## Find the conjuages of a given algebraic element over QQ

Find the minimal polynomial and conjugates of each of the following elements over $\mathbb{Q}$: $\sqrt{3}$, 4, $\sqrt{-3}/5$, and $\sqrt[3]{5}$.

Note that $p(x) = x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3})$ has $\sqrt{3}$ as a root. $p(x)$ is Eisenstein at 3, hence irreducible over $\mathbb{Q}$. Thus $p(x)$ is the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}$. The conjugates of $\sqrt{3}$ are $\pm \sqrt{3}$.

Note that $q(x) = x - 4$ has 4 as a root. Since $q(x)$ is linear, it is irreducible, and so is the minimal polynomial of 4 over $\mathbb{Q}$. The only conjugate of 4 is 4 itself.

Note that $t(x) = 25x^2 + 3 = (5x + \sqrt{3})(5x - \sqrt{3})$ has $\sqrt{-3}/5$ as a root. $t(x)$ is Eisenstein at 3, hence irreducible over $\mathbb{Q}$. Thus $t(x)$ is the minimal polynomial of $\sqrt{-3}/5$. The conjugates of $\sqrt{-3}/5$ are $\pm \sqrt{-3}/5$.

Note that $s(x) = x^3-5 = (x - \sqrt[3]{5})(x - \sqrt[3]{5}(\frac{-1}{2} + \frac{\sqrt{3}}{2}i))(x - \sqrt[3]{5}(\frac{-1}{2} - \frac{\sqrt{3}}{2}i))$ has $\sqrt[3]{5}$ as a root.$s(x)$ is Eisenstein at 5, hence irreducible over $\mathbb{Q}$. The conjugates of $\sqrt[3]{5}$ are $\sqrt[3]{5}$, $\sqrt[3]{5}(\frac{-1}{2} + \frac{\sqrt{3}}{2}i)$, and $\sqrt[3]{5}(\frac{-1}{2} - \frac{\sqrt{3}}{2}i)$.