Polygraph common ideal bases and freeness of every alternating-ideal power

Let $B=\mathbb{C}[y_1,\dots ,y_n]$, $E={\mathbb{A}}^2$, and let $x_j,y_j$ ($j\in [n]$) and $a_i,b_i$ ($i\in [l]$) be coordinates on $E^n\times E^l$. Let $f:[l]\to [n]$ range over all functions.

$$\begin{gathered} W_f=V(a_i-x_{f(i)},b_i-y_{f(i)}:i \\ in[l]) \end{gathered}$$, $R(n,l)=C[x,y,a,b]\mathrm{/}{\bigcap }_f$ $I(W_f)$.

For $0\le r\le n$, $0\le k\le l$ and integer m let Y(m,r,k) be the reduced union of $$\begin{gathered} V(x_j:j \\ inT)\cap W_f \end{gathered}$$ over pairs (T,f) satisfying $\mathrm{\mid }T\cap [r]\setminus f([k])\mathrm{\mid }\ge m$, and let I(m,r,k) be its ideal in R(n,l). The cases $m\le 0$ give $I=0$ and $m>r$ give $I=R$; also $I(n,n,k)=R$ if $k>0$. Only finitely many nontrivial ideals are involved at each (n,l).

We construct an x-homogeneous B-basis of R(n,l) with a subset spanning every I(m,r,k). The construction is by induction on (n,l), with a secondary induction on x-degree. The x-degree counts x,a and the y-degree counts y,b. Each x-degree component is finite over B: the monic relations ${\prod }_j(a_i-x_j)={\prod }_j(b_i-y_j)=0$ make R finite over C[x,y]. Coordinate rings of all reduced arrangements used below embed in direct sums of their component polynomial rings, so they are torsion-free over B.

Let $U_1\subset Spec$ B be the distinct-y locus and $U_2$ the locus with at most one equality among the $y_j$. The complement of $U_2$ has codimension at least two. At a point above $U_1$ only one $W_f$ occurs. At a point above $U_2\setminus U_1$, with the sole equal pair indexed $N=\{p,q\}$, the local arrangement is a product of a polynomial affine space with R(N,L), where L consists of those i whose assigned y-value equals $y_p=y_q$. Indeed assignments outside this pair are fixed and eliminate their $a_i,b_i$, while assignments inside it are exactly the two choices p,q. This is an equality of local ideals, since the unwanted component ideals contain an invertible y-difference.

Here are the coefficient-extension principles used throughout. If a globally specified subset T of a torsion-free B-module is a basis after localization at every point of $U_2$, it is a global B-basis. Expand an element at the generic point; its finitely many coefficients are the same in every localization and have no pole at any height-one prime. Polynomial unique factorization implies they lie in B. Torsion-freeness gives the resulting equality globally. Likewise two free submodules of one torsion-free module which coincide on $U_2$ coincide globally. Finally, if T is a B-basis of M and T∩J spans J over C(y), it spans J over B, since the unique basis coefficients outside T∩J must vanish. Local equality of ideals at all points above $U_i$ implies equality after B-localization at $U_i$, by testing all prime localizations of the latter module. The global elements in the first principle are essential: mere local freeness off codimension two does not imply freeness.

Over $U_1$ the ring is a sum over f of C(y)[x]. In a coordinate-subspace union, standard square-free monomial ideals therefore show that the x-degree Hilbert function of its generic fiber counts pairs (e,f), $$\begin{gathered} e \\ in{N}^n \end{gathered}$$, with the corresponding monomial $x^e$ not killed on at least one component. For Y(m,r,k), this condition is

|[r] minus $(\{j:e_j>0\}\cup f([k]))\mathrm{\mid }\ge m$.

For R(n,l) the generic Hilbert series is $n^l\mathrm{/}(1-t{)}^n$. This counting statement follows by embedding into the component rings, localizing to separate their distinct y-values, and using ordinary monomial bases in each surviving coordinate-subspace union. In a fixed x-degree, homogeneous generators modulo (y) lift to generators by graded Nakayama; if their number equals this generic rank, they are a free basis, since a relation would vanish generically in a free source.

