## 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 $\alpha$.