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I'm trying to find a basis for $\mathbb{Q}(i,\sqrt{2}+i,\sqrt{3}+i)$ and $\mathbb{Q}(\sqrt[4]{3},i)$ over $\mathbb{Q}.$ For $\mathbb{Q}(i,\sqrt{2}+i,\sqrt{3}+i)$, I think the basis is $\{1,i,\sqrt{2},\sqrt{3},\sqrt{6},\sqrt{2}i,\sqrt{3}i,\sqrt{6}i\}$. But I'm not sure and I don't know how to find a basis for this.

For $\mathbb{Q}(\sqrt[4]{3},i)$, I think the basis is $\{1,i,\sqrt[4]{3},\sqrt[3]{3},\sqrt{3},\sqrt[4]{3}i,\sqrt[3]{3}i,\sqrt{3}i\}$. Again, I'm not sure and I want to know in both cases, how can I find a basis.

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  • $\begingroup$ Can you describe the $\mathbb{Q}(\sqrt[4]{3},i)$ set? $\endgroup$ – user261263 Apr 8 '16 at 3:37
  • $\begingroup$ $\mathbb{Q}(\sqrt[4]{3},i)=\{a+b\sqrt[4]{3}+ci\})$, right? $\endgroup$ – Kelan Apr 8 '16 at 3:39
  • $\begingroup$ Right. $\sqrt[3]{3}$ doesn't seem to be in $\mathbb{Q}(\sqrt[4]{3},i)$ $\endgroup$ – user261263 Apr 8 '16 at 3:42
  • $\begingroup$ Yes, I noticed that. But I'm pretty sure the basis contains 8 elements. I couldn't find the other 2 elements. $\endgroup$ – Kelan Apr 8 '16 at 3:46
  • $\begingroup$ You want to replace cube root of 3 with 3^(3/4). $\endgroup$ – Vik78 Apr 8 '16 at 4:35
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You are right, if you are sure that the basis contains 8 elements, because

$$ [\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}]=[\mathbb{Q}(i)(\sqrt [4]{3}):\mathbb{Q}(i)]\cdot [\mathbb{Q}(i):\mathbb{Q}]$$

And for $i$ there is $f (X)=X^2+1$ the minimal polynomial over $\mathbb{Q}$, because it is irreducible. $g (X)=X^4-3$ is the minimal polynomial of $\sqrt [4]{3} $ over $\mathbb {Q}(i) $, since it is irreducible with prime element $p=3$ in Eisenstein and $\mathbb{Z}[i]$ is PID with $Quot (\mathbb{Z}[i])=\mathbb{Q}(i)$. So

$$ dim_\mathbb{Q}(\mathbb{Q}(\sqrt[4]{3},i))=[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}]=[\mathbb{Q}(i)(\sqrt [4]{3}):\mathbb{Q}(i)]\cdot [\mathbb{Q}(i):\mathbb{Q}]=\deg(g)\cdot \deg(f)=4\cdot 2=8$$

If we find 8 elements which are linear independent, then we are done. This 8 elements are

$$\{1,i,\sqrt[4]{3},\sqrt{3},3^\frac{3}{4},i\sqrt[4]{3}, i\sqrt {3}, i3^\frac{3}{4}\}.$$

The linear independence is just a matter of linear algebra.

In the same way you can show with the minimal polynomials $f (X)=X^2+1$, $g (X)=X^2-2$ and $h (X)=X^2-3$ that $$[\mathbb{Q}(i,\sqrt {2},\sqrt {3}):\mathbb{Q}]=8.$$ Just use the multiplicativity formula for degrees of extensions two times. Now you have to find 8 linear independent elements in $\mathbb{Q}(i,\sqrt{2},\sqrt{3})$, namely

$$\{1,i,\sqrt{2},\sqrt{3},\sqrt{6}, \sqrt{2}i,\sqrt{3}i,\sqrt{6}i\}.$$

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