Complete local modular systems and integral idempotent lifting

Put $B=A_{\mathfrak{p}}$. It is a nonzero Noetherian local domain by [F1]. Its residue is $\mathrm{Frac}(A\mathrm{/}\mathfrak{p})$: modulo $\mathfrak{p}$, the inverted elements are exactly the nonzero elements of $A\mathrm{/}\mathfrak{p}$. This identifies that field with $k_0\subseteq k$. Localizing a domain is injective. The elementary prime correspondence used below follows directly from fractions: a prime disjoint from the denominators extends to a prime, and contraction recovers it because a fraction belongs to the extended ideal exactly when its numerator does.

If $\mathfrak{p}\cap \mathbb{Z}=0$, every nonzero integer is a unit in $B$, so fraction arithmetic identifies $B$ with a prime localization of $\mathbb{Q}[v,v^{-1}]$. The latter is a PID: ideals in its localization are generated by the image of a generator of their contracted ideal in $\mathbb{Q}[v]$. Thus $B$ is a field or has a nonzero principal maximal ideal $(a)$. In the second case [F2] and the chain $(0)⊊(a)$ give dimension one. The class of $a$ modulo $(a{)}^2$ is nonzero, since $a=a^{2}b$ in the domain would imply $1=ab$ with $a$ a nonunit. Hence its embedding dimension is one. In the field case both dimensions are zero. Thus $B$ is regular in both cases.

Otherwise $\mathfrak{p}\cap \mathbb{Z}=(\mathrm{\ell })$ for a prime integer $\mathrm{\ell }$. The prime $\mathfrak{p}\mathrm{/}(\mathrm{\ell })$ in the PID ${\mathbb{F}}_{\mathrm{\ell }}[v,v^{-1}]$ is zero or $(f)$ for an irreducible Laurent polynomial $f$. If it is zero, $\mathfrak{p}=(\mathrm{\ell })$ and the principal-maximal-ideal argument of step 2.1 proves regularity. Otherwise choose a Laurent lift $G$ of $f$; then the maximal ideal of $B$ is $\mathfrak{m}=(\mathrm{\ell },G)$. The chain $(0)⊊(\mathrm{\ell })B⊊\mathfrak{m}$ and [F2] give $\dim B=2$. The quotient $B\mathrm{/}(\mathrm{\ell })$ is the domain ${\mathbb{F}}_{\mathrm{\ell }}[v,v^{-1}{]}_{(f)}$. If $\mathrm{\ell }=r{\mathrm{\ell }}^2+s\mathrm{\ell }G+t{G}^2$ in $B$, reduction modulo $\mathrm{\ell }$ gives $\bar{t}{f}^2=0$, hence $t=\mathrm{\ell }{t}^{\mathrm{\prime }}$. Cancellation of $\mathrm{\ell }$ then gives $1=r\mathrm{\ell }+sG+t^{\mathrm{\prime }}{G}^2\in \mathfrak{m}$, impossible. Thus $\mathrm{\ell }\mathrm{\notin }{\mathfrak{m}}^2$. If $a\mathrm{\ell }+bG\in {\mathfrak{m}}^2$, reducing modulo $\mathrm{\ell }$ gives $\bar{b}f\in (f^2)$, so $b\in \mathfrak{m}$. It follows that $a\mathrm{\ell }\in {\mathfrak{m}}^2$, and $\mathrm{\ell }\mathrm{\notin }{\mathfrak{m}}^2$ forces $a\in \mathfrak{m}$, since otherwise $a$ is a unit. The two displayed generators therefore give a basis of $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$; its dimension is two. Thus $B$ is regular in this final case.

By [F3], $R=\widehat{B}$ is complete separated Noetherian local with residue $k_0$, is regular, and is a domain. Faithful flatness makes $B\to R$ injective: if $b$ maps to zero, flat tensoring the inclusion $bB\hookrightarrow B$ gives an injection $bB{\otimes }_{B}R\hookrightarrow R$ whose image is zero; faithfulness forces $bB=0$, so $b=0$. Therefore $A\hookrightarrow R$ induces $\mathbb{Q}(v)\hookrightarrow F$, with the required residue map. If $\mathfrak{p}=0$, this also covers the field $B=R=\mathbb{Q}(v)$ with zero maximal ideal.

For a finite projective module $P$ over any local ring, lift a basis of $P\mathrm{/}\mathfrak{m}P$ to $r$ elements of $P$. Nakayama on the finite cokernel makes $R^r\to P$ surjective. Projectivity splits it as $R^r=P\oplus Q$, with $Q$ finite. Reduction of this direct sum shows $Q\mathrm{/}\mathfrak{m}Q=0$, so Nakayama gives $Q=0$. Hence $P$ is free of rank $r$. Over our domain this rank equals both ${\dim }_F(F{\otimes }_{R}P)$ and ${\dim }_{k_0}(k_0{\otimes }_{R}P)$.

Let $x_0\in C$ lift an idempotent modulo ${\mathfrak{m}}_{R}C$. Define ${\delta }_i=x_i^2-x_i$ and $x_{i+1}=x_i+(1-2x_i){\delta }_i$. All factors here are polynomials in $x_i$, so integer-polynomial expansion gives ${\delta }_{i+1}={\delta }_i^2(4{\delta }_i-3)$. Thus ${\delta }_i\in {\mathfrak{m}}_R^{2^i}C$ and $x_{i+1}-x_i\in {\mathfrak{m}}_R^{2^i}C$. The finite-module completeness in [F4] gives a limit $q\equiv x_0$ modulo ${\mathfrak{m}}_{R}C$. Multiplication is continuous, since these ideal powers are two-sided ideals, so separatedness gives $q^2=q$. This construction involves no division and works in every characteristic.

For any right $C$-module $X$, the $R$-linear decomposition $X=Xq\oplus X(1-q)$ shows that projection $X\mapsto Xq$ commutes with every scalar extension. It is exact: under a surjection $X\to Y$, lift $y\in Y$ to $x$, and then $xq$ lifts $yq$; kernels and injections follow by restriction of the same projection. If $0\to I_{i+1}\to I_i\to B_i\to 0$ is a filtration sequence, projection gives $0\to I_{i+1}q\to I_{i}q\to B_{i}q\to 0$. When $B_i$ is finite free, $B_{i}q$ is finite projective and hence free by step 5.1; the sequence splits as $R$-modules. It therefore retains its exact layer after scalar extension, including reduction to $k_0$.

Finally let $J_R={⨁}_E{M}_{d_E}(R)$ act on an $R$-finite-free module $X$. With matrix units $e_{ij}^E$, put $Q_E=e_{11}^{E}X$. It is a direct summand of $X$, hence finite free by step 5.1. The map $R^{d_E}{\otimes }_R{Q}_E\to e_{E}X$ sending $u_i\otimes x$ to $e_{i1}^{E}x$ has inverse $y\mapsto {\sum }_i{u}_i\otimes e_{1i}^{E}y$, by the matrix-unit identities. Therefore the generic and residue multiplicities of the $E$-standard simple are both ${\mathrm{rank}}_R{Q}_E$. This proves the final assertion and completes the lemma.