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.

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