Let $K$ be a field extension of $F$ and $L$ be also a field extension of F. If $\alpha \in K$ and $\beta \in L$ are both algebraic over $F$, then show that there is an isomorphism of fields: $\mu : F(\alpha) \to F(\beta)$, which is the identity on the subfield $F$ and which maps $\alpha \mapsto \beta$ iff the monic irreducible polynomials for $\alpha$ and $\beta$ over $F$ are the same.
Suppose that $f (x)$ is the minimal polynomial of both $\alpha$ and $\beta$ over $F$. Then there exist two isomorphisms $\mu : F[x]/(f (x)) \to F(\alpha)$ and $\phi: F[x]/(f (x)) \to F(\beta)$. Now take the the composite isomorphism $ψ = \phi \mu^{−1} : F(\alpha) \to F(\beta)$. This shows that $F(\alpha)$ and $F(\beta)$ are isomorphic under $\psi$.
But how to prove the other direction that is $\psi:F(\alpha) \to F(\beta)$ is an isomorphism such that the given conditions hold then $\alpha,\beta$ has same minimal polynomial.
Let $f_{1},f_{2}$ be the minimal polynomials of $\alpha,\beta$ respectively. Suppose that we have an isomorphism $\Psi:F(\alpha)\rightarrow F(\beta)$ such that $\Psi(\alpha)=\beta$.
Since $f_{1}(\alpha)=0$, the image $\Psi(f_{1}(\alpha))=f_{1}(\Psi(\alpha))=f_{1}(\beta)$ also vanishes, which implies that $f_{2}$ divides $f_{1}$. On the other hand, $0=f_{2}(\beta)=f_{2}(\Psi(\alpha))=\Psi(f_{2}(\alpha))$. By the injectivity of $\Psi$, $f_{2}(\alpha)=0$, thus $f_{1}|f_{2}$. Since $f_{1},f_{2}$ are both monic, we have $f_{1}=f_{2}$.