Exercise 5.1.23 (Coprime relation)

Solution: Let $M:K$ be a field extension and ${\alpha }_1,...,{\alpha }_n\in M$. If $\mathit{gcd}(\mathit{de}{g}_K({\alpha }_1),...,\mathit{de}{g}_K({\alpha }_n))=1$ (i.e., coprime), then we have that, $[K({\alpha }_1,...,{\alpha }_n):K]=[K({\alpha }_1):K]...[K({\alpha }_n):K]$

We prove the following generalization:

Proposition Let $M:K$be a field extension, and $\alpha ,\beta \in M$be algebraic over $K$.

Assume that $[K(\alpha ):K]=m$and $[K(\beta ):K]=n$, where $\gcd (n,m)=1$. Then $[K(\alpha ,\beta ):K]=\mathit{nm}.$

Note: the generalization proposed by Hassaan Naeem seems not true. To give a counterexample, take $\alpha =\sqrt[6]{2},\beta =\sqrt[15]{2},\gamma =\sqrt[10]{2}$. Then $[\mathbb{Q}(\alpha ):\mathbb{Q}]=6,[\mathbb{Q}(\beta ):\mathbb{Q}]=15,[\mathbb{Q}(\gamma ):\mathbb{Q}]=10$ (because $t^6-2,t^{15}-2,t^{10}-2$ are irreducible by Eisenstein’s Criterium).

Moreover $\gcd (6,15,10)=\gcd (2\times 3,3\times 5,2\times 5)=1$.

But $\mathbb{Q}(\alpha ,\beta ,\gamma )=\mathbb{Q}(\sqrt[6]{2},\sqrt[15]{2},\sqrt[10]{2})\subset \mathbb{Q}(\sqrt[30]{2})$, where $30=\mathrm{gcm}(6,15,10)$, because $\sqrt[6]{2}={(\sqrt[30]{2})}^5$, etc...

The converse is true: since $\frac{1}{30}=-\frac{1}{6}+\frac{3}{15}$, $\sqrt[30]{2}={(\sqrt[6]{2})}^{-1}{(\sqrt[15]{2})}^3\in \mathbb{Q}(\alpha ,\beta ,\gamma )$, thus

Therefore

If we want to generalize, we must assume that $\gcd ({\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->}_k({\alpha }_i),{\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->}_k({\alpha }_j))=1$ if $i\ne j$.