Finding the degree $[\mathbb{Q}(\sqrt[4]{3},\sqrt[5]{3}):\mathbb{Q}]$

Question

What I tried:

Consider the field extensions: $$\mathbb{Q}\subset\mathbb{Q} (\sqrt[4]{3})\subset\mathbb{Q} (\sqrt[4]{3},\sqrt[5]{3}) $$

The polynomial $f(x)=x^4-3\in\mathbb{Q}[x]$ has $\sqrt[4]{3}$ as root, and is irreducible over $\mathbb{Q}[x]$, so $$[\mathbb{Q}(\sqrt[4]{3}):\mathbb{Q}]=\deg(f)=4.$$

Also, I can considerer the polynomial $g(x)=x^4-\sqrt[20]{3}\in\mathbb{Q}(\sqrt[4]{3})[x]$, which has $\sqrt[5]{3}$ as root, so if this polynomial is irreducible, so $$[\mathbb{Q}(\sqrt[4]{3},\sqrt[5]{3}):\mathbb{Q} (\sqrt[4]{3}) ] \implies [\mathbb{Q}(\sqrt[4]{3},\sqrt[5]{3}):\mathbb{Q}]=4\cdot4=16.$$ Am I right?

Answer 1

Consider two extension towers \begin{align*} \Bbb Q &\subset \Bbb Q(\sqrt [5] 3) \subset \Bbb Q (\sqrt [5] 3, \sqrt [4] 3), \\ \Bbb Q &\subset \Bbb Q(\sqrt [4] 3) \subset \Bbb Q (\sqrt [5] 3, \sqrt [4] 3), \end{align*} and let the degree be $N$, then the multiplicative property indicates that $4 \mid N$ and $5 \mid N$, where $4,5$ are degrees of extensions at the beginning of the tower. Hence $20 \mid N$ and then $2 0 \leqslant N$. Cf. M. Artin, Algebra, 2nd edition, Corollary 15.3.8, $N \leqslant 4 \times 5 = 20 $ as well. Therefore the degree shall be $\boldsymbol {20}$.

Corollary 15.3.8 Let $\mathcal K$ be an extension field of a field $F$, let $K$ and $F'$ be subfields of $\mathcal K$ that are finite extensions of $F$, and let $K'$ denote the subfield of $\mathcal K$ generated by the two fields $K$ and $F'$ together. Let $[K': F] = N, [K: F] = m$ and $[F': F] = n$. Then $m$ and $n$ divide $N$, and $N\leqslant mn$.