equivalent statements of Lindemann-Weierstrass theorem
Write for each . Suppose , where . Moving to the LHS and by multiplying a common denominator we can assume that where . We want to show that . We induct on . The case when is trivial because is never 0 and therefore forces .
By induction hypothesis, suppose the statement is true when . Now suppose . If are linearly independent over , are algebraically independent and certainly linearly independent over . So suppose are not linearly independent over . Without loss of generality, we can assume , where and . By multiplying a common denominator we can further assume that and . Then
Since , we get
If we define
then and By the induction hypothesis, . It is not hard to see that and therefore are linearly independent.
We first need two lemmas:
Lemma 1. Given 2., if is algebraic over , then is transcendental over .
Suppose where . Then we have
Since , are all distinct, are linearly independent by the hypothesis. Thus, and we have , which means that is transcendental over . ∎
Lemma 2. Given 2., if and are linearly independent and algebraic over , then is transcendental over .
Let and . Suppose where . We want to show that . Write
where each . Now, . So
where each . Now, the exponents in the above equation are all distinct, or else we would end up with and being linearly dependent, contrary to the assumption. Therefore, by 2 (Lindemann-Weierstrass Version 2), all are linearly independent, which means each . This implies that . But and , we must have . ∎
Now onto the main problem. We proceed by induction on the number of linearly independent algebraic elements over . The case when is covered in Lemma 1, since a linearly independent singleton is necessarily non-zero. So suppose are linearly independent and algebraic over . Then each pair are independent and algebraic over , . Thus is transcendental over for all . This means that is transcendental over
Now let for all . Suppose where . To show the algebraic independence of the , we need to show that . Rearranging terms of and we have
If we let , we see that and . Since is transcendental over , we must have . This implies that each . But then are algebraically independent by the induction hypothesis, we must have each . This means that . ∎
|Title||equivalent statements of Lindemann-Weierstrass theorem|
|Date of creation||2013-03-22 18:05:18|
|Last modified on||2013-03-22 18:05:18|
|Last modified by||CWoo (3771)|