Let $\mathbb Q ⊆ L$ be fields. Assume that $L$ is a normal extension of $\mathbb Q$ and $|L : \mathbb Q| = 81$. Show that there exist intermediate fields $K_1$, $K_2$, and $K_3$ with $K_0 = \mathbb Q ⊆ K_1 ⊆ K_2 ⊆ K_3 ⊆ K_4 = L$ , where, for $i = 1, . . . , 4$ , $|K_i : K_{i−1}| = 3$.
$|X : Y|$ means degree of an extension.
Generaly let R be a commutative ring with identity, and let $F ⊆ R$ be a field. Consider $R$ a vector space over $F$. The dimension of this vector space is denoted by $|R : F|$ or $dim_F (R)$. If $F ⊆ E$ are
fields, then $|E : F|$ is called the degree of the field extension.
This is a homework question of mine: Exercise 26.1.7, from Shahriari, Algebra in Action.
I don't know how to start. I think I Should be use the tower rule; But how?!
Tower rule: Let F, K, and E be fields with F ⊆ K ⊆ E. Assume that |K : F| and |E : K| are finite. Then |E : F| = |E : K| |K : F|.
If the degree of the extension is $81=3^4$ and it's Galois then the Galois group has order 81. So the second Sylow's theorem guarantees that there exist some subgroups $H_1,H_2,H_3$ whose orders are $3,9,27$ respectively, which verifies that $H_i\trianglelefteq H_{i+1}$ . The second fundamental theorem of Galois theory also guarantees that there exist some Galois extensions $K_1|\mathbb{Q},K_2|\mathbb{Q}, K_3|\mathbb{Q}$ whose degrees are $3,9,27$ respectively. So you have the extensions tower $$K_0 = \mathbb Q ⊆ K_1 ⊆ K_2 ⊆ K_3 ⊆ K_4 = L$$ , where, for $i = 1, . . . , 4$ , $|K_i : K_{i−1}| = 3$.