Fdtxjr

Why is the US ranked as #45 in Press Freedom ratings, despite its extremely permissive free speech laws?

Do I need to write [sic] when a number is less than 10 but isn't written out?

Why do remote US companies require working in the US?

How to get from Geneva Airport to Metabief, Doubs, France by public transport?

If Nick Fury and Coulson already knew about aliens (Kree and Skrull) why did they wait until Thor's appearance to start making weapons?

How is this set of matrices closed under multiplication?

Where does this common spurious transmission come from? Is there a quality difference?

TikZ: How to reverse arrow direction without switching start/end point?

Why don't programming languages automatically manage the synchronous/asynchronous problem?

Would a completely good Muggle be able to use a wand?

Writing differences on a blackboard

How did people program for Consoles with multiple CPUs?

What steps are necessary to read a Modern SSD in Medieval Europe?

Legal workarounds for testamentary trust perceived as unfair

Is wanting to ask what to write an indication that you need to change your story?

What does "Its cash flow is deeply negative" mean?

Why doesn't UK go for the same deal Japan has with EU to resolve Brexit?

Is French Guiana a (hard) EU border?

Is it convenient to ask the journal's editor for two additional days to complete a review?

What happened in Rome, when the western empire "fell"?

Bartok - Syncopation (1): Meaning of notes in between Grand Staff

Which one is the true statement?

Why, when going from special to general relativity, do we just replace partial derivatives with covariant derivatives?

How to prove a simple equation?

Maximum number of algebraically independent elements of a transcendental extension

2

$begingroup$

As part of a course in Algebraic Geometry, I am trying to prove a corollary of a given lemma:

Lemma: If $K$ is a field and $alpha_1,dots, alpha_m$ are algebraically independent in $K$ and $beta in K$ is algebraically dependent of $alpha_1, dots alpha_m$ but not on $alpha_1, dots alpha_m-1$, then $alpha_m$ is algebraically dependent on $alpha_1,dots, alpha_m-1, beta$.

Corollary: If $K$ is a field and $alpha_1, dots,alpha_n$ are algebraically independent, and $beta_1, dots, beta_n$ are algebraically independent (both in $K$) and $beta_i in K(alpha_1,dots, alpha_n)$ then $m leq n$.

My notes say "Use the (Exchange) lemma to replace the $alpha_i$ by $beta_i$ one at a time".

I've tried applying the lemma to say:

Re-label if necessary so that $beta_1$ is algebraically dependent on $alpha_1,dots,alpha_r$ but not on $alpha_1,dots, alpha_r-1$ and replace $alpha_r$ with $beta_1$. Relabel again so that we now have $alpha_1,dots,alpha_n-1,beta_1$, algebraically independent since in any polynomial containing $beta_1$ and the $alpha_i$ we can sub in the expression for $beta_1$ as an element of $K(alpha_1,dots, alpha_n)$ to get an algebraic dependence on the $alpha_i$. By the lemma the removed $alpha_n$ is algebraically dependent on $alpha_1,dots,alpha_n-1,beta_1$ I'm not sure where to go from here or even if this is the right line of approach, any help would be much appreciated.

number-theory algebraic-geometry extension-field transcendental-numbers transcendence-theory

share|cite|improve this question

edited Mar 19 at 11:16

Brout

2,6061431

asked Feb 10 '14 at 14:52

Tom OldfieldTom Oldfield

9,55312058

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $F = K(alpha_1,ldots,alpha_m-1)$ and $alpha_m$ transcendental over $F$. Then the assumption is that for some non-constant polynomials $P,Q in F[t]$, $P(beta) = Q(alpha_m)$ which implies $F(alpha_m)/ F(P(beta))$ is algebraic so $F(alpha_m)/F(beta)$ is algebraic.
  $endgroup$
  – reuns
  Mar 19 at 12:13
  
  

  
  
  
  

add a comment | 

2

$begingroup$

As part of a course in Algebraic Geometry, I am trying to prove a corollary of a given lemma:

Lemma: If $K$ is a field and $alpha_1,dots, alpha_m$ are algebraically independent in $K$ and $beta in K$ is algebraically dependent of $alpha_1, dots alpha_m$ but not on $alpha_1, dots alpha_m-1$, then $alpha_m$ is algebraically dependent on $alpha_1,dots, alpha_m-1, beta$.

