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 $K=\mathbb{Q}(\tau)$.


1 Answer 1


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 $K=\mathbb{Q}(\tau)$.


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