Prove that $Q(u)$ contains a subfield that has degree $3$ over $Q.$

Question

Let $F$ be a splitting field of an irreducible polynomial over $\mathbb{Q}$, and $u$ be a root of the irreducible polynomial. Suppose that $[F:Q]=27$. Prove that $\mathbb{Q(u)}$ contains a subfield that has degree 3 over $\mathbb{Q}$.

Answer 1

We have $[F:Q]=[F:Q(u)][Q(u):Q]$. If $[Q(u):Q]=3$ done.

if $Q(u)=F$ The extension is normal since $F$ is the splitting field and is separable since the characteristic of $Q$ is zero, thus Galoisian $Gal(F:Q)$ is a $3^3$ group Sylow theorem implies it has a $3^2$-subgroup $H$, Artin implies that $[F:F^H]=3^2$, since $[F:Q]=[F:F^H][F^H:Q]$, we deduce that $[F^H:Q]=3$.

if $[Q(u):Q]=9$ since the extension is Galois, $Q(u)=F^H$ where $H$ has order 3, Sylow implies that $H$ is contained in $L$ of cardinal 9 and as above we obtain that $[F^L:Q]=3$.