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If $ b in K $ is algebraic of degree n over $F$ then $ [F (b):F ]=n $. Is the converse true?

Infinite Degree Algebraic Field ExtensionsA confusion regarding the nature of elements in a field extension.Separability of polynomialsProving that the algebraic reals form an infinite field extension over $mathbb Q$If $a,bin K$ are algebraic over $k$, then $apm b, ab$ and $ab^-1$ are also algebraic.The degree of an algebraic element over a field extensionA field extension with same transcendental degree is algebraicIs there a correlation between the degree of an extension and it being an algebraic extension?if $L/F$ a field extension, if b is algebraic over $F(a)$ and $b$ not algebraic over $F$, then $a$ is algebraic over $F(b)$. Example?Indeterminate is Algebraic over Transcendental Extension

0

$begingroup$

Let $F$ be a field and $K$ be an extension of $F$ if $ bin K $ is algebraic of degree $n$ over $F$ then $ [F (b):F]=n $ .

I want to know if I am given that $ [F (b):F] =n $ then can I tell that b is algebraic of degree n over F?

If not can someone give me an example please?

Thank you in advance.

Edit : I have started studying "extension fields" and I have so many doubts , the above question was one of my doubts .

I have this doubt because there is a theorem in my textbook that " The element $ ain K $ is algebraic over F if and only if F (a) is a finite extension over F"

and the next theorem was " If $ ain K $ is algebraic of degree n over F, then [F (a):F]=n"

I didn't understand why Mr. Herstein didn't use " if and only if " term in the second case .

In my opinion the "if" part is also true but I thought I should ask the experts.

field-theory

share|cite|improve this question

edited 2 days ago

suchanda adhikari

asked 2 days ago

suchanda adhikarisuchanda adhikari

807

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I understand what you said ? but I did not understand what is the relation with my question ? well I just started to study field theory so please elaborate a little more?
  $endgroup$
  – suchanda adhikari
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK I understood what you said in your statement but why the converse is not true ? I think I should edit my question.Why my argument is wrong?
  $endgroup$
  – suchanda adhikari
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't understand yet why the converse is not true? can you give me an example.? Sorry for disturbing
  $endgroup$
  – suchanda adhikari
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In your example $ [mathbb Q(sqrt2):mathbb Q] $ = 2 and $ sqrt 2 $ is algebraic of degree 2 over $ mathbb Q $ so how this example contradicts the converse?
  $endgroup$
  – suchanda adhikari
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes but I didn't ask for that. ...and you stated in your comment "the converse is not exactly true ". so I am asking for a counter example.
  $endgroup$
  – suchanda adhikari
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $F$ be a field and $K$ be an extension of $F$ if $ bin K $ is algebraic of degree $n$ over $F$ then $ [F (b):F]=n $ .

I want to know if I am given that $ [F (b):F] =n $ then can I tell that b is algebraic of degree n over F?

If not can someone give me an example please?

Thank you in advance.

Edit : I have started studying "extension fields" and I have so many doubts , the above question was one of my doubts .

I have this doubt because there is a theorem in my textbook that " The element $ ain K $ is algebraic over F if and only if F (a) is a finite extension over F"

and the next theorem was " If $ ain K $ is algebraic of degree n over F, then [F (a):F]=n"

I didn't understand why Mr. Herstein didn't use " if and only if " term in the second case .

In my opinion the "if" part is also true but I thought I should ask the experts.

field-theory

share|cite|improve this question

edited 2 days ago

suchanda adhikari

asked 2 days ago

suchanda adhikarisuchanda adhikari

807

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I understand what you said ? but I did not understand what is the relation with my question ? well I just started to study field theory so please elaborate a little more?
  $endgroup$
  – suchanda adhikari
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK I understood what you said in your statement but why the converse is not true ? I think I should edit my question.Why my argument is wrong?
  $endgroup$
  – suchanda adhikari
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't understand yet why the converse is not true? can you give me an example.? Sorry for disturbing
  $endgroup$
  – suchanda adhikari
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In your example $ [mathbb Q(sqrt2):mathbb Q] $ = 2 and $ sqrt 2 $ is algebraic of degree 2 over $ mathbb Q $ so how this example contradicts the converse?
  $endgroup$
  – suchanda adhikari
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes but I didn't ask for that. ...and you stated in your comment "the converse is not exactly true ". so I am asking for a counter example.
  $endgroup$
  – suchanda adhikari
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Let $F$ be a field and $K$ be an extension of $F$ if $ bin K $ is algebraic of degree $n$ over $F$ then $ [F (b):F]=n $ .

