Scheme effectivity and flatness for marked Artin lifts

By [F1] the section sequence $0\to N\to C\to B\to 0$ is exact. Its kernel squares to zero, so its prime ideals correspond to those of $B$ and have the same topology. Each stalk $E_x$ is local: an element lifting a unit of ${\mathcal{O}}_{X,x}$ is a unit, since after lifting an inverse their product is $1+n$, whose inverse is $1-n$ for $n^2=0$. Hence $(U,E{\mathrm{\mid }}_U)$ is a locally ringed space. Define the map to $\mathrm{Spec}C$ by sending $x$ to the inverse image in $C$ of the maximal ideal of $E_x$; this is the preceding homeomorphism. On each basic open $D(c)$, the section $c$ is invertible in every stalk and hence as a section of $E$ there, so localization gives $C_c\to \mathrm{\Gamma }(D(c),E)$. These maps agree under further localization and define the sheaf map of a morphism of locally ringed spaces. At corresponding primes it induces a map between exact sequences $0\to N_{\mathfrak{p}}\to C_{\mathfrak{p}}\to B_{\mathfrak{p}}\to 0$ and $0\to M_x\to E_x\to {\mathcal{O}}_{X,x}\to 0$. The outside maps are isomorphisms by [F1]; the middle is therefore an isomorphism by a direct kernel and quotient chase. Thus the locally ringed space is affine. These identifications are restrictions of the same sheaf $E$, so the affine charts agree on intersections and give the asserted scheme, including its exact kernel.

A flat module $V$ over a local Artin ring $(R,\mathfrak{m},\kappa )$ is free, allowing arbitrary rank. Choose a $\kappa$-basis of $V\mathrm{/}\mathfrak{m}V$, lift it to $V$, and map the free module $F=R^{(T)}$ to $V$. The map on associated graded modules for the finite $\mathfrak{m}$-adic filtrations is an isomorphism. Indeed tensor exactness identifies ${\mathfrak{m}}^{j}V\mathrm{/}{\mathfrak{m}}^{j+1}V$ with $({\mathfrak{m}}^j\mathrm{/}{\mathfrak{m}}^{j+1}){\otimes }_{R}V$, which in turn is $({\mathfrak{m}}^j\mathrm{/}{\mathfrak{m}}^{j+1}){\otimes }_{\kappa }(V\mathrm{/}\mathfrak{m}V)$; for $F$ the same formula holds. The map between these modules is the identity on the first factor and the chosen basis isomorphism on the second. Since ${\mathfrak{m}}^n=0$ for some $n$, induction up this finite filtration shows $F\to V$ is an isomorphism. This uses no finite-generation assumption on $V$.

On the affine chart in step 1.1, $B$ is flat over $A$ and hence free by step 1.2. Lift an $A$-basis $(b_t)$ to elements of $C$ and form $F^{\mathrm{\prime }}=(A^{\mathrm{\prime }}{)}^{(T)}\to C$. Its reduction modulo $I$ is the isomorphism $A^{(T)}\to B$, because the multiplication hypothesis says $IC=N$ and $C\mathrm{/}IC=B$. Its cokernel $Q$ satisfies $Q=IQ$, whence $Q=I^{2}Q=0$. Its kernel lies in $I{F}^{\mathrm{\prime }}$, since its reduction is injective. But $I{F}^{\mathrm{\prime }}\to N$ is the isomorphism $I^{(T)}\to I{\otimes }_{A}B\to N$ supplied by the same basis and the prescribed kernel map. The kernel is zero. Thus $C$ is free over $A^{\mathrm{\prime }}$ and is flat. In the reverse direction, if $C$ is flat over $A^{\mathrm{\prime }}$ and $C\mathrm{/}IC=B$, tensor $0\to I\to A^{\mathrm{\prime }}\to A\to 0$ with $C$: this gives $0\to I{\otimes }_{A^{\mathrm{\prime }}}C\to C\to B\to 0$. Since $I^2=0$, $I{\otimes }_{A^{\mathrm{\prime }}}C=I{\otimes }_{A}B$, with its canonical multiplication map. Thus the marking of the kernel is forced, rather than an extra optional identification. The same calculation on stalks proves the converse without assuming that a lift is affine in advance.

For local finite presentation, choose an affine chart where $B$ is finitely presented over $A$, and choose finite algebra generators $b_1,\dots ,b_r$ and lifts $z_1,\dots ,z_r$ in $C$. Every $c\in C$ is, modulo $N$, a polynomial in the $z_j$ with coefficients lifted from $A$. Its remaining part lies in $N=I{\otimes }_{A}B$ and is a finite sum of elements $ip(z_1,\dots ,z_r)$. Hence $A^{\mathrm{\prime }}[Z_1,\dots ,Z_r]\to C$ is surjective. The ring $A^{\mathrm{\prime }}$ is Noetherian because its ideals are $k$-subspaces of a finite-dimensional vector space; [F2] makes this polynomial ring Noetherian, so the kernel is finitely generated. This is a finite presentation of $C$. Base change gives exactly $B$ since $IC=N$. Steps 1.1 and 2.1 consequently prove flatness and local finite presentation on an open cover of the constructed scheme. Finally, a sheaf-algebra isomorphism fixes the underlying space and is local on stalks by step 1.1, hence is a scheme isomorphism, and conversely a marked scheme isomorphism gives precisely that sheaf map. The specified reduction of $X$ is retained by composing its unchanged marking; no new reduction is chosen.