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I have a query that arose regarding this exercise that I already tried. I have already proven that $L=\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha)=\mathbb{Q}(\alpha)$ with $\alpha=\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$

So given an element in $L$, I can consider this element as:

i) $a+b\alpha$ with $a,b\in\mathbb{Q}$ (here I consider $L=\mathbb{Q}(\alpha)$

ii) $a+b\alpha$ with $a,b\in \mathbb{Q}(\sqrt{2},\sqrt{3})$ (here I consider $\mathbb{Q}(\sqrt{2},\sqrt{3})(\alpha)$

that is, can I consider either of the two options according to my convenience?

I ask it because I am trying to prove that $\varphi:L\to L$ with $\varphi(\alpha)=\sqrt{(2-\sqrt{2})(3+\sqrt{3})}$ and $\varphi(q)=q$ for all $q\in\mathbb{Q}$ is a automorphism (i.e. $\varphi\in \text{Aut}(L/\mathbb{Q})$), and when taking an element in $L$ this question arose

auto-correction 1. The above should say that

i) $a_0+a_1\alpha+a_2\alpha^2+\cdots +\alpha_7 \alpha^7$ with $a_i\in\mathbb{Q}$ (here I consider $L=\mathbb{Q}(\alpha)$

ii) $a_0+a_1\alpha+a_2\alpha^2+\cdots +\alpha_7 \alpha^7$ with $\in \mathbb{Q}(\sqrt{2},\sqrt{3})$ (here I consider $\mathbb{Q}(\sqrt{2},\sqrt{3})(\alpha)$ Because $\left\{1,\alpha,\alpha^2,\ldots, \alpha^7\right\}$ is the basis of the extension

auto-correction 2. The above should say that

i) $a_0+a_1\alpha+a_2\alpha^2+\cdots +\alpha_7 \alpha^7$ with $a_i\in\mathbb{Q}$ (here I consider $L=\mathbb{Q}(\alpha)$

ii) $a_0+a_1\alpha$ with $a_i\in \mathbb{Q}(\sqrt{2},\sqrt{3})$ (here I consider $L=\mathbb{Q}(\sqrt{2},\sqrt{3})(\alpha)$ Because $\left\{1,\alpha\right\}$ is the basis of the extension

Now, I undertands!

eraldcoil
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  • Yes, you can express an element of $L$ in either form. But why is $\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha) = \mathbb{Q}(\alpha)$? – Pranav Chinmay Nov 29 '20 at 01:21
  • I proves that $\sqrt{2},\sqrt{3}\in \mathbb{Q}(\alpha)$ it is possible to write $\sqrt{2},\sqrt{3}$ in terms of $\alpha, \alpha^2$, etc – eraldcoil Nov 29 '20 at 01:24
  • I meant to ask: how do you prove that? It is not immediate to my eyes. – Pranav Chinmay Nov 29 '20 at 01:28
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    @PranavChinmay The element $\alpha^2$ lives in the biquadratic extension $\Bbb Q(\sqrt 2, \sqrt 3)$. It is clear that $\Bbb Q(\alpha^2)$ is none of $\Bbb Q, \Bbb Q(\sqrt 2), \Bbb Q(\sqrt 3), \Bbb Q(\sqrt 6)$, hence it must be the whole $\Bbb Q(\sqrt 2, \sqrt 3)$. This shows that $\sqrt 2, \sqrt 3 \in \Bbb Q(\alpha^2) \subseteq \Bbb Q(\alpha)$. – WhatsUp Nov 29 '20 at 02:28
  • @WhatsUp Thank you. I suppose you show that $\mathbb{Q}(\alpha^2)$ is none of $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$, $\mathbb{Q}(\sqrt{6})$ by noting that $\alpha^2$ is permuted nontrivially by all three of the nontrivial automorphisms in $Gal(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q})$, while $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{6}$ are all fixed by at least one of the nontrivial automorphisms? – Pranav Chinmay Nov 29 '20 at 02:33
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    @PranavChinmay If $(2 + \sqrt 2)(3 + \sqrt 3)$ is in $\Bbb Q(\sqrt 2)$, then $3 + \sqrt 3$ is in $\Bbb Q(\sqrt 2)$, which is impossible. For $\Bbb Q(\sqrt 6)$ one can rewrite the product as $6 + \sqrt 6 + \sqrt 2(3 + \sqrt 6)$. – WhatsUp Nov 29 '20 at 02:53
  • Linking the other question. – Jyrki Lahtonen Nov 29 '20 at 04:28
  • The extension $\mathbb Q(\alpha)/\mathbb Q(\sqrt{2},\sqrt{3})$ is of degree $2,$ not degree $8$ like $\mathbb Q(\alpha)/\mathbb Q.$ – spaceisdarkgreen Nov 29 '20 at 05:20
  • I understood now. It should have written $a_0+a_1\alpha$ in the case $L=\mathbb{Q}(\sqrt{2},\sqrt{3})(\alpha)$ because $\left{1,\alpha\right}$ is the basis in this case. – eraldcoil Nov 29 '20 at 10:06

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