Show that the set $\{a+b\omega:\omega^3=1\}$ is a field

Question

Show that the set $\{a+b\omega:\omega^3=1\}$, where a and b are rational numbers, is a field with respect to addition and multiplication.

I considered $\omega$ to be a cube root of unity. We can easily see that the set is an abelian group under addition. But I am wondering how it can be closed under multiplication.

Let $x=a+b\omega$, $\;y=c+d\omega$ be any two elements of $F$. Then the set $F$ is closed if $x.y\in F$ i.e. $(a+b\omega$).($c+d\omega)\in F$

But $(a+b\omega$)($c+d\omega) = (ac+bd\omega^2)+(ad+bc)\omega \notin F$ as $ac+bd\omega^2$ is not a real number.

Can some one please throw some hint?

Answer 1

Hint:

Note that $$1+\omega + \omega^2=0$$

Hence $$\omega^2 = -(1+\omega)$$

Answer 2

This is a field over $\Bbb{Q}$, namely the cyclotomic field $\Bbb{Q}(\zeta_3)$. It has degree $\phi(3)=2$, because $1+\zeta_3+\zeta_3^2=0$. Here $\omega=\zeta_3$ is a primitive third root of unity. For a proof see this post:

Proving existence of a multiplicative inverse in a field