# equivalent statements of Lindemann-Weierstrass theorem

###### Proposition 1.

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

1. 1.

If $\alpha_{1},\ldots,\alpha_{n}$$\mathbb{Q}$, then $e^{\alpha_{1}},\ldots,e^{\alpha_{n}}$ are algebraically independent over $\mathbb{Q}$.

2. 2.

If $\alpha_{1},\ldots,\alpha_{n}$ are distinct algebraic numbers over $\mathbb{Q}$, then $e^{\alpha_{1}},\ldots,e^{\alpha_{n}}$ are linearly independent over $\mathbb{Q}$.

###### Proof.

$(1\Longrightarrow 2).$ Write $A_{i}=e^{\alpha_{i}}$ for each $i=1,\dots,n$. Suppose $0=r_{1}A_{1}+\cdots+r_{n}A_{n}$, where $r_{i}\in\mathbb{Q}$. Moving $r_{1}A$ to the LHS and by multiplying a common denominator we can assume that $r_{1}A_{1}=r_{2}A_{2}+\cdots+r_{n}A_{n}$ where $r_{i}\in\mathbb{Z}$. We want to show that $r_{1}=\cdots=r_{n}=0$. 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 $n. Now suppose $n=k$. If $\alpha_{1},\ldots,\alpha_{k}$ are linearly independent over $\mathbb{Q}$, $A_{1},\ldots,A_{k}$ are algebraically independent and certainly linearly independent over $\mathbb{Q}$. So suppose $\alpha_{1},\ldots,\alpha_{k}$ are not linearly independent over $\mathbb{Q}$. Without loss of generality, we can assume $s_{1}\alpha_{1}=s_{2}\alpha_{2}+\cdots+s_{k}\alpha_{k}$, where $s_{i}\in\mathbb{Q}$ and $s_{1}\neq 0$. By multiplying a common denominator we can further assume that $s_{i}\in\mathbb{Z}$ and $s_{1}>0$. Then

 ${A}_{1}^{\phantom{1}s_{1}}=e^{s_{1}\alpha_{1}}=e^{s_{2}\alpha_{2}+\cdots+s_{k}% \alpha_{k}}=e^{s_{2}\alpha_{2}}\cdots e^{s_{k}\alpha_{k}}={A}_{2}^{\phantom{2}% s_{2}}\cdots{A}_{k}^{\phantom{k}s_{k}}.$

Since $r_{1}A_{1}=r_{2}A_{2}+\cdots+r_{n}A_{n}$, we get

 $\displaystyle(r_{1}^{s_{1}})({A}_{2}^{\phantom{2}s_{2}}\cdots{A}_{k}^{\phantom% {k}s_{k}})$ $\displaystyle=$ $\displaystyle(r_{1}^{s_{1}}){A}_{1}^{\phantom{1}s_{1}}$ $\displaystyle=$ $\displaystyle(r_{1}A_{1})^{s_{1}}$ $\displaystyle=$ $\displaystyle(r_{2}A_{2}+\cdots+r_{k}A_{k})^{s_{1}}$ $\displaystyle=$ $\displaystyle g(A_{2},\ldots,A_{k}),$

where $g(x_{2},\ldots,x_{k})=(r_{2}x_{2}+\cdots+r_{k}x_{k})^{s_{1}}\in\mathbb{Q}[x_{2% },\ldots,x_{k}].$ Partition the numbers ${s}_{i}^{\phantom{i}{}^{\prime}}s$ into non-negative and negative ones, so that, say, $s_{c(1)},\ldots,s_{c(l)}$ are non-negative and $s_{d(1)},\ldots,s_{d(m)}$ are negative, then

 $(r_{1}^{s_{1}}){A}_{c(1)}^{\phantom{c(1)}s_{c(1)}}\cdots{A}_{c(i)}^{\phantom{c% (i)}s_{c(i)}}=g(A_{2},\ldots,A_{k}){A}_{d(1)}^{\phantom{d(1)}s_{d(1)}}\cdots{A% }_{d(j)}^{\phantom{d(j)}s_{d(j)}}.$

If we define

 $f(x_{2},\ldots,x_{k})=r_{1}^{s_{1}}{x}_{c(1)}^{\phantom{c(1)}s_{c(1)}}\cdots{x% }_{c(i)}^{\phantom{c(i)}s_{c(i)}}-g(x_{2},\ldots,x_{k}){x}_{d(1)}^{\phantom{d(% 1)}s_{d(1)}}\cdots{x}_{d(j)}^{\phantom{d(j)}s_{d(j)}},$

then $f(x_{2},\ldots,x_{k})\in\mathbb{Q}[x_{2},\ldots,x_{k}]$ and $f(A_{2},\ldots,A_{k})=0.$ By the induction hypothesis, $f=0$. It is not hard to see that $r_{1}=\cdots=r_{k}=0$ and therefore $A_{1},\ldots,A_{k}$ are linearly independent.

