Generic freeness over a Noetherian domain

We prove by induction on $n\ge 0$ the following claim $P(n)$: for every finitely generated $A$-algebra $B=A[b_1,\dots ,b_n]$, every finitely generated $B$-module $M$ admits $0\mathrm{\ne }a\in A$ with $M_a$ free over $A_a$. Since every finitely generated $A$-algebra has such a presentation by [F2], $P(n)$ for the $n$ belonging to $B$ yields the statement. [F2, F9] 1.2 The case $n=0$: here $B$ is the image of $A$ and $M$ is a finitely generated $A$-module. Choose a prime filtration $0=M_0\subset \cdots \subset M_r=M$ with quotients $A\mathrm{/}{\mathfrak{p}}_i$ as in [F3] (the empty filtration when $M=0$). For each $i$ with ${\mathfrak{p}}_i\mathrm{\ne }0$ choose $0\mathrm{\ne }{a}_i\in {\mathfrak{p}}_i$; if all ${\mathfrak{p}}_i=0$ put $a=1$. Otherwise put $a={\prod }_{i:{\mathfrak{p}}_i\mathrm{\ne }0}{a}_i$, a product of finitely many nonzero elements of the domain $A$, hence nonzero. For each $i$ exactness of localisation [F4] identifies $(M_i\mathrm{/}{M}_{i-1}{)}_a$ with $(M_i{)}_a\mathrm{/}(M_{i-1}{)}_a$, and $(A\mathrm{/}{\mathfrak{p}}_i{)}_a=A_a\mathrm{/}{\mathfrak{p}}_i{A}_a$ equals $0$ when ${\mathfrak{p}}_i\mathrm{\ne }0$ (because $a\in {\mathfrak{p}}_i{A}_a$) and equals $A_a$ when ${\mathfrak{p}}_i=0$. So the localised filtration has quotients $0$ or $A_a$. A short exact sequence $0\to E^{\mathrm{\prime }}\to E\to E^{\mathrm{\prime }\mathrm{\prime }}\to 0$ of $A_a$-modules with $E^{\mathrm{\prime }},E^{\mathrm{\prime }\mathrm{\prime }}$ free is split: for a basis $(e_j)$ of $E^{\mathrm{\prime }\mathrm{\prime }}$ choose preimages $u_j\in E$, one for each basis element (simultaneously for all $j$, by AC [F8]), and the resulting map $E^{\mathrm{\prime }\mathrm{\prime }}\to E$ is a section. Hence $E\cong E^{\mathrm{\prime }}\oplus E^{\mathrm{\prime }\mathrm{\prime }}$ is free, with the union of bases as a basis [F1]. Applying this successively from $M_0$ up the finite filtration shows that $M_a$ is free over $A_a$. [F1, F3, F4, F8] 1.3 Assume $n\ge 1$ and $P(n-1)$, and let $B=A[b_1,\dots ,b_n]$ and $M$ be as in $P(n)$. Put $A^{\mathrm{\prime }}=A[b_1,\dots ,b_{n-1}]$, so that $B=A^{\mathrm{\prime }}[b_n]$ by [F2], and put $x=b_n$. Choose $m_1,\dots ,m_t\in M$ generating $M$ as a $B$-module [F1], and set $M_0=A^{\mathrm{\prime }}{m}_1+\cdots +A^{\mathrm{\prime }}{m}_t\subseteq M$ and $M_k={\sum }_{j=0}^k{x}^j{M}_0$ for $k\ge 0$. Each $M_k$ is a finitely generated $A^{\mathrm{\prime }}$-module and $M_0\subseteq M_1\subseteq \dots$ is an ascending chain; the union is $M$, because $M={\sum }_{i}B{m}_i={\sum }_i{A}^{\mathrm{\prime }}[x]m_i={\sum }_{j\ge 0}{x}^j{M}_0$. [F1, F2] 1.4 For each $k\ge 0$ the $A^{\mathrm{\prime }}$-linear map ${\phi }_k\text{\hspace{0.17em}⁣}:M_0\to M_{k+1}\mathrm{/}{M}_k$, $m\mapsto x^{k+1}m+M_k$, is