Take $n=2$ and $l>0$. Index basis candidates p[e,f] by $$\begin{gathered} e=(e_1,e_2) \\ in{N}^2 \end{gathered}$$ and f:[l]→{1,2}. For $e=0$ define

$p[0,f]={\prod }_{j>1,f(j)\mathrm{\ne }f(1)}(b_j-b_1)$ · $(b_1-y_{2}iff(1)=1,else1)$.

For $e=(0,h)$, $h>0$, split $f^{-1}(1)=S\bigsqcup T$ with S its first $min(h,\mathrm{\mid }{f}^{-1}(1)\mathrm{\mid })$ indices, and define

$$\begin{gathered} p[(0,h),f]=x_2^{h-\mathrm{\mid }S\mathrm{\mid }}{\prod }_{i \\ inS}(a_i-x_1-x_2){\prod }_{j \\ inT}(b_j-y_2) \end{gathered}$$.

Let θ exchange $x_1,x_2$ and $y_1,y_2$, fix a,b, and compose f with the transposition. Set $p[(h,0),f]=x_1$ θp[(0,h−1),θf]. If both $e_j>0$, set $p[e,f]=(x_1{x}_2{)}^q$ p[e−(q,q),f], $q=min(e_1,e_2)$.

These elements form a B-basis. To verify spanning, first reduce modulo $y,x_1{x}_2$. It is enough that the p[(0,h),f] span further modulo $x_1$; multiplication by $x_1$ and the transposition then cover the complementary summand. Vanishing on the $W_f$ gives the relations $a_i^2=a_i{x}_2$, $a_i{b}_i=b_i^2=x_2{b}_i=0$, and $a_j{b}_i=a_i{b}_j$ for $i<j$ in this quotient. Thus it is spanned by $$\begin{gathered} x_2^q{\prod }_{i \\ inS}{a}_i{\prod }_{j \\ inT}{b}_j \end{gathered}$$, where S,T are disjoint, every index in S precedes every index in T, and $q=0$ if T is nonempty. If S,T are both nonempty, choose $h=\mathrm{\mid }S\mathrm{\mid }$ and $f^{-1}(1)=S\cup T$; the displayed candidate reduces exactly to the monomial because all extra $x_2{b}_j$ terms vanish. If T is empty, use $h=q+\mathrm{\mid }S\mathrm{\mid }$ and induct down from larger q, since expanding $\prod (a_i-x_2)$ leaves the desired term and terms with larger $x_2$ exponent. If S is empty, the $h=0$ candidates with $f(1)=1$ give all square-free b-monomials containing $b_1$; those with $f(1)=2$ give the remaining square-free b-monomials modulo those already obtained. The constant is included. This proves spanning in every fixed degree; multiplying back by powers of $x_1{x}_2$ and the $x_1$-transposed candidates proves spanning modulo (y). The degree enumerator of all candidates is $2^l\mathrm{/}(1-t{)}^2$, exactly the generic Hilbert series. The coefficient-count principle of the argument beginning at step 1.1 proves freeness and independence.

The candidates in I(m,r,k) are exactly the ones with

|[r] minus $(\{j:e_j>0\}\cup f([k]))\mathrm{\mid }<m$.

Here are explicit equations sufficient to check membership, including all nontrivial cases:

$I(2,2,0)=(x_1,x_2,a_1,\dots ,a_l)$, $I(1,2,k)=(x_1{x}_2,a_i-x_1-x_2,b_i-b_1:i\le k)$, $I(1,1,k)=(x_1,a_i-x_2,b_i-y_2:i\le k)$.

For $k=0$ the empty sums are omitted. Each right-hand ideal vanishes on its defining arrangement. Substitution of the displayed products puts each selected p in the corresponding right-hand ideal. The one exceptional-looking product is $x_1(b_1-y_1)$ in I(1,2,k), $k>0$; the relation $(a_1-x_2)(b_1-y_1)=0$ makes it equal to $-(a_1-x_1-x_2)(b_1-y_1)$. All other factors either include $x_1{x}_2$, include a chosen $a_i-x_1-x_2$ or $b_i-b_1$, or vanish after setting $x_1=0,a_i=x_2,b_i=y_2$ for $i\le k$. This exhausts the three listed cases and the trivial ideals. The generic count from the argument beginning at step 1.1 gives exactly the complementary basis elements, so the selected elements span the entire ideal, not just a subideal. Thus the equations above and the common-basis assertion hold scheme-theoretically.

