $K[a_0,a_1,dots,a_n]$, which in turn has finite degree over $K$. Hence $K[b]$ has finite degree over $K$ and so $bin A$.

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answered Jan 23 at 21:59

egregegreg

184k1486205

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah! That's also a very nice way of solving the problem! :) Thanks!
  $endgroup$
  – C. Brendel
  Jan 23 at 22:19
  

  
  
  
  

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Suppose $L$ is algebraically closed. Then if $f(x) in A[x]$ is a monic polynomial, we may view it as a polynomial $f(x) in L[x]$. Since $L$ is algebraically closed, this has a root $alpha in L$. We want to show that actually $alpha in A subset L$. But since $alpha$ is the root of a polynomial with coefficients in $A$, we immediately have that $alpha$ is algebraic over $A$, and you said you've already shown that this implies that $alpha$ is algebraic over $K$. But then by definition of $A$, we have that $alpha in A$. Thus $A$ is algebraically closed.

The reason that $K subset A$ is algebraic is simply by definition: any element $alpha in A$ is by definition algebraic over $K$.

share|cite|improve this answer

answered Jan 23 at 21:37

Ashwin IyengarAshwin Iyengar

1,181615

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah yes! That's very nice and elegantly simple! Thanks a lot! :)
  $endgroup$
  – C. Brendel
  Jan 23 at 22:17
  

  
  
  
  

add a comment | 

1

$begingroup$

Suppose $L$ is algebraically closed. Then if $f(x) in A[x]$ is a monic polynomial, we may view it as a polynomial $f(x) in L[x]$. Since $L$ is algebraically closed, this has a root $alpha in L$. We want to show that actually $alpha in A subset L$. But since $alpha$ is the root of a polynomial with coefficients in $A$, we immediately have that $alpha$ is algebraic over $A$, and you said you've already shown that this implies that $alpha$ is algebraic over $K$. But then by definition of $A$, we have that $alpha in A$. Thus $A$ is algebraically closed.

The reason that $K subset A$ is algebraic is simply by definition: any element $alpha in A$ is by definition algebraic over $K$.

share|cite|improve this answer

answered Jan 23 at 21:37

Ashwin IyengarAshwin Iyengar

1,181615

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah yes! That's very nice and elegantly simple! Thanks a lot! :)
  $endgroup$
  – C. Brendel
  Jan 23 at 22:17
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

Suppose $L$ is algebraically closed. Then if $f(x) in A[x]$ is a monic polynomial, we may view it as a polynomial $f(x) in L[x]$. Since $L$ is algebraically closed, this has a root $alpha in L$. We want to show that actually $alpha in A subset L$. But since $alpha$ is the root of a polynomial with coefficients in $A$, we immediately have that $alpha$ is algebraic over $A$, and you said you've already shown that this implies that $alpha$ is algebraic over $K$. But then by definition of $A$, we have that $alpha in A$. Thus $A$ is algebraically closed.

The reason that $K subset A$ is algebraic is simply by definition: any element $alpha in A$ is by definition algebraic over $K$.

share|cite|improve this answer

answered Jan 23 at 21:37

Ashwin IyengarAshwin Iyengar

1,181615

$endgroup$

Suppose $L$ is algebraically closed. Then if $f(x) in A[x]$ is a monic polynomial, we may view it as a polynomial $f(x) in L[x]$. Since $L$ is algebraically closed, this has a root $alpha in L$. We want to show that actually $alpha in A subset L$. But since $alpha$ is the root of a polynomial with coefficients in $A$, we immediately have that $alpha$ is algebraic over $A$, and you said you've already shown that this implies that $alpha$ is algebraic over $K$. But then by definition of $A$, we have that $alpha in A$. Thus $A$ is algebraically closed.

