# “Non-equality” of distinct polynomials with the same transcendental argument

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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You should probably make explicit the assumption that the coefficients $a_i,b_i$ are algebraic numbers, since otherwise the assertion is not true. – Pete L. Clark Feb 21 '11 at 12:43
@Peter: of course, you're right. But I assume the OP meant transcendental over $\mathbb Q$, which is what unadorned "transcendental" usually means. – lhf Feb 22 '11 at 23:56

That follows directly from the definition of transcendental number: if $p(\alpha)=q(\alpha)$ then $p(\alpha)-q(\alpha)=0$.
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 $\rm\:F\:$.