Degree of a finite field extension. How to find?

Question

What is the degree of the extension $[\mathbb Q(\sqrt2 + \sqrt3 + \sqrt5) : \mathbb Q ]$? Can you explain, what I must to do in this example?

Answer 1

$[\mathbb Q(\sqrt2 + \sqrt3 + \sqrt5) : \mathbb Q ]=[\mathbb Q(\sqrt2 , \sqrt3 , \sqrt5) : \mathbb Q ]=[\mathbb Q(\sqrt2 ,\sqrt3 , \sqrt5) : \mathbb Q (\sqrt2 , \sqrt3)].[\mathbb Q(\sqrt2 ,\sqrt3 ) : \mathbb Q (\sqrt2 )].[\mathbb Q(\sqrt2 ) : \mathbb Q ]=2.2.2=8$

$\mathbb Q(\sqrt2 + \sqrt3 + \sqrt5) $ is a $8 $ degree extension over $Q$.And basis are $1,\sqrt2,\sqrt3,\sqrt5,\sqrt6,\sqrt{10},\sqrt{15},\sqrt{30}$

Answer 2

Hint: The stabilizer of $ \sqrt{2} + \sqrt{3} + \sqrt{5} $ in $ \textrm{Gal}(\mathbf Q(\sqrt 2, \sqrt 3, \sqrt 5)/\mathbf Q) $ is trivial.

Answer 3

An elementary solution to show that at least $8$ dimensions needed

(i.e. the degree is at least $8$)

Our field has to be closed under multiplication. Let's see what will go wrong if we consider only four dimensions with the basis $(1,\sqrt2,\sqrt3,\sqrt5).$

The good news is that

$$\sqrt2+\sqrt3+\sqrt4=0\times1+1\times\sqrt2+1\times \sqrt3+1\times \sqrt4.$$

However, let's compute the following product

$$(a+b\sqrt2+c\sqrt3+d\sqrt4)\times(a+b\sqrt2+c\sqrt3+d\sqrt4).$$

Unfortunately the following irrational numbers will appear: $\sqrt6,\sqrt{10},\sqrt{15}$.

Let's try with the following basis then $$(1,\sqrt2,\sqrt3,\sqrt5,\sqrt6,\sqrt{10},\sqrt{15}).$$

To make it short: we fail again because a new irrationality shows up if we multiply $2$ elements given in this basis: $\sqrt{30}$ emerges.

However the basis $$(1,\sqrt2,\sqrt3,\sqrt5,\sqrt6,\sqrt{10},\sqrt{15},\sqrt{30}).\tag 1$$

works. Consider the following matrix of the possible products of the elements of the basis

$$\begin{matrix} \times&\sqrt2&\sqrt3&\sqrt5&\sqrt6&\sqrt{10}&\sqrt{15}&\sqrt{30}\\ \\ \sqrt2 & 2&\sqrt6&\sqrt{10}&2\sqrt3&2\sqrt5&\sqrt{30}&2\sqrt{15}\\ \sqrt3 &\sqrt6&3&\sqrt{15}&3\sqrt2&\sqrt{30}&3\sqrt5&3\sqrt{10}\\ \sqrt5 &&&\ 5\\ \sqrt6&&&&\ 6\\ \sqrt{10}&&&&&\ 10&&\ \cdots\\ \sqrt{15}&&&&&&\ 15\\ \sqrt{30}&&&\cdots&&&&\ 30. \end{matrix}$$

Anybody could go on and check that no new "irrationalities" will show up in this multiplication table.

So our final basis is $(1)$: That is we have $8$ dimensions as the high level algebraic argumentation show in the other examples.

The weakness of this elementary solution is twofold. (1) It cannot be generalized for more complicated cases. (2) So far, it did not prove anything else but that the multiplication works. Of course one can find $1=1+0\times\sqrt 2+\cdots$ and $0$ easily.

So I proved only that at least $8$ dimensions are needed.