surjective, since $M_{k+1}=M_k+x^{k+1}{M}_0$; hence $Q_k:=M_{k+1}\mathrm{/}{M}_k$ is a quotient of the finitely generated $A^{\mathrm{\prime }}$-module $M_0$. The kernels $K_k=\ker {\phi }_k$ form an ascending chain of $A^{\mathrm{\prime }}$-submodules of $M_0$. The ring $A^{\mathrm{\prime }}$ is finitely generated over the Noetherian ring $A$, hence is a quotient of a polynomial ring $A[x_1,\dots ,x_{n-1}]$, hence Noetherian by [F6] and [F7]. Therefore $M_0$ is a Noetherian $A^{\mathrm{\prime }}$-module by [F5], so the chain $(K_k)$ stabilizes: there is $N\ge 0$ with $K_k=K_N$ for all $k\ge N$. Consequently $Q_k\cong M_0\mathrm{/}{K}_N=:Q$ as $A^{\mathrm{\prime }}$-modules for every $k\ge N$. [F5, F6, F7] 1.5 Apply the induction hypothesis $P(n-1)$ to the finitely generated $A^{\mathrm{\prime }}$-modules $M_0$, $Q_0,\dots ,Q_{N-1}$ and $Q$: there are nonzero elements $a_{\ast },a_0,\dots ,a_{N-1},a_{\mathrm{\infty }}\in A$ after which all these modules become free over the corresponding localisation of $A$. Put $a=a_{\ast }\left({\prod }_{k=0}^{N-1}{a}_k\right)a_{\mathrm{\infty }}\mathrm{\ne }0$, a product of finitely many nonzero elements of the domain $A$. Then $(M_0{)}_a$ and each $(Q_k{)}_a$ for $k<N$ and $Q_a$ are free $A_a$-modules, and for $k\ge N$ exactness of localisation [F4] turns the isomorphism $Q_k\cong Q$ into $(Q_k{)}_a\cong Q_a$, so $(Q_k{)}_a$ is free over $A_a$ for every $k\ge 0$. [F4] 1.6 Finally $0\to (M_k{)}_a\to (M_{k+1}{)}_a\to (Q_k{)}_a\to 0$ is exact by [F4], and $(Q_k{)}_a$ is free, so for each $k$ we may choose an $A_a$-linear section $s_k\text{\hspace{0.17em}⁣}:(Q_k{)}_a\to (M_{k+1}{)}_a$ of the projection, i.e. a simultaneous choice of preimages of the elements of a basis of $(Q_k{)}_a$ (AC [F8]). Define ${\theta }_0=\mathrm{i}\mathrm{d}$ and ${\theta }_{k+1}\text{\hspace{0.17em}⁣}:(M_0{)}_a\oplus (Q_0{)}_a\oplus \cdots \oplus (Q_k{)}_a\to (M_{k+1}{)}_a$ by ${\theta }_{k+1}={\theta }_k\oplus s_k$; each ${\theta }_k$ is an isomorphism by the splitting just described, and ${\theta }_{k+1}$ restricts to ${\theta }_k$, so the ${\theta }_k$ glue to an $A_a$-module isomorphism $M_a={\bigcup }_k(M_k{)}_a\cong (M_0{)}_a\oplus {⨁}_{k\ge 0}(Q_k{)}_a$. A direct sum of free modules is free, a basis being the union of bases of the summands [F1], so $M_a$ is a free $A_a$-module, of rank the cardinality of that basis. This completes the induction and the proof. [F1, F8] The Axiom of Choice is inherited by the prime-filtration input in step 1.2 and is used in the splitting arguments of steps 1.2 and 1.6 to choose basis preimages and the family of sections simultaneously. The remaining displayed choices are finite. The empty module is free of rank zero, and for $M=0$ the prime filtration is empty so the base case produces $a=1$. $\mathrm{\square }$