A flat local map splits regular sequences into base and fibre parts

For any ideal $I\subset R$ and $R\mathrm{/}I$-module $L$, balanced maps give $L{\otimes }_{R\mathrm{/}I}(S\mathrm{/}IS)\cong L{\otimes }_{R}S$, by $l\otimes (s+IS)\mapsto l\otimes s$, with inverse $l\otimes s\mapsto l\otimes (s+IS)$. These formulas are independent of representatives because $IL=0$, by Universal property of the tensor product for balanced maps into abelian groups. Hence $S\mathrm{/}IS$ is flat over $R\mathrm{/}I$. The quotient identification is also $M{\otimes }_{R}R\mathrm{/}I\cong M\mathrm{/}IM$ naturally.

First establish the lifting assertion for a single $y\in \mathfrak{n}$ which is a nonzerodivisor on $S\mathrm{/}\mathfrak{m}S$. Flatness identifies ${\mathfrak{m}}^{a}S\mathrm{/}{\mathfrak{m}}^{a+1}S$ with $({\mathfrak{m}}^a\mathrm{/}{\mathfrak{m}}^{a+1}){\otimes }_{R}S$, by tensoring the inclusion and quotient sequences of these ideals. The latter is $({\mathfrak{m}}^a\mathrm{/}{\mathfrak{m}}^{a+1}){\otimes }_k(S\mathrm{/}\mathfrak{m}S)$, by the same balanced formulas as step 1.1. The first factor is finite dimensional over $k$, since $R$ is Noetherian; a finite basis identifies this with finitely many copies of the fibre. Multiplication by $y$ is therefore injective on each graded piece.

The exact sequences $0\to {\mathfrak{m}}^{a}S\mathrm{/}{\mathfrak{m}}^{a+1}S\to S\mathrm{/}{\mathfrak{m}}^{a+1}S\to S\mathrm{/}{\mathfrak{m}}^{a}S\to 0$ now show inductively that $y$ acts injectively on every $S\mathrm{/}{\mathfrak{m}}^{a}S$: an element killed by $y$ maps to zero in the last quotient and then is zero in the first term. If $ys=0$ in $S$, this implies $s\in {\bigcap }_a{\mathfrak{m}}^{a}S=0$ by The Krull intersection is the $(1-a)$-torsion submodule, and it vanishes in the Jacobson-radical case, applied to the finite $S$-module $S$ and the ideal $\mathfrak{m}S\subseteq \mathfrak{n}=J(S)$. This invocation uses that theorem's stated AC boundary.

For every proper ideal $I\subset R$, step 1.1 gives a flat local map $R\mathrm{/}I\to S\mathrm{/}IS$ of Noetherian local rings, with the same closed fibre. Applying steps 2.1–3.1 to this map proves that $y$ is also a nonzerodivisor on $S\mathrm{/}IS$. Thus $ys\in IS$ implies $s\in IS$.

Put $T=S\mathrm{/}yS$. Right exactness (Tensoring is right exact) applied to $S\stackrel{y}{\to }S\to T\to 0$ identifies $I{\otimes }_{R}T$ with $IS\mathrm{/}yIS$, because flatness identifies $I{\otimes }_{R}S$ with $IS$. The multiplication map $I{\otimes }_{R}T\to T$ is injective: if $u\in IS$ equals $ys$, step 4.1 gives $s\in IS$, so $u\in yIS$. For $I=R$ it is the identity. The ideal criterion Flatness is equivalent to preserving injections and to the ideal and finitely generated ideal tests proves $T$ flat over $R$.

Flat tensoring preserves injectivity of each base regular element $x_i$ on the successive quotients. Those quotients remain flat over the corresponding base quotient by step 1.1. The ideals generated by the images of $x_i$ are contained in $\mathfrak{n}$, so their quotients are nonzero. Thus the base sequence is $S$-regular, in agreement with Localisation And Faithfully Flat Base Change Of Regular Sequences. Set $R^{\mathrm{\prime }}=R\mathrm{/}(x_1,\dots ,x_r)R$ and $S^{\mathrm{\prime }}=S\mathrm{/}(x_1,\dots ,x_r)S$; this is flat local with unchanged closed fibre.

Apply steps 2.1–5.1 repeatedly to $R^{\mathrm{\prime }}\to S^{\mathrm{\prime }}$ and the successive $y_j$ quotients. Fibre regularity supplies the nonzerodivisor hypothesis at each stage; the conclusion supplies both the lifted nonzerodivisor and flatness needed at the next stage. The final quotient is nonzero because all generators lie in $\mathfrak{n}$. This proves regularity in the convention of Regular Sequence On A Module and every asserted quotient-flatness statement, including empty sequences.