equivalent statements of Lindemann-Weierstrass theorem

Proposition 1.

The following versions of the Lindemann-Weierstrass Theorem are equivalent:

1. If ${\alpha }_1,\mathrm{\dots },{\alpha }_n$ are linearly independent algebraic numbers over $\mathbb{Q}$, then $e^{{\alpha }_1},\mathrm{\dots },e^{{\alpha }_n}$ are algebraically independent over $\mathbb{Q}$.

2. If ${\alpha }_1,\mathrm{\dots },{\alpha }_n$ are distinct algebraic numbers over $\mathbb{Q}$, then $e^{{\alpha }_1},\mathrm{\dots },e^{{\alpha }_n}$ are linearly independent over $\mathbb{Q}$.

Proof.

$(1⟹2).$ Write $A_i=e^{{\alpha }_i}$ for each $i=1,\mathrm{\dots },n$. Suppose $0=r_1{A}_1+\mathrm{\cdots }+r_n{A}_n$, where $r_i\in \mathbb{Q}$. Moving $r_{1}A$ to the LHS and by multiplying a common denominator we can assume that $r_1{A}_1=r_2{A}_2+\mathrm{\cdots }+r_n{A}_n$ where $r_i\in \mathbb{Z}$. We want to show that $r_1=\mathrm{\cdots }=r_n=0$. We induct on $n$. The case when $n=1$ is trivial because $A_1$ is never 0 and therefore $0=r_1{A}_1$ forces $r_1=0$.

By induction hypothesis, suppose the statement is true when . Now suppose $n=k$. If ${\alpha }_1,\mathrm{\dots },{\alpha }_k$ are linearly independent over $\mathbb{Q}$, $A_1,\mathrm{\dots },A_k$ are algebraically independent and certainly linearly independent over $\mathbb{Q}$. So suppose ${\alpha }_1,\mathrm{\dots },{\alpha }_k$ are not linearly independent over $\mathbb{Q}$. Without loss of generality, we can assume $s_1{\alpha }_1=s_2{\alpha }_2+\mathrm{\cdots }+s_k{\alpha }_k$, where $s_i\in \mathbb{Q}$ and $s_1\ne 0$. By multiplying a common denominator we can further assume that $s_i\in \mathbb{Z}$ and $s_1>0$. Then

$$A_1^{s_1}=e^{s_1{\alpha }_1}=e^{s_2{\alpha }_2+\mathrm{\cdots }+s_k{\alpha }_k}=e^{s_2{\alpha }_2}\mathrm{\cdots }{e}^{s_k{\alpha }_k}=A_2^{s_2}\mathrm{\cdots }{A}_k^{s_k}.$$

Since $r_1{A}_1=r_2{A}_2+\mathrm{\cdots }+r_n{A}_n$, we get

$(r_1^{s_1})(A_2^{s_2}\mathrm{\cdots }{A}_k^{s_k})$ $=$ $(r_1^{s_1})A_1^{s_1}$

$=$ ${(r_1{A}_1)}^{s_1}$

$=$ ${(r_2{A}_2+\mathrm{\cdots }+r_k{A}_k)}^{s_1}$

$=$ $g(A_2,\mathrm{\dots },A_k),$

where $g(x_2,\mathrm{\dots },x_k)={(r_2{x}_2+\mathrm{\cdots }+r_k{x}_k)}^{s_1}\in \mathbb{Q}[x_2,\mathrm{\dots },x_k].$ Partition the numbers $s_i^{{}^{\prime }}s$ into non-negative and negative ones, so that, say, $s_{c(1)},\mathrm{\dots },s_{c(l)}$ are non-negative and $s_{d(1)},\mathrm{\dots },s_{d(m)}$ are negative, then

$$(r_1^{s_1})A_{c(1)}^{s_{c(1)}}\mathrm{\cdots }{A}_{c(i)}^{s_{c(i)}}=g(A_2,\mathrm{\dots },A_k)A_{d(1)}^{s_{d(1)}}\mathrm{\cdots }{A}_{d(j)}^{s_{d(j)}}.$$

If we define

$$f(x_2,\mathrm{\dots },x_k)=r_1^{s_1}{x}_{c(1)}^{s_{c(1)}}\mathrm{\cdots }{x}_{c(i)}^{s_{c(i)}}-g(x_2,\mathrm{\dots },x_k)x_{d(1)}^{s_{d(1)}}\mathrm{\cdots }{x}_{d(j)}^{s_{d(j)}},$$

then $f(x_2,\mathrm{\dots },x_k)\in \mathbb{Q}[x_2,\mathrm{\dots },x_k]$ and $f(A_2,\mathrm{\dots },A_k)=0.$ By the induction hypothesis, $f=0$. It is not hard to see that $r_1=\mathrm{\cdots }=r_k=0$ and therefore $A_1,\mathrm{\dots },A_k$ are linearly independent.

$(1⟸2).$ We first need two lemmas:

Lemma 1. Given 2., if $\alpha \ne 0$ is algebraic over $\mathbb{Q}$, then $e^{\alpha }$ is transcendental over $\mathbb{Q}$.

Proof.

