Simple extensions being well-defined.

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Let $F$ be a field and let $\alpha\notin F$

We define $$ F(\alpha) = \{f(\alpha)/g(\alpha) : f,g\in F[x], \hspace{0.15cm} g(\alpha) \neq 0\} $$ as the smallest field containing both $F$ and $\alpha$.

How can we morally allow $f, g$ to take inputs from outside its domain? Say I have $$ f(x) = c_0 + c_1x + \ldots + c_nx^n. $$ Thus, $$ f(\alpha) = c_0 + c_1\alpha + \ldots + c_n\alpha^n. $$

How can we evaluate this expression if $\alpha\notin F$?

In the case where $F = \mathbb{R}$ and $\alpha = i$, I understand how this works, but mainly since $i$ is defined in terms of radicals and integers, for which we already know how they interact with $i$.

But I can't understand why this holds in general? Like what if I had $\mathbb{R}(T)$, where $T$ is some binary tree. I don't see how a polynomial could take a graph as an input?

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If you write $F(\alpha)$, then $\alpha$ must be an element of some bigger field (or at least ring) containing $F$ and $\alpha$, so that it makes sense to write $c_0+c_1α +\dots +c_n α^n$.

The expression "the smallest field containing both $F$ and $α$" does not really make sense in itself, it must be "the smallest subfield of $R$" where $R$ is some ring with $F\subset R$ and $\alpha\in R$ (and $\alpha$ must then be invertible in $R$, or else you can only define $F[\alpha]$).

Or else $\alpha$ is an indeterminate, and then $F(\alpha)$ is defined formally (as a field of rational functions in the indeterminate $\alpha$).