# Thread: Degree of an extension - easier way?

1. ## Degree of an extension - easier way?

My book gives an example in which they calculate the degree of a field extension using $[K:F]=[K:E][E:F]$. In particular it shows that $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=12$. 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 $\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) = \mathbb{Q}(\sqrt[3]{2})(\sqrt[4]{3})$, can we just say that since the minimal polynomial of $\sqrt[4]{3}$ is $x^4-3$, $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})]=4$, then similar to obtain $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$, which gives us $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})][\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3\cdot 4 = 12$?

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

Thanks.

2. ## Re: Degree of an extension - easier way?

Originally Posted by AlexP
My book gives an example in which they calculate the degree of a field extension using $[K:F]=[K:E][E:F]$. In particular it shows that $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=12$. 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 $\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) = \mathbb{Q}(\sqrt[3]{2})(\sqrt[4]{3})$, can we just say that since the minimal polynomial of $\sqrt[4]{3}$ is $x^4-3$, $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})]=4$, then similar to obtain $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$, which gives us $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})][\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3\cdot 4 = 12$?

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

Thanks.
by Eisenstein's criterion, the minimal polynomial of $\sqrt[4]{3}$ over $\mathbb{Q}$ is $x^4 - 3$ but $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})]$ is the degree of the minimal polynomial of $\sqrt[4]{3}$ over $\mathbb{Q}(\sqrt[3]{2})$ not $\mathbb{Q}$. so, the problem is that you didn't prove that $x^4-3$ is irreducible over $\mathbb{Q}(\sqrt[3]{2}).$
your textbook's approach is avoiding the above problem and i think it's a good approach.
the reason that $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})] \le 4$ is that $\sqrt[4]{3}$ is a zero of the polynomial $x^4-3 \in \mathbb{Q}(\sqrt[3]{2})[x]$ and thus the degree of the minimal polynomial of $\sqrt[4]{3}$ over $\mathbb{Q}(\sqrt[3]{2})$ is at most $\deg (x^4-3)=4.$

3. ## Re: Degree of an extension - easier way?

To clarify... Is it the fact that the minimal polynomial of $\sqrt[4]{3}$ has degree 4 in $\mathbb{Q}$ that gives us an upper bound of degree 4 on the minimal polynomial of $\sqrt[4]{3}$ in $\mathbb{Q}(\sqrt[3]{2})$?