It follows that every sum/intersection in the lattice generated by these ideals and independent coordinate ideals is radical. Indeed the common basis makes every quotient by such a lattice ideal free over B; over C(y) the arrangement is a union of coordinate subspaces, hence reduced, and a torsion-free algebra embedding in its reduced generic fiber is reduced. After the $U_2$ product reduction, the same radical assertion holds locally on $U_2$ for the lattice generated by I(m,r,k) for arbitrary n. Sums and intersections remain computed by the corresponding subsets of the common basis in the two-coordinate factor and monomials in the other x variables.

Two additional local facts will be used. Setting any x or a coordinate to zero in a reduced union of $W_f$ is reduced: simultaneous translation of all x,a coordinates by minus that coordinate defines a retraction of the quotient into the original reduced ring. For $n=2$ the ideal $I(1,1,t-1)+(a_t)$ is radical as well. Eliminate the fixed pairs $a_i=x_2,b_i=y_2$ for $i<t$, reducing to $(x_1,a_1)$. Its radical is the union of the $f(1)=1$ component with $x_1=0$ and the $f(1)=2$ component with $x_1=x_2=0$. Modulo $b_1-y_1$ the ideal is already radical; its colon by $b_1-y_1$ is $(x_1,x_2,a_1,\dots ,a_l,b_1-y_2)$. The inclusions follow from $(b_1-y_1)(b_1-y_2)=0$, $(b_1-y_1)(a_1-x_2)=0$, and the determinant with rows $(a_i,b_i,1),(a_1,b_1,1),(x_1,y_1,1)$, which reduces to $a_i(b_1-y_1)$. Equality of this colon and its radical's colon, together with equality modulo $b_1-y_1$, gives equality with the radical by the modular law. The $U_2$ product reduction gives the same local assertion for all n.

For a point of the polygraph, its admissible functions f form one equivalence class: f(i),g(i) are equivalent when their (x,y) pairs coincide there. The pointwise maximal f in that class maximizes |T∩[r]\f([k])| simultaneously for all m,r,k, where T is the set of zero x-coordinates. Consequently intersections of the Y(m,r,k) are reduced unions with the same f and both cardinality conditions imposed. This proves that their coordinate rings are B-torsion-free. It also proves all set-theoretic cardinality manipulations below without choosing incompatible assignments for two different conditions.

Use the cyclic shift $\theta (j)=j+1$ mod n on x,y and fix a,b. Put $J_t=I(1,1,t)$. If π forgets $a_t,b_t$ and $k<t$, then

Y(1,1,t)∪(Y(m,r,k)∩Y(1,1,t−1)) $=Y(1,1,t)\cup ({\pi }^{-1}\theta Y(m-1,r-1,k)\cap Y(1,1,t-1))$;

if $k\ge t$ replace (m−1,r−1,k) on the right by (m,r−1,k−1). To prove this restrict outside Y(1,1,t). In Y(1,1,t−1), the maximal admissible f then has $f(t)=1$ and 1 absent from f([t−1]); otherwise a larger index with the same point would remove 1 from every f([t]) and put the point in Y(1,1,t). For $k<t$ the removed index 1 contributes exactly one to the cardinality; for $k\ge t$ it is already in f([k]) and contributes zero. Removing that index and the t-th coordinate gives the asserted formulas. Adding back Y(1,1,t) gives the full unions.

