Let $a,b \in \mathbb{Q}$ and $d \neq 0,1$ be a square free integer.
Define $\overline{a + b\sqrt{d}} = a - b\sqrt{d}$
If $f \in \mathbb{Z}[x]$ show that: $f(\overline{\alpha}) = \overline{f(\alpha)}$
The notation may be somewhat confusing. Prior to this we had defined a set: \begin{equation*} A = \{a + b\sqrt{d} : a,b \in \mathbb{Q}\} \cap \overline{\mathbb{Z}} \end{equation*} where $\overline{\mathbb{Z}}$ is the algebraic closure of $\mathbb{Z}$.
Here is my working so far:
A polynomial $f$ in $\mathbb{Z}[x]$ has the form:
$ f(x) = \sum_{i=0}^n a_ix^i $
I think we might require that $f$ be monic, so let $|a_n| = 1$.
Now, by the binomial theorem:
$(a - b\sqrt{d})^i = \sum_{k=0}^i \binom{i}{k} a^{i-k}(-b\sqrt{d})^k$
and as such we may write:
$ f(\overline{\alpha}) = \sum_{i=0}^n a_i\sum_{k=0}^i \binom{i}{k} a^{i-k}(-b\sqrt{d})^k $
But this becomes real messy real fast! I don't even know how to begin calculating $\overline{f(\alpha)}$. Could someone provide some advice for showing $f(\overline{\alpha}) = \overline{f(\alpha)}$?