Suppose we are given two simple algebraic extensions $F(\alpha), F(\beta)$ (over a field $F$) with an $F$-isomorphism $\phi$ of extensions (i.e. fixing $F$) such that $\phi(\alpha)=\beta$.

Then we want to prove that $\alpha, \beta$ have the same minimal polynomial.

I'd like to argue that $F(\alpha) \simeq F[x]/(f(x))$ and $F(\beta) \simeq F[y]/(g(y))$ where $f,g$ are the minimal polys of $\alpha, \beta$ and from this I should get an isomorphism $F[x] \simeq F[y]$ sending $x$ to $y$ that is induced by $\phi$. But I'm not sure if this argument is actually beating around the bush, or if there is a more direct or obvious argument?


We have $ 0 = \phi(0) = \phi(f(\alpha)) = f(\phi(\alpha)) = f(\beta) $ implies that $g$ divides $f$.

By symmetry, $f$ divides $g$ and so $f=g$ (since both are monic).


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