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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.

Myshkin
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  • as $\sqrt{2},\sqrt{3}\in \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ we should have $\sqrt{6}\in \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$... do you think one can get $\sqrt{6}$ from ${1,\sqrt{2},\sqrt{3},\sqrt{5}}$ when considered as basis.... do you see there is some thing wrong??? –  Sep 14 '13 at 06:47
  • the simplest basis is $\sqrt{2}^{n_2}\sqrt{3}^{n_3}\sqrt{5}^{n_5}$ for all $n_2,n_3,n_5\in{0,1}$ – user8268 Sep 14 '13 at 06:54
  • a related question (possibly duplicate): http://math.stackexchange.com/questions/113689/proving-that-left-mathbb-q-sqrt-p-1-dots-sqrt-p-n-mathbb-q-right-2n-f?rq=1 – user8268 Sep 14 '13 at 06:58
  • Also: http://math.stackexchange.com/questions/30687/the-square-roots-of-different-primes-are-linearly-independent-over-the-field-of – Belgi Sep 14 '13 at 19:44
  • My experience has been that deriving the order of these extensions is very interesting, especially the last. Worth the trouble to work it out. – Bill Kleinhans Sep 14 '13 at 19:06

3 Answers3

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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.

Orat
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davidlowryduda
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  • Thank you for the answer but I don't know why degree $4$, why not $3$? – Myshkin Sep 14 '13 at 06:57
  • Use the tower property. $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ has subfields of $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ which are degree $2$ extensions, so we know that $2$ divides the degree of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}$ and that also it is at most $4$. But then, since $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$ (you should check this as practice!) we have that $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ is a proper extension of $\mathbb{Q}(\sqrt{2})$ and so the full extension is degree $4$. – Ryan Sullivant Sep 14 '13 at 07:10
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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.

Orat
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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

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