I have been given this problem to prove:
Let $\alpha \in \mathbb{C}$ and $\mathbb{Z}[\alpha]$ be the intersection of all unital subrings of $\mathbb{C}$ containing $\alpha$. Show that $\mathbb{Z}[\alpha] = \{f(\alpha):f \in \mathbb{Z}[x]\}$
My question is "$\mathbb{Z}[\alpha]$ be the intersection of all unital subrings of $\mathbb{C}$" an equivilent definition to $\mathbb{Z}[\alpha] = \{a+\alpha b | a,b \in \mathbb{Z}\}$?
My initial approach was to assume that $\mathbb{Z}[\alpha] = \{a+\alpha b | a,b \in \mathbb{Z}\}$? the show that an element $a$ in that ring equals $a_0+ a_1\alpha \cdots +a_n\alpha^n$. Then naturally, we have that $\mathbb{Z}[\alpha] = \{f(\alpha):f \in \mathbb{Z}[x]\}$. But I am assuming there are major holes my approach.
Thanks for any help!
Let $\cal F$ be any family of sets. If $X\in{\cal F}$ is such that $X\subseteq Y$ for all $Y\in\cal F$, then $\bigcap{\cal F}=X$. This is a purely set-theoretic fact that you should try to prove.
Here this works with $X=\{f(\alpha):f(x)\in\Bbb Z[x]\}$ and $\cal F$ the family of all unital subrings containing the element $\alpha$. First you need to show $X$ is a unital subring (what does that mean?) containing $\alpha$, then argue it is contained in any other unital subring of $\Bbb C$ containing $\alpha$.
Assuming $\Bbb Z[\alpha]=\{a+b\alpha:a,b\in\Bbb Z\}$ is pretty much assuming a false version of the very conclusion you're trying to prove, so that approach is very wrong. Also, probably you've been prejudiced by quadratic extensions - in general elements of $\Bbb Z[\alpha]$ do not look like $a+b\alpha$, it only works out that nicely in quadratic extensions. Generally you need higher powers of $\alpha$ too.