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**AlexP** My book gives an example in which they calculate the degree of a field extension using $\displaystyle [K:F]=[K:E][E:F]$. In particular it shows that $\displaystyle [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=12$. I feel like I might have an easier way to do it than the book does, and I was hoping someone could confirm that this works, or if it does not, why it does not.

Since $\displaystyle \mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) = \mathbb{Q}(\sqrt[3]{2})(\sqrt[4]{3})$, can we just say that since the minimal polynomial of $\displaystyle \sqrt[4]{3}$ is $\displaystyle x^4-3$, $\displaystyle [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})]=4$, then similar to obtain $\displaystyle [\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$, which gives us $\displaystyle [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})][\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3\cdot 4 = 12$?

The book first points out that we know that both 3 and 4 divide $\displaystyle [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]$ so it's at least 12, but then the minimum polynomial of $\displaystyle \sqrt[4]{3}$ is $\displaystyle x^4-3$ so $\displaystyle [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})] \le 4$ (*when is it less?*) and so $\displaystyle [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}] \le 12$, yielding of course $\displaystyle [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=12$.

Thanks.