Separable residue elements adjoin across a maximal subfield

Let $\bar{p}(T)\in \rho (K)[T]$ be the minimal polynomial of $\bar{u}$. Since $\bar{u}$ is separable over $\rho (K)$, one has ${\bar{p}}^{\mathrm{\prime }}(\bar{u})\mathrm{\ne }0$. Lift the coefficients of $\bar{p}$ through the residue isomorphism $\rho :K\to \rho (K)$ to a monic polynomial $p(T)\in K[T]\subseteq A[T]$.

By [L1], the simple residue root $\bar{u}$ of $\bar{p}$ lifts uniquely to some $u\in A$ with $p(u)=0$. Then $K[u]$ is an integral domain finite over $K$, and its fraction field $K^{\mathrm{\prime }}:=K(u)$ sits inside $A$ because every nonzero element of $K[u]$ has nonzero residue, hence is a unit in the local ring $A$. The residue image of $K^{\mathrm{\prime }}$ contains both $\rho (K)$ and $\bar{u}$.

The residue map is injective on $K^{\mathrm{\prime }}$: if $x\mathrm{/}y\in K^{\mathrm{\prime }}$ has zero residue, then $\rho (x)=0$, so $x=0$ because $K[u]\to k$ remains injective on polynomials of degree smaller than the minimal polynomial of $\bar{u}$. Moreover, $u\mathrm{\notin }K$ because $\bar{u}\mathrm{\notin }\rho (K)$. Thus $K^{\mathrm{\prime }}$ is a strictly larger residue-injective subfield containing a lift of $\bar{u}$.

Therefore every separable residue element adjoins across a maximal residue-injective subfield. The role of [L2] is to show exactly why this contradicts maximality in the later corollary.