If a is a nonzero algebraic element over F, then all of its conjugates are nonzero

Let F be a field and let \alpha be a nonzero algebraic element over F. Prove that the conjugates of \alpha are nonzero.


Let p(x) be the minimal polynomial of \alpha over F. Recall that the conjugates of \alpha are precisely the roots of p(x). If p(x) is linear, then the only conjugate of \alpha is \alpha itself. Suppose p(x) has degree at least 2. If one of the roots of p is 0, then in fact p(x) = xp^\prime(x); note that \alpha is a root of p^\prime and that p^\prime has degree strictly smaller than that of p, violating the minimalness of p(x). So 0 is not a conjugate of \alpha.

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