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Let $\alpha$ be a root of $f = x^4 - 7$, hence $\alpha$ is a fourth root of $7$.

Consider the number field $F = \mathbb{Q}(\sqrt[4]{7})$, which is of degree $4$.

Since it is of degree $4$, there are $4$ complex embeddings $\sigma_1, \sigma_2, \sigma_3, \sigma_4 : F \rightarrow \mathbb{C}$.

The conjugates of $\alpha$ are the roots of $f_{\alpha} = f$, which are $$\alpha, \alpha \epsilon, \alpha \epsilon^2, \alpha \epsilon^3$$ where $\epsilon = e^{\frac{2 \pi i}{4}}$, a primitive fourth root of unity.

Hence, the complex embeddings are given by $\sigma_1(\alpha) = \alpha, \sigma_2 (\alpha) = \alpha \epsilon, \sigma_3 (\alpha) = \alpha \epsilon^2, \sigma_4 (\alpha) = \alpha \epsilon^3 $.

Now, I am trying to find $\sigma_{i} (\sqrt{7}) $ for $i = 1,2,3,4.$

Since $\alpha^4 = \sqrt[4]{7}$, we get that $\sqrt{7} = \pm \alpha^2$

Hence, $\sigma_1 (\sqrt{7}) = \sigma_1 (\pm \alpha^2) = \pm \sigma_1 (\alpha)^2 = \pm \alpha^2 $ and similarly, $\sigma_{2} (\sqrt{7}) = \pm \sigma_2 (\alpha)^2 = \pm \alpha^2 \epsilon^2 = \mp \alpha^2, \sigma_3 (\sqrt{7}) = \pm \alpha^2 \epsilon^4 = \pm \alpha^2$ and $\sigma_4 (\sqrt{7}) = \pm \alpha^2 \epsilon^6 = \mp \alpha^2$.

Now, how do I decide on the sign of these complex embeddings? Which ones are $+$ and which ones are $-$?

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  • $\begingroup$ I think your signs depend on the choose of $\alpha$, why don't you take $\alpha=\sqrt[4]{7}$?. Or at least express $\alpha=\sqrt[4]{7}i^{k}$ where $k\in\{0,1,2,3\}$. Also, why don't you use $i$ instead of $\epsilon$? $\endgroup$ – Julian Mejia May 27 at 15:45

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