Field extension degree theorem

Question

Calculating the degree of this extension $[\mathbb Q(\sqrt[3]{2},\sqrt2):\mathbb Q]=n $

With the irreducible polynomials I know:

$[\mathbb Q(\sqrt[3]{2}):\mathbb Q]=3$, $[\mathbb Q(\sqrt2):\mathbb Q]=2.$

So $2|n$ and $3|n$ that means $n\geq6.$

On the other hand $n\leq2\times3=6.$ So $n$ must be $6.$

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What I want to know is what result or theorem is applied here for doing this last part.

Answer 1

Let $L,K$ be finite field extensions of a field $k$, with $[L:k]=l$ and $[K:k]=k$. Then $[LK:k]\leq [L:k]\cdot [K:k]$. In particular, if $(l,k)=1$, then $[LK:k]=l \cdot k$

In your case $k=\mathbb{Q}$,$L=\mathbb{Q}(\sqrt[3]{2})$, $l=3$ and $K=\mathbb{Q}(\sqrt{2})$, $k=2$, so $[\mathbb{Q}(\sqrt[3]{2},\sqrt{2}):\mathbb{Q}]=6$.

Answer 2

Note that $\sqrt[3]{2}$ satisfies $x^3- 2$ over $\mathbb Q (\sqrt 2)$. Hence $[\mathbb Q (\sqrt[3] 2):\mathbb Q (\sqrt 2)]$ is at most $3$.