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The first part follows from the fact that for any algebraic elements $a_i$ the following holds:

$$F(a_1,...,a_n)=(F(a_1,...,a_k))(a_{k+1},...,a_n)$$

where $F$ is the field generated by the $a_i$. The second part is giving me trouble. The $a$ has to be such that for every $\lambda_i$ there exist $\mu_j$ such that the following holds (and vice versa since the fields must be the same):

$$\lambda_0+\lambda_1a+...+\lambda_3a^{3}=\mu_0+\mu_1\sqrt{3}+\mu_2i+\mu_3\sqrt{3}i$$

I don't know where to go from here. Any help would be greatly appreciated.

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In many cases $a=\sqrt{3}+i$ is a good guess. Let's check that this $a$ works. Obviously $a\in\mathbb{Q}(\sqrt{3},i)$ and hence $\mathbb{Q}(a)\subseteq\mathbb{Q}(\sqrt{3},i)$. For the other direction note that:

$\frac{1}{a}=\frac{1}{\sqrt{3}+i}=\frac{\sqrt{3}-i}{4}$

This is an element in $\mathbb{Q}(a)$. if we multiply by $4$ we get $\sqrt{3}-i\in\mathbb{Q}(a)$. Hence $\sqrt{3}=\frac{(\sqrt{3}+i)+(\sqrt{3}-i)}{2}\in\mathbb{Q}(a)$, and then $i=\sqrt{3}-(\sqrt{3}-i)\in\mathbb{Q}(a)$. So $\mathbb{Q}(\sqrt{3},i)\subseteq\mathbb{Q}(a)$.

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  • $\begingroup$ Nice ...........+1 $\endgroup$ – Aqua Aug 17 at 14:50

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