Let $R = \{a + b\alpha + c\alpha^2|\ a,b,c \in \mathbb{Z}\}\subseteq \mathbb{C}$ where $\alpha $ is a zero of $ x^3 - 4x +2 = 0$
Is $R$ an integral domain? From my research, I understand that a way to show this is an integral domain is by verifying that it's a subring of C. But am unsure how to proceed.
On
Exactly as you wrote, it is enough to show that $R$ is a subring of $\mathbb{C}$ (since $\mathbb{C}$ is an integral domain).
Note that $0\in R$ so you need to verify that $R$ is closed under addition and multiplication: For $a+b_1a+c_1a^2,a+b_2a+c_2a^2\in R$,
$\bullet$ $(a+b_1a+c_1a^2)+(a+b_2a+c_2a^2)=a+(b_1+b_2-1)a+(c_1+c_2)a^2\in R$
$\bullet$ $(a+b_1a+c_1a)(a+b_2a+c_2a)= l_4a^4+l_3a^3+l_2a^2+l_1a$ for some $l_i\in\mathbb{Z}$ (that we can compute-but don't really need to). I leave it to you to prove that $l_4x^4+l_3x^3+l_2x^2+l_1x=(x^3-4x+2)f(x)$ for some $f\in\mathbb{Z}$ and that $f(a)\in R$ (why will this imply that the product is an element of $R$?)
You have basically said how to do it: as R is a subset of $\mathbb{C}$ all that is left is to show that R is a ring. It contains the unit element 1, so you now need to show that it is closed under addition and multiplication. Addition should be obvious: for multiplication see what happens when you multiply $a+b\alpha+c\alpha^2$ by $d+e\alpha+f\alpha^2$. You can eliminate the $\alpha^3$ and $\alpha^4$ terms by using the fact that $\alpha^3-4\alpha+2= 0$. Then, as you say, it is a subring of $\mathbb{C}$ so it is a domain.