I want to know if I am given that $ [F (b):F] =n $ then can I tell that b is algebraic of degree n over F?

If not can someone give me an example please?

Thank you in advance.

Edit : I have started studying "extension fields" and I have so many doubts , the above question was one of my doubts .

I have this doubt because there is a theorem in my textbook that " The element $ ain K $ is algebraic over F if and only if F (a) is a finite extension over F"

and the next theorem was " If $ ain K $ is algebraic of degree n over F, then [F (a):F]=n"

I didn't understand why Mr. Herstein didn't use " if and only if " term in the second case .

In my opinion the "if" part is also true but I thought I should ask the experts.

field-theory

share|cite|improve this question

edited 2 days ago

suchanda adhikari

asked 2 days ago

suchanda adhikarisuchanda adhikari

807

$endgroup$

share|cite|improve this question

edited 2 days ago

suchanda adhikari

asked 2 days ago

suchanda adhikarisuchanda adhikari

807

share|cite|improve this question

edited 2 days ago

suchanda adhikari

asked 2 days ago

suchanda adhikarisuchanda adhikari

807

asked 2 days ago

suchanda adhikarisuchanda adhikari

807

asked 2 days ago

suchanda adhikarisuchanda adhikari

807

1 Answer
1

active

oldest

votes

1

$begingroup$

Yes. In this case $b^n = p(b)$ where $p$ is a polynomial of degree $n-1$, so $b$ is a root of the degree $n$ polynomial $x^n - p(x)$. If $b$ were a root of a lower degree polynomial, then the elements $1,b,dotsc,b^n-1$ would not be linearly independent, so the extension would not be degree $n$.

It's been a while since I studied field theory, so I'm not sure whether the assertion that $1,b,dotsc,b^n-1$ form a basis needs more justification.

share|cite|improve this answer

answered 2 days ago

Rylee LymanRylee Lyman

1879

$endgroup$

add a comment | 

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1

$begingroup$

Yes. In this case $b^n = p(b)$ where $p$ is a polynomial of degree $n-1$, so $b$ is a root of the degree $n$ polynomial $x^n - p(x)$. If $b$ were a root of a lower degree polynomial, then the elements $1,b,dotsc,b^n-1$ would not be linearly independent, so the extension would not be degree $n$.

It's been a while since I studied field theory, so I'm not sure whether the assertion that $1,b,dotsc,b^n-1$ form a basis needs more justification.

share|cite|improve this answer

answered 2 days ago

Rylee LymanRylee Lyman

1879

$endgroup$

add a comment | 

1

$begingroup$

Yes. In this case $b^n = p(b)$ where $p$ is a polynomial of degree $n-1$, so $b$ is a root of the degree $n$ polynomial $x^n - p(x)$. If $b$ were a root of a lower degree polynomial, then the elements $1,b,dotsc,b^n-1$ would not be linearly independent, so the extension would not be degree $n$.

It's been a while since I studied field theory, so I'm not sure whether the assertion that $1,b,dotsc,b^n-1$ form a basis needs more justification.

share|cite|improve this answer

answered 2 days ago

Rylee LymanRylee Lyman

1879

$endgroup$

add a comment | 

$begingroup$

Yes. In this case $b^n = p(b)$ where $p$ is a polynomial of degree $n-1$, so $b$ is a root of the degree $n$ polynomial $x^n - p(x)$. If $b$ were a root of a lower degree polynomial, then the elements $1,b,dotsc,b^n-1$ would not be linearly independent, so the extension would not be degree $n$.

It's been a while since I studied field theory, so I'm not sure whether the assertion that $1,b,dotsc,b^n-1$ form a basis needs more justification.

share|cite|improve this answer

answered 2 days ago

Rylee LymanRylee Lyman

1879

$endgroup$

share|cite|improve this answer

answered 2 days ago

Rylee LymanRylee Lyman

1879

answered 2 days ago

Rylee LymanRylee Lyman

1879

answered 2 days ago

Rylee LymanRylee Lyman

1879