For a finite one-variable algebra $S=R[b]\mathrm{/}J$ define, for $m\ge 1$, $E_m(S)=\{c\in R:c{b}^{m-1}+{\sum }_{i=0}^{m-2}{c}_i{b}^i\in J\text{ for some }{c}_i\in R\}$, with the sum empty for $m=1$, and put $E_0(S)=0$. Thus the coefficient is that of the specified term $b^{m-1}$; multiplying a lower-degree relation by a power of $b$ supplies its later-stage leading coefficient. We also write $e_m(S)=E_m(S)$. With $F_m=R⟨1,\dots ,b^{m-1}⟩\subset S$ and $F_0=0$, the cyclic map $R\to F_m\mathrm{/}{F}_{m-1}$ has exactly this kernel, so $F_m\mathrm{/}{F}_{m-1}=R\mathrm{/}{E}_m(S)$. The ideals increase, and a monic relation gives $F_d=S$ and $E_m(S)=R$ for all $m>d$. Their zero sets are the loci where the fiber length is at least $m$. At a base prime $P$, a fiber algebra of vector-space dimension $q=0$ is the zero algebra; Nakayama then gives $S_P=0$, so every $E_m(S{)}_P=R_P$ for $m\ge 1$. For $q>0$, over a residue field the finite one-variable quotient has basis $1,\dots ,b^{q-1}$, where $q$ is its dimension, and Nakayama makes $F_q$ the whole localized algebra but $F_{q-1}$ a proper submodule. Hence $E_q$ is proper there and $E_{q+1}$ is the unit ideal. Partial elimination commutes with flat scalar extension: intersect the relation module $J$ with the finite-degree polynomial module, use flat exactness, and then take the image of the coefficient map. For a union defined by $J_1\cap J_2$ the next inclusion follows.

$E_s(R[b]\mathrm{/}{J}_1)$ $E_t(R[b]\mathrm{/}{J}_2)$ ⊂ $E_{s+t-1}(R[b]\mathrm{/}(J_1\cap J_2))$,

because multiplying two relations adds their leading degrees and multiplies their leading coefficients.

Suppose R has a common B-basis partitioned into disjoint sets $B_j$ so that $E_m$ is spanned by the $B_j$ with $j<m$, and suppose A is B-torsion-free. It is enough to know this elimination-ideal equality locally over $U_2$. The filtration quotients then show that

${\bigcup }_{m\ge 1}$ $B_m$ $\{1,b,\dots ,b^{m-1}\}$

Write $S$ for the finite algebra called $A$ in step 30.1, let $e_m(S)$ denote its actual partial elimination ideal, and put $D_m=B⟨{\bigcup }_{j<m}{\mathcal{B}}_j⟩$. Let $F_m=R⟨1,b,\dots ,b^{m-1}⟩\subset S$, with $F_0=0$. For every base prime $Q$ above which the $y$-tuple lies in $U_2$, equality $e_m(S)=D_m$ after localization at every prime of $R_Q$ implies equality as ideals of $R_Q$: a nonzero quotient of the two ideals would have nonempty support and a prime localization detecting it. Therefore $(F_m\mathrm{/}{F}_{m-1}{)}_Q\cong (R\mathrm{/}{D}_m{)}_Q$ has basis $b^{m-1}{\bigcup }_{j\ge m}{\mathcal{B}}_j$. Since $b$ satisfies a monic relation, $F_d=S$ for some finite $d$; these finitely many successive quotient bases lift to a basis of $S_Q$. The union of their lifts is precisely the set in step 31.1. These are globally specified elements, not locally chosen frames. They are a basis at the generic point, and for every $s\in S$ its finite generic expansion in them has coefficients regular at every height-one prime of $B$, because every such prime is in $U_2$. Unique factorization gives coefficients in $B$; torsion-freeness of $S$ makes this expansion an equality in $S$. Thus the displayed set is a global basis. To prove ideal compatibility, let $I\subset R$ be spanned by a subset $\mathcal{A}\subset \mathcal{B}$ and assume that the pullback to $R$ of $e_m(S\mathrm{/}IS)$ equals $I+e_m(S)$ locally over $U_1$. Work at such a localization. The quotient-filtration identity gives $F_m\mathrm{/}(F_{m-1}+IS\cap F_m)\cong R\mathrm{/}{e}_m(S\mathrm{/}IS)$. On the other hand the modular law and the definition of $e_m(S)$ give $F_m\mathrm{/}(F_{m-1}+b^{m-1}I)\cong R\mathrm{/}(e_m(S)+I)$. Both isomorphisms are induced by the same cyclic generator $b^{m-1}$, so their equal kernels imply $F_{m-1}+IS\cap F_m=F_{m-1}+b^{m-1}I$. Subtracting the indicated element of $b^{m-1}I\subset IS$ shows that multiplication by $b^{m-1}$ gives the actual isomorphism $I\mathrm{/}(I\cap e_m(S))\cong (IS\cap F_m)\mathrm{/}(IS\cap F_{m-1})$. The subset $b^{m-1}{\bigcup }_{j\ge m}(\mathcal{A}\cap {\mathcal{B}}_j)$ is its basis. Induction through the finite filtration therefore shows that ${\mathcal{A}}^{\mathrm{\prime }}=(\mathcal{A}\{1,b,b^2,\dots \})\cap {\mathcal{B}}^{\mathrm{\prime }}$ spans $IS$ generically. Every element of ${\mathcal{A}}^{\mathrm{\prime }}$ already lies in $IS$, and ${\mathcal{B}}^{\mathrm{\prime }}$ is now a global basis of $S$; expansion in that basis shows that every coefficient outside ${\mathcal{A}}^{\mathrm{\prime }}$ vanishes. Hence ${\mathcal{A}}^{\mathrm{\prime }}$ spans $IS$ over $B$. This argument proves ideal compatibility without assuming in advance that $S\mathrm{/}IS$ is torsion-free or reduced.

