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.

  • $\begingroup$ 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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.