How can I show for some transcendental number $\alpha \in \mathbb{C}$, that two distinct polynomials of the form $b_{n}\alpha^n + \cdot \cdot \cdot + b_{2}\alpha^2 + b_{1}\alpha$ and $a_{m}\alpha^m + \cdot \cdot \cdot + a_{2}\alpha^2 + a_{1}\alpha$ can never equal each other?
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That follows directly from the definition of transcendental number: if $p(\alpha)=q(\alpha)$ then $p(\alpha)-q(\alpha)=0$. |
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HINT $\ \ $ By definition, $\rm\ \alpha\:$ is transcendental over $\rm\:F\ \iff\ F[x] \cong F[\alpha]\ $ via evaluation $\rm\ x\to \alpha\:.\ $ For example, $\rm\ \mathbb Q[\pi] \cong \mathbb Q[x]\:.\:$ Therefore, in particular, transcendental evaluation is always injective. Thus any element transcendental over a field $\rm\:F\:$ serves equally well as an "indeterminate" over $\rm\:F\:$. |
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