00:43
$a_n (\alpha + \beta)^n = a_n (\alpha^n + \alpha^{n-1}\beta + \alpha^{n-2}\beta^2 + \cdots + \alpha\beta^{n-1}+\beta^n)$
Let $s=(\alpha+\beta), p=\alpha\beta$
$a_n (\alpha + \beta)^n = a_n (\alpha^{n-1}(\alpha + \beta) + \alpha^{n-3}\beta^2 (\alpha + \beta) + \cdots \alpha\beta^{n-1}(\alpha+\beta))$
=A(TA+TA+...+TA)=A(T+T+...+T)=AT=T
$a_n (\alpha + \beta)^n = a_n (\alpha^{n-1}(\alpha + \beta) + \alpha^{n-3}\beta^2 (\alpha + \beta) + \cdots \alpha\beta^{n-1}(\alpha+\beta) + \beta^n)$
Let $p(x)=\sum_{n=1}^ma_nx^n$
Let $s=(\alpha + \beta), k = \alpha\beta$ where $\alpha,\beta$ transcendental
$p(s) = \sum_{n} a_n s^n = 0$
$a_ns^n = a_n (\alpha^n + \alpha^{n-1}\beta + \alpha^{n-2}\beta^2 + \cdots + \alpha\beta^{n-1}+\beta^n)$
$a_ns^n = a_n (\alpha^{n-1}(\alpha + \beta) + \alpha^{n-3}\beta^2 (\alpha + \beta) + \cdots \alpha\beta^{n-1}(\alpha+\beta))$
= A(TA+TA+...+TA)=A(T+T+...+T)=AT=T
Hence $\sum_n a_ns^n= \sum_n T = T$
but this is a contradiction since $0=A$
$a_n (\alpha + \beta)^n = a_n (\alpha^{n-1}(\alpha + \beta) + \alpha^{n-3}\beta^2 (\alpha + \beta) + \cdots \alpha\beta^{n-1}(\alpha+\beta) + \beta^n)$
=A(TA+TA+...+TA+T)=A(T+...+T)=AT=T
but this is a contradiction since $0=A$
$p(s) = \sum_{n} a_n s^n = \sum_{n} a_n\alpha^k\beta^k = 0$
=$\sum_n AA^k=\sum_n A = A$
There is no contradiction
Suppose $s,k$ both transcendental. Then:
7 mins ago, by
Secret = A(TA+TA+...+TA)=A(T+T+...+T)=AT=T
Acutally, this step is wrong
= A(TA+TA+...+TA)=A(T+T+...+T)=AA=A
Lindlemann Weierstrass theorem showed that the set $\{e^{\alpha_n}\}$ is algebraically independent for algebraic numbers $\alpha$. i.e.
$$\sum_{n}a_ne^{\alpha_n} \neq 0$$
However, while the transcendence of sums and products of arbitrary transcendentals is an open question, is the above nonzer...
