# Sign of these complex embeddings

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 $$-$$?

• 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 $\epsilon$? – Julian Mejia May 27 at 15:45