For an increasingly ordered auxiliary label set $L$, the notation $R(n,L)$ uses those labels and $I_{n,L}(m,r,k^{\mathrm{\prime }})$ counts its first $k^{\mathrm{\prime }}$ labels; after deleting a label the remaining labels are always ordered increasingly. For $1\le k\le l$ and $0\le r\le n$ let Z'(r,k) be the reduced union of $V(a_k)\cap W_f$ over f with f(k)∈[r]\f([k−1]). It is finite over Z(n,L), $L=[l]\setminus \{k\}$, generated by the one remaining coordinate $b_k$; ${\prod }_j(b_k-y_j)=0$. Its partial elimination ideals satisfy

$E_m(Z^{\mathrm{\prime }}(r,k))$ ⊂ $I_{n,L}(m,r,k-1)$

everywhere, with equality locally on $U_2$. The inclusion follows from fiber counts on $U_1$: above an assignment f the allowed points are precisely the indices j∈[r]\f([k−1]) with $x_j=0$. Their $count\ge m$ defines the arrangement on the right, densely on each component, so leading coefficients vanish there.

For the reverse inclusion on $U_1$, the defining ideal of the arrangement contains $$\begin{gathered} {\prod }_{j \\ inT}(x_j,b_k-y_j) \end{gathered}$$, $T=[r]\setminus f([k-1])$. For $1\le m\le \mathrm{\mid }T\mathrm{\mid }+1$, its leading coefficients include every square-free monomial of degree |T|+1−m in those $x_j$, exactly the ideal of the locus with at least m zero coordinates. For $m>\mathrm{\mid }T\mathrm{\mid }$, the monic relation $$\begin{gathered} {\prod }_{j \\ inT}(b_k-y_j) \end{gathered}$$ makes $E_m$ the unit ideal, as required for the empty locus; this also covers T empty. At a $U_2$ point split the indices into the unique pair N and its complement. The components with f(k) outside N give this same coordinate calculation; those with f(k) in N give the two-coordinate special arrangement. In that two-coordinate arrangement, $E_1$ is the image ideal I(1,r,k−1); all $E_m$ for $m>2$ are the unit ideal; for $r=1$ or $k>1$, already $E_m=1$ for $m>1$. The only remaining case is $m=r=2,k=1$, where $$\begin{gathered} I(2,2,0)=(x_1,x_2,a_i:i \\ inL) \end{gathered}$$. These generators are the $b_1$-leading coefficients of

$(b_1-y_2)x_1$, $(b_1-y_1)x_2$, $(b_1-y_1)a_i+(b_i-y_2)x_1$.

They vanish on Z'(2,1), the last by the three-row determinant relation already used in the argument beginning at step 10.1. This proves the two-coordinate equality. In the product reduction, the product inclusion for elimination ideals gives the sum over $s+t=m+1$ of the tensor products of the outside-coordinate ideal for s and the pair ideal for t. The sum is radical by the common two-coordinate basis, and its zero set is the locus where the two counts add to at least m. It is therefore exactly I(m,r,k−1) locally. This verifies the required $U_2$ scheme equality.

