A base-field isomorphism extends to an isomorphism between splitting fields of corresponding polynomials

Let $P(n)$ assert the theorem for every such datum of degree $n$. If $n=0$, both root sets are empty, so $E=F$ and $E^{\mathrm{\prime }}=F^{\mathrm{\prime }}$; the required extension is $\sigma$.

Let $n\ge 1$ and assume $P(m)$ for all $m<n$. Choose a root $\alpha \in E$ of $f$, and let $p$ be its minimal polynomial over $F$. By [F2], $p\mid f$. Thus $p^{\mathrm{\prime }}={\sigma }_{\ast }p$ divides $f^{\mathrm{\prime }}$ in $E^{\mathrm{\prime }}[x]$. Since $f^{\mathrm{\prime }}$ is a product of linear factors there, unique factorisation in $E^{\mathrm{\prime }}[x]$ from [F5] makes $p^{\mathrm{\prime }}$ split over $E^{\mathrm{\prime }}$; choose a root $\beta \in E^{\mathrm{\prime }}$ of $p^{\mathrm{\prime }}$.

By [F3], $\sigma$ extends to an isomorphism ${\sigma }_1:F(\alpha )\to F^{\mathrm{\prime }}(\beta )$. Factor $f=(x-\alpha )g$ in $F(\alpha )[x]$. Applying coefficient transport by ${\sigma }_1$ gives $f^{\mathrm{\prime }}=(x-\beta )g^{\mathrm{\prime }}$ with $g^{\mathrm{\prime }}=({\sigma }_1{)}_{\ast }g$, and $\mathrm{deg}g=n-1$.

The field $E$ is a splitting field of $g$ over $F(\alpha )$: it splits $g$, and it is generated by $\alpha$ together with the other roots of $f$, which are the roots of $g$. Similarly, $E^{\mathrm{\prime }}$ is a splitting field of $g^{\mathrm{\prime }}$ over $F^{\mathrm{\prime }}(\beta )$. The induction hypothesis extends ${\sigma }_1$ to an isomorphism $E\to E^{\mathrm{\prime }}$.

The base case and inductive step establish $P(n)$ for all $n$ by [F1].