Corollary: If $K$ is a field and $alpha_1, dots,alpha_n$ are algebraically independent, and $beta_1, dots, beta_n$ are algebraically independent (both in $K$) and $beta_i in K(alpha_1,dots, alpha_n)$ then $m leq n$.

My notes say "Use the (Exchange) lemma to replace the $alpha_i$ by $beta_i$ one at a time".

I've tried applying the lemma to say:

Re-label if necessary so that $beta_1$ is algebraically dependent on $alpha_1,dots,alpha_r$ but not on $alpha_1,dots, alpha_r-1$ and replace $alpha_r$ with $beta_1$. Relabel again so that we now have $alpha_1,dots,alpha_n-1,beta_1$, algebraically independent since in any polynomial containing $beta_1$ and the $alpha_i$ we can sub in the expression for $beta_1$ as an element of $K(alpha_1,dots, alpha_n)$ to get an algebraic dependence on the $alpha_i$. By the lemma the removed $alpha_n$ is algebraically dependent on $alpha_1,dots,alpha_n-1,beta_1$ I'm not sure where to go from here or even if this is the right line of approach, any help would be much appreciated.

number-theory algebraic-geometry extension-field transcendental-numbers transcendence-theory

share|cite|improve this question

edited Mar 19 at 11:16

Brout

2,6061431

asked Feb 10 '14 at 14:52

Tom OldfieldTom Oldfield

9,55312058

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $F = K(alpha_1,ldots,alpha_m-1)$ and $alpha_m$ transcendental over $F$. Then the assumption is that for some non-constant polynomials $P,Q in F[t]$, $P(beta) = Q(alpha_m)$ which implies $F(alpha_m)/ F(P(beta))$ is algebraic so $F(alpha_m)/F(beta)$ is algebraic.
  $endgroup$
  – reuns
  Mar 19 at 12:13
  
  

  
  
  
  

add a comment | 

$begingroup$

As part of a course in Algebraic Geometry, I am trying to prove a corollary of a given lemma:

Lemma: If $K$ is a field and $alpha_1,dots, alpha_m$ are algebraically independent in $K$ and $beta in K$ is algebraically dependent of $alpha_1, dots alpha_m$ but not on $alpha_1, dots alpha_m-1$, then $alpha_m$ is algebraically dependent on $alpha_1,dots, alpha_m-1, beta$.

Corollary: If $K$ is a field and $alpha_1, dots,alpha_n$ are algebraically independent, and $beta_1, dots, beta_n$ are algebraically independent (both in $K$) and $beta_i in K(alpha_1,dots, alpha_n)$ then $m leq n$.

My notes say "Use the (Exchange) lemma to replace the $alpha_i$ by $beta_i$ one at a time".

I've tried applying the lemma to say:

Re-label if necessary so that $beta_1$ is algebraically dependent on $alpha_1,dots,alpha_r$ but not on $alpha_1,dots, alpha_r-1$ and replace $alpha_r$ with $beta_1$. Relabel again so that we now have $alpha_1,dots,alpha_n-1,beta_1$, algebraically independent since in any polynomial containing $beta_1$ and the $alpha_i$ we can sub in the expression for $beta_1$ as an element of $K(alpha_1,dots, alpha_n)$ to get an algebraic dependence on the $alpha_i$. By the lemma the removed $alpha_n$ is algebraically dependent on $alpha_1,dots,alpha_n-1,beta_1$ I'm not sure where to go from here or even if this is the right line of approach, any help would be much appreciated.

number-theory algebraic-geometry extension-field transcendental-numbers transcendence-theory

share|cite|improve this question

edited Mar 19 at 11:16

Brout

2,6061431

asked Feb 10 '14 at 14:52

Tom OldfieldTom Oldfield

9,55312058

$endgroup$

share|cite|improve this question

edited Mar 19 at 11:16

Brout

2,6061431

asked Feb 10 '14 at 14:52

Tom OldfieldTom Oldfield

9,55312058

share|cite|improve this question

edited Mar 19 at 11:16

Brout

2,6061431

asked Feb 10 '14 at 14:52

Tom OldfieldTom Oldfield

9,55312058

edited Mar 19 at 11:16

Brout

2,6061431

edited Mar 19 at 11:16

Brout

2,6061431

asked Feb 10 '14 at 14:52

Tom OldfieldTom Oldfield

9,55312058

asked Feb 10 '14 at 14:52

Tom OldfieldTom Oldfield

9,55312058