# Proving equivalence between finite field extensions.

Given $$K=\mathbb{Q}(\omega)$$ where $$\omega$$ is a complex cube root of 1. The minimal polynomial of $$\omega$$ is of degree 2 and we have that $$\tau = (a+b\omega)\in K$$ ($$a,b\in \mathbb{Q}$$) has minimal polynomial of degree 2 also. I am trying to show that $$K=\mathbb{Q}(\tau)$$.

I have found an explanation that says since the minimal polynomial has degree 2, the field extension $$\mathbb{Q}(\tau)$$ has degree 2 over $$\mathbb{Q}$$ (which I am fine with as I have proved $$\mathbb{\tau}$$ to be algebraic and used that $$[F(\omega) : F]$$ is finite if and only if $$\omega$$ is algebraic over $$F$$. Moreover, in this case, the degree of the extension is precisely the degree of the minimal polynomial of $$\omega$$ over $$F$$.) However it then says that ... and the field extension is contained in $$K$$. Hence $$K=\mathbb{Q}(\tau)$$. I am struggling to understand,

1) How does the minimal polynomial having degree 2 imply that the field extension $$\mathbb{Q}(\tau)$$ is contained in $$K$$.

2) How does the field extension $$\mathbb{Q}(\tau)$$ having degree 2 over $$\mathbb{Q}$$ and is contained in $$K$$ imply $$K=\mathbb{Q}(\tau)$$.

The minimal polynomial is irrelevant for the first part, instead observe that $$\mathbb{Q}(\tau)$$ is contained in $$K=\mathbb{Q}(\omega)$$ because $$\tau=a+b\omega\in K$$ and $$K$$ is a field.
And then since $$K$$ has degree $$2$$, by the tower rule we have $$2=[K:\mathbb{Q}]=[K:\mathbb{Q}(\tau)][\mathbb{Q}(\tau):\mathbb{Q}]=2[K:\mathbb{Q}(\tau)]$$ and so $$[K:\mathbb{Q}(\tau)]=1$$ and $$K=\mathbb{Q}(\tau)$$.