A field isomorphism transports polynomials coefficientwise and carries roots, factorizations, and splitting to roots, factorizations, and splitting

Apply [F1] to $\sigma$ and the indeterminate $x\in F^{\mathrm{\prime }}[x]$. This gives ${\sigma }_{\ast }$ with the displayed coefficient formula. Applying the same construction to ${\sigma }^{-1}$ gives its inverse, so ${\sigma }_{\ast }$ is an isomorphism.

For $f={\sum }_i{a}_i{x}^i$, the homomorphism laws give $\tau (f(\alpha ))={\sum }_i\sigma (a_i)\tau (\alpha {)}^i=({\sigma }_{\ast }f)(\tau (\alpha ))$. Thus $f(\alpha )=0$ if and only if $({\sigma }_{\ast }f)(\tau (\alpha ))=0$, because $\tau$ is injective.

Applying the construction of step 1.1 to $\tau$ gives a coefficientwise isomorphism ${\tau }_{\ast }:E[x]\to E^{\mathrm{\prime }}[x]$ extending ${\sigma }_{\ast }$. It transports every product factorisation and, in particular, a linear factorisation of $f$ to one of ${\sigma }_{\ast }f$. Since $\tau$ is bijective, step 1.2 gives a bijection of root sets, and $\tau (F(\text{roots of }f))=F^{\mathrm{\prime }}(\text{roots of }{\sigma }_{\ast }f)$. The splitting-field claim follows from [F3].