## The conjugates of an element over a given field need not lie in the field

Let $F(\theta)$ be an algebraic extension of a field $F$ and let $\alpha \in F(\theta)$. Must the conjugates of $\alpha$ over $F(\theta)$ be in $F(\theta)$?

Let $\theta = \sqrt[4]{5}$. We computed in a previous exercise that $\beta = 3 + 2i \sqrt[4]{5}$ is a conjugate of $\alpha = 3 + 2\sqrt[4]{5}$ over $\mathbb{Q}(\theta)$. If $\beta \in \mathbb{Q}(\theta)$, then surely $i \in \mathbb{Q}(\theta)$.

Let $\delta = a + b\theta + c\theta^2 + d\theta^3$ and suppose $\delta^2 = -1$. Comparing coefficients, we have (1) $a^2 + 5c^2 + 10bd = -1$, (2) $2ab + 10cd = 0$, (3) $2ac + b^2 + 5d^2 = 0$, and (4) $2ad + 2bc = 0$. From (2), we have $a = -5cd/b$. Substituting into (4), we have $c(b^2 - 5d^2) = 0$. If $c = 0$, then $a = 0$. From (1), we have $b = -1/10d$. Substituting into (3), we have $1/100d^2 + 5d^2 = 0$; this is a contradiction since the left hand side of this equation must be positive. So $c \neq 0$, and we have $b^2 = 5d^2$. Combined with (3), we have $ac + b^2 = 0$; substituting for $a$, we have $d = b^3/5c^2$, and indeed $a = -b^2/c$. Substituting into (4), we have $-b(b^4 - 5c^3) = 0$. If $b = 0$, then $a = d = 0$, and (1) yields $5b^2 = -1$, a contradiction since the left hand side is positive. So $b \neq 0$, and $b^4 = 5c^3$. Substituting $a$ and $d$ into (3), we have $b^6 = 5b^2c^4$, so that $b^4 = 5c^4$. Thus $c = 1$. Substituting $a$, $c$, and $d$ into (1), we have $3b^4 + 5 = -1$. This is a contradiction since the left hand side is positive.

In particular, $i \notin \mathbb{Q}(\theta)$; so $\beta \notin \mathbb{Q}(\theta)$. So the conjugates of an element $\alpha$ over some field need not lie in the field.