$[E:\mathbb{Q}]$, the degree of $E$ over $\mathbb{Q}$

Question

Algebraic Extension $E=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$, I need to find the $[E:\mathbb{Q}]$, the degree of $E$ over $\mathbb{Q}$

I know degree of $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ is $2$ by which I mean $\mathbb{Q}(\sqrt{2})$ is a vector space over $\mathbb{Q}$ of dimension $2$ with basis $\{1,\sqrt{2}$} but here what I have to calculate? what will be the intermediate steps? will $E$ be a vector space over $\mathbb{Q}$ of dimension $4$ with basis $\{1,\sqrt{2},\sqrt{3},\sqrt{5}\}$? any detail explanaion will be appreciated ..Thanks a lot.

Answer 1

Can $\sqrt 3$ be written in $\mathbb{Q}(\sqrt 2)$? If not, then $\mathbb{Q}(\sqrt 2, \sqrt 3)$ has degree $4$ over $\mathbb{Q}$ (do you know why $4$ and not, say, $3$?).

Once you've decided that, you should decide if $\sqrt 5$ can be written in $\mathbb{Q}(\sqrt 2, \sqrt 3)$.

As a related example, let's examine $\mathbb{Q}(\sqrt 2, \sqrt 7)$. I know $\mathbb{Q}(\sqrt 2)$ and $\mathbb{Q}(\sqrt 7)$ are each degree 2 over $\mathbb{Q}$, so I know that $\mathbb{Q}(\sqrt 2, \sqrt 7)$ will be of degree $2$ or $4$. To see if it's of degree $2$, I want to see if $\sqrt 7 \in \mathbb{Q}(\sqrt 2)$.

Suppose it is. Then $c\sqrt 7 = a + b\sqrt 2$ for some $a,b,c \in \mathbb{Z}$. Squaring both sides, we see that $7c^2 = a^2 + 2b^2 + 2ab\sqrt 2$, or rather that $\sqrt 2 = \dfrac{7c^2 - a^2 - 2b^2}{2ab}$. But this is impossible, as $\sqrt 2$ is irrational. Thus $\sqrt 7 \not \in \mathbb{Q}(\sqrt 2)$, and so this extension is of degree 4.

Answer 2

Degree of field extension is multiplicative. Hence $[E:\mathbb{Q}]$ is decomposed into $$ [E:\mathbb{Q}(\sqrt{2}, \sqrt{3})] [\mathbb{Q}(\sqrt{2}, \sqrt{3}):\mathbb{Q}(\sqrt{2})] [\mathbb{Q}(\sqrt{2}):\mathbb{Q}]. $$ As you already know, $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2$. Similar reasoning works to the rest of degrees and you can conclude that $[E:\mathbb{Q}] = 2^3 = 8$.

Explicitly, a $\mathbb{Q}$-basis of $E$ is $$ \{ 1, \sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \sqrt{10}, \sqrt{15}, \sqrt{30} \} $$ for example.

Answer 3

Hint : take your own time and convince yourself that :

$\mathbb{Q}(\sqrt{2}) \subsetneq \mathbb{Q}(\sqrt{2},\sqrt{3})\subsetneq \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$

then, you can use

$[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}(\sqrt{2},\sqrt{3})].[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})].[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]$

then we have $[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}]=8$ as

$[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}(\sqrt{2},\sqrt{3})]=[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})]=[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$