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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?

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marked as duplicate by Dietrich Burde, Shaun, tilper, Arnaud D., MPW Jan 27 '17 at 20:11

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    $\begingroup$ If you know about the primitive element theorem, you should find that this field is actually equal to $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})$ if I remember correctly, and then finding its degree through a "telescopic basis" argument shouldn't be too hard. If you don't know about the first theorem I mentionned, you can try to prove my claim and then conlude anyway $\endgroup$ – Max Jan 27 '17 at 12:39
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$[\mathbb Q(\sqrt2 + \sqrt3 + \sqrt5) : \mathbb Q ]=[\mathbb Q(\sqrt2 , \sqrt3 , \sqrt5) : \mathbb Q ]=[\mathbb Q(\sqrt2 ,\sqrt3 , \sqrt5) : \mathbb Q (\sqrt2 , \sqrt3)].[\mathbb Q(\sqrt2 ,\sqrt3 ) : \mathbb Q (\sqrt2 )].[\mathbb Q(\sqrt2 ) : \mathbb Q ]=2.2.2=8$

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

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    $\begingroup$ But how do you know that $\mathbb{Q}(\sqrt{2}+\sqrt{3}+\sqrt{5})=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ ? $\endgroup$ – Rene Schipperus Jan 27 '17 at 13:16
  • $\begingroup$ @ReneSchipperus:I am writing my ans. in two different comments. $\endgroup$ – MatheMagic Jan 27 '17 at 13:54
  • $\begingroup$ Since, $\sqrt2+\sqrt{3}+\sqrt[]{5}\in \mathbb{Q}(\sqrt2,\sqrt{3},\sqrt[]{5})$ then it must be algebraic. Assume, $|\mathbb{Q}(\sqrt2+\sqrt{3}+\sqrt[]{5}):\mathbb Q|=2$ which means that $\sqrt2+\sqrt{3}+\sqrt[]{5}$ is a root of second degree polynomial $x^2+bx+c$, by this you can easily reach contradiction.Similarly it can not be $4$. Now, you know that $\mathbb{Q}(\sqrt2+\sqrt{3}+\sqrt[]{5})\subseteq \mathbb{Q}(\sqrt2,\sqrt{3},\sqrt[]{5})$ and $|\mathbb{Q}(\sqrt2,\sqrt{3},\sqrt[]{5}):\mathbb Q|=8$ $\endgroup$ – MatheMagic Jan 27 '17 at 13:54
  • $\begingroup$ then we must have $|\mathbb{Q}(\sqrt2+\sqrt{3}+\sqrt[]{5}):\mathbb Q|=8$ as it cannot be two or four, since their dimensions are equal, we must have equality, $$\mathbb{Q}(\sqrt2,\sqrt{3},\sqrt[]{5})=\mathbb{Q}(\sqrt2+\sqrt{3} + \sqrt[]{5})$$ $\endgroup$ – MatheMagic Jan 27 '17 at 13:55
  • $\begingroup$ Actually, at the same time I found this other current question about a the same topic math.stackexchange.com/questions/2116309/… see the comments to the answer there. $\endgroup$ – Rene Schipperus Jan 27 '17 at 14:01
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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.

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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.

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