$(1\Longleftarrow 2).$ We first need two lemmas:

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

###### Proof.

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

 $\displaystyle 0$ $\displaystyle=$ $\displaystyle r_{0}+r_{1}e^{\alpha}+\cdots+r_{n}(e^{\alpha})^{n}$ $\displaystyle=$ $\displaystyle r_{0}e^{0}+r_{1}e^{\alpha}+\cdots+r_{n}e^{n\alpha}.$

Since $\alpha\neq 0$, $0,\alpha,\ldots,n\alpha$ are all distinct, $1,e^{\alpha},\ldots,e^{n\alpha}$ are linearly independent by the hypothesis. Thus, $r_{0}=r_{1}=\ldots=r_{n}=0$ 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

 $f(x)=r_{0}(B)+r_{1}(B)x+\cdots+r_{n}(B)x^{n},$

where each $r_{i}(x)=p_{i}(x)/q_{i}(x)$ with $p_{i}(x)$, $q_{i}(x)\neq 0\in\mathbb{Q}[x]$. Let $Q(x)=q_{1}(x)\cdots q_{n}(x)$. So $Q(B)$, being the product of the denominators $q_{i}(B)\neq 0$, is non-zero. Multiply $f(x)$ by $Q(B)$ we get a new polynomial $g(x)$ such that

 $g(x)=R_{0}(B)+R_{1}(B)x+\cdots+R_{n}(B)x^{n},$

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

 $\displaystyle 0$ $\displaystyle=$ $\displaystyle R_{0}(B)+R_{1}(B)A+\cdots+r_{n}(B)A^{n}$ $\displaystyle=$ $\displaystyle\sum_{j=0}^{m_{0}}a_{0j}B^{j}+\sum_{j=0}^{m_{1}}a_{1j}B^{j}A+% \cdots+\sum_{j=0}^{m_{n}}a_{nj}B^{j}A^{n}$ $\displaystyle=$ $\displaystyle\sum_{j=0}^{m_{0}}a_{0j}e^{j\beta}+\sum_{j=0}^{m_{1}}a_{1j}e^{j% \beta+\alpha}+\cdots+\sum_{j=0}^{m_{n}}a_{nj}e^{j\beta+n\alpha},$

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 $Q(B)\neq 0$, 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 $\alpha_{1},\ldots,\alpha_{k}$ are linearly independent and algebraic over $\mathbb{Q}$. Then each pair $\alpha_{k},\alpha_{i}$ are independent and algebraic over $\mathbb{Q}$, $i\neq k$. Thus $e^{\alpha_{k}}$ is transcendental over $\mathbb{Q}(e^{\alpha_{i}})$ for all $i\neq k$. This means that $e^{\alpha_{k}}$ is transcendental over $\mathbb{Q}(e^{\alpha_{1}},\ldots,e^{\alpha_{k-1}}).$

Now let $A_{i}=e^{\alpha_{i}}$ for all $i=1,\ldots,k$. Suppose $f(A_{1},\ldots,A_{k})=0$ where $f\in\mathbb{Q}[x_{1},\ldots,x_{k}]$. 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

 $0=f(A_{1},\ldots,A_{k})=\sum_{j=0}^{m}g_{j}(A_{1},\ldots,A_{k-1})(A_{k})^{j}.$

If we let $g(x)=f(A_{1},\ldots,A_{k-1},x)$, we see that $g(x)\in\mathbb{Q}(A_{1},\ldots,A_{k-1})[x]$ and $g(A_{k})=f(A_{1},\ldots,A_{k-1},A_{k})=0$. Since $e^{\alpha_{k}}$ is transcendental over $\mathbb{Q}(A_{1},\ldots,A_{k-1})$, we must have $g(x)=0$. This implies that each $g_{j}(A_{1},\ldots,A_{k-1})=0$. But then $A_{1},\ldots,A_{k-1}$ 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 EquivalentStatementsOfLindemannWeierstrassTheorem 2013-03-22 18:05:18 2013-03-22 18:05:18 CWoo (3771) CWoo (3771) 4 CWoo (3771) Result msc 12D99 msc 11J85