The reason that $K subset A$ is algebraic is simply by definition: any element $alpha in A$ is by definition algebraic over $K$.

share|cite|improve this answer

answered Jan 23 at 21:37

Ashwin IyengarAshwin Iyengar

1,181615

share|cite|improve this answer

share|cite|improve this answer

answered Jan 23 at 21:37

Ashwin IyengarAshwin Iyengar

1,181615

answered Jan 23 at 21:37

Ashwin IyengarAshwin Iyengar

1,181615

answered Jan 23 at 21:37

Ashwin IyengarAshwin Iyengar

1,181615

1,181615

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah yes! That's very nice and elegantly simple! Thanks a lot! :)
  $endgroup$
  – C. Brendel
  Jan 23 at 22:17
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah yes! That's very nice and elegantly simple! Thanks a lot! :)
  $endgroup$
  – C. Brendel
  Jan 23 at 22:17
  

  
  
  
  

$begingroup$
Ah yes! That's very nice and elegantly simple! Thanks a lot! :)
$endgroup$
– C. Brendel
Jan 23 at 22:17

$begingroup$
Ah yes! That's very nice and elegantly simple! Thanks a lot! :)
$endgroup$
– C. Brendel
Jan 23 at 22:17

add a comment | 

1

$begingroup$

Let $f(x)=a_0+a_1x+dots+a_nx^nin A[x]$ have positive degree. Then $f(x)$ has a root $b$ in $L$.

Now note that $b$ has finite degree over $K[a_0,a_1,dots,a_n]$, which in turn has finite degree over $K$. Hence $K[b]$ has finite degree over $K$ and so $bin A$.

share|cite|improve this answer

answered Jan 23 at 21:59

egregegreg

184k1486205

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah! That's also a very nice way of solving the problem! :) Thanks!
  $endgroup$
  – C. Brendel
  Jan 23 at 22:19
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $f(x)=a_0+a_1x+dots+a_nx^nin A[x]$ have positive degree. Then $f(x)$ has a root $b$ in $L$.

Now note that $b$ has finite degree over $K[a_0,a_1,dots,a_n]$, which in turn has finite degree over $K$. Hence $K[b]$ has finite degree over $K$ and so $bin A$.

share|cite|improve this answer

answered Jan 23 at 21:59

egregegreg

184k1486205

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah! That's also a very nice way of solving the problem! :) Thanks!
  $endgroup$
  – C. Brendel
  Jan 23 at 22:19
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

Let $f(x)=a_0+a_1x+dots+a_nx^nin A[x]$ have positive degree. Then $f(x)$ has a root $b$ in $L$.

Now note that $b$ has finite degree over $K[a_0,a_1,dots,a_n]$, which in turn has finite degree over $K$. Hence $K[b]$ has finite degree over $K$ and so $bin A$.

share|cite|improve this answer

answered Jan 23 at 21:59

egregegreg

184k1486205

$endgroup$

Let $f(x)=a_0+a_1x+dots+a_nx^nin A[x]$ have positive degree. Then $f(x)$ has a root $b$ in $L$.

Now note that $b$ has finite degree over $K[a_0,a_1,dots,a_n]$, which in turn has finite degree over $K$. Hence $K[b]$ has finite degree over $K$ and so $bin A$.

share|cite|improve this answer

answered Jan 23 at 21:59

egregegreg

184k1486205

share|cite|improve this answer

share|cite|improve this answer

answered Jan 23 at 21:59

egregegreg

184k1486205

answered Jan 23 at 21:59

egregegreg

184k1486205

answered Jan 23 at 21:59

egregegreg

184k1486205

184k1486205

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah! That's also a very nice way of solving the problem! :) Thanks!
  $endgroup$
  – C. Brendel
  Jan 23 at 22:19
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah! That's also a very nice way of solving the problem! :) Thanks!
  $endgroup$
  – C. Brendel
  Jan 23 at 22:19
  

  
  
  
  

$begingroup$
Ah! That's also a very nice way of solving the problem! :) Thanks!
$endgroup$
– C. Brendel
Jan 23 at 22:19

$begingroup$
Ah! That's also a very nice way of solving the problem! :) Thanks!
$endgroup$
– C. Brendel
Jan 23 at 22:19

add a comment | 

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