Suppose $f(e^{\alpha })=0$ where $f(x)=r_0+r_{1}x+\mathrm{\cdots }+r_n{x}^n\in \mathbb{Q}[x]$. Then we have

$0$ $=$ $r_0+r_1{e}^{\alpha }+\mathrm{\cdots }+r_n{(e^{\alpha })}^n$

$=$ $r_0{e}^0+r_1{e}^{\alpha }+\mathrm{\cdots }+r_n{e}^{n\alpha }.$

Since $\alpha \ne 0$, $0,\alpha ,\mathrm{\dots },n\alpha$ are all distinct, $1,e^{\alpha },\mathrm{\dots },e^{n\alpha }$ are linearly independent by the hypothesis. Thus, $r_0=r_1=\mathrm{\dots }=r_n=0$ and we have $f(x)=0$, which means that $e^{\alpha }$ is transcendental over $\mathbb{Q}$. ∎

Lemma 2. Given 2., if $\alpha$ and $\beta$ are linearly independent and algebraic over $\mathbb{Q}$, then $e^{\alpha }$ is transcendental over $\mathbb{Q}(e^{\beta })$.

Proof.

Let $A=e^{\alpha }$ and $B=e^{\beta }$. Suppose $f(A)=0$ where $f(x)\in \mathbb{Q}(B)[x]$. We want to show that $f(x)=0$. Write

$$f(x)=r_0(B)+r_1(B)x+\mathrm{\cdots }+r_n(B)x^n,$$

where each $r_i(x)=p_i(x)/q_i(x)$ with $p_i(x)$, $q_i(x)\ne 0\in \mathbb{Q}[x]$. Let $Q(x)=q_1(x)\mathrm{\cdots }{q}_n(x)$. So $Q(B)$, being the product of the denominators $q_i(B)\ne 0$, is non-zero. Multiply $f(x)$ by $Q(B)$ we get a new polynomial $g(x)$ such that

$$g(x)=R_0(B)+R_1(B)x+\mathrm{\cdots }+R_n(B)x^n,$$

where each $R_i(x)=r_i(x)Q(x)=p_i(x)Q(x)/q_i(x)\in \mathbb{Q}[x]$. Now, $g(A)=f(A)Q(B)=0$. So

$0$ $=$ $R_0(B)+R_1(B)A+\mathrm{\cdots }+r_n(B)A^n$

$=$ ${\displaystyle \sum _{j=0}^{m_0}}{a}_{0j}{B}^j+{\displaystyle \sum _{j=0}^{m_1}}{a}_{1j}{B}^{j}A+\mathrm{\cdots }+{\displaystyle \sum _{j=0}^{m_n}}{a}_{nj}{B}^j{A}^n$

$=$ ${\displaystyle \sum _{j=0}^{m_0}}{a}_{0j}{e}^{j\beta }+{\displaystyle \sum _{j=0}^{m_1}}{a}_{1j}{e}^{j\beta +\alpha }+\mathrm{\cdots }+{\displaystyle \sum _{j=0}^{m_n}}{a}_{nj}{e}^{j\beta +n\alpha },$

where each $a_{ij}\in \mathbb{Q}$. Now, the exponents in the above equation are all distinct, or else we would end up with $\alpha$ and $\beta$ being linearly dependent, contrary to the assumption. Therefore, by 2 (Lindemann-Weierstrass Version 2), all $e^{i\beta +j\alpha }$ are linearly independent, which means each $a_{ij}=0$. This implies that $g(x)=0$. But $g(x)=f(x)Q(B)$ and $Q(B)\ne 0$, we must have $f(x)=0$. ∎

Now onto the main problem. We proceed by induction on the number of linearly independent algebraic elements over $\mathbb{Q}$. The case when $n=1$ is covered in Lemma 1, since a linearly independent singleton is necessarily non-zero. So suppose ${\alpha }_1,\mathrm{\dots },{\alpha }_k$ are linearly independent and algebraic over $\mathbb{Q}$. Then each pair ${\alpha }_k,{\alpha }_i$ are independent and algebraic over $\mathbb{Q}$, $i\ne k$. Thus $e^{{\alpha }_k}$ is transcendental over $\mathbb{Q}(e^{{\alpha }_i})$ for all $i\ne k$. This means that $e^{{\alpha }_k}$ is transcendental over $\mathbb{Q}(e^{{\alpha }_1},\mathrm{\dots },e^{{\alpha }_{k-1}}).$

Now let $A_i=e^{{\alpha }_i}$ for all $i=1,\mathrm{\dots },k$. Suppose $f(A_1,\mathrm{\dots },A_k)=0$ where $f\in \mathbb{Q}[x_1,\mathrm{\dots },x_k]$. To show the algebraic independence of the $A_i^{\prime }s$, we need to show that $f=0$. Rearranging terms of $f$ and we have

$$0=f(A_1,\mathrm{\dots },A_k)=\sum _{j=0}^m{g}_j(A_1,\mathrm{\dots },A_{k-1}){(A_k)}^j.$$

If we let $g(x)=f(A_1,\mathrm{\dots },A_{k-1},x)$, we see that $g(x)\in \mathbb{Q}(A_1,\mathrm{\dots },A_{k-1})[x]$ and $g(A_k)=f(A_1,\mathrm{\dots },A_{k-1},A_k)=0$. Since $e^{{\alpha }_k}$ is transcendental over $\mathbb{Q}(A_1,\mathrm{\dots },A_{k-1})$, we must have $g(x)=0$. This implies that each $g_j(A_1,\mathrm{\dots },A_{k-1})=0$. But then $A_1,\mathrm{\dots },A_{k-1}$ are algebraically independent by the induction hypothesis, we must have each $g_j=0$. This means that $f=0$. ∎

Title	equivalent statements of Lindemann-Weierstrass theorem
Canonical name	EquivalentStatementsOfLindemannWeierstrassTheorem
Date of creation	2013-03-22 18:05:18
Last modified on	2013-03-22 18:05:18
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	4
Author	CWoo (3771)
Entry type	Result
Classification	msc 12D99
Classification	msc 11J85