Assume inductively that $R=R(n,L)$ has its common basis, put $S=\mathcal{O}(Z^{\mathrm{\prime }}(r,k))$, and set $I=I(m^{\mathrm{\prime }},r^{\mathrm{\prime }},k^{\mathrm{\prime }})\subset R$. The arrangement embedding makes $S$ torsion-free over $B$, and steps 34.1--38.1 give the partial elimination equality over $U_2$, so the first part of step 32.1 produces a global basis of $S$. For its compatibility with $IS$, put $Y^{\mathrm{\prime }}=\mathrm{Spec}(S\mathrm{/}IS)$ and first establish the quotient partial-elimination equality without supposing $IS$ radical. At a point of $U_1$ the base has its unique component $W_f$; the distinct possible values $b_k=y_j$ separate the special-arrangement branches, indexed by $j\in [r]\setminus f([k-1])$ with $x_j=0$. If the base point also belongs to $Y(m^{\mathrm{\prime }},r^{\mathrm{\prime }},k^{\mathrm{\prime }})$, all these fiber points remain in the reduced preimage $Y_{\mathrm{r}\mathrm{e}\mathrm{d}}^{\mathrm{\prime }}$. Consequently its fiber has at least $m$ points at every point of $Y(m^{\mathrm{\prime }},r^{\mathrm{\prime }},k^{\mathrm{\prime }})\cap Y(m,r,k-1)\cap U_1$. These intersections are scheme-theoretically reduced: on the unique component their defining ideals are sums of square-free coordinate-monomial ideals, and such a sum is again a square-free monomial ideal. The fiber-length criterion in step 27.1 therefore implies $e_m(Y_{\mathrm{r}\mathrm{e}\mathrm{d}}^{\mathrm{\prime }})\subset I+I(m,r,k-1)$ locally over $U_1$. If $J\subset R[b_k]$ defines $S$, then $J+IR[b_k]\subset \sqrt{J+IR[b_k]}$, so the leading-coefficient definition gives $e_m(Y^{\mathrm{\prime }})\subset e_m(Y_{\mathrm{r}\mathrm{e}\mathrm{d}}^{\mathrm{\prime }})$. Conversely $I\subset e_m(Y^{\mathrm{\prime }})$ for every $m>0$, since multiplying a constant relation by $b_k^{m-1}$ supplies its leading coefficient, and $e_m(S)\subset e_m(Y^{\mathrm{\prime }})$ because $J\subset J+IR[b_k]$. Using $e_m(S)=I(m,r,k-1)$ over $U_1$ proves the exact scheme identity $e_m(Y^{\mathrm{\prime }})=I+e_m(S)$ there. The second part of step 32.1 now proves that the lifted subset with base coefficient in $I$ spans $IS$. Thus $S\mathrm{/}IS$ is free over $B$. Its generic fiber is reduced: after inverting all $y$-differences, the branches are disjoint, and on the branch $b_k=y_j$ the quotient adds $x_j=0$ to a square-free coordinate-monomial ideal. A square-free monomial ideal remains radical after adding a coordinate variable. Finally $S\mathrm{/}IS$ injects into its generic fiber because it is free over the domain $B$; a nilpotent must have zero image and hence be zero. This proves that $IS$ is globally radical and supplies its common basis, in that order.

For $1\le r\le n$, put $S=\mathcal{O}(Z^{\mathrm{\prime }}(r,k))$, $S^{\mathrm{\prime }\mathrm{\prime }}=\mathcal{O}(Z^{\mathrm{\prime }}(r-1,k))$, and $K=\ker (S\to S^{\mathrm{\prime }\mathrm{\prime }})$. For a base basis vector $p$ define $m(p)$ by $p\in I(m(p)+1,r,k-1)$ and $p\mathrm{\notin }I(m(p),r,k-1)$, and define $m^{\mathrm{\prime }}(p)$ in the same way with $r-1$. The defining counts give the ideal inclusions $I(a,r-1,k-1)\subset I(a+1,r,k-1)\subset I(a+1,r-1,k-1)$ and therefore $m^{\mathrm{\prime }}(p)\le m(p)\le m^{\mathrm{\prime }}(p)+1$. The compatible bases supplied by step 39.1 are ${\mathcal{B}}_1=\{p{b}_k^j:0\le j<m(p)\}$ for $S$ and ${\mathcal{B}}_1^{\mathrm{\prime }}=\{p{b}_k^j:0\le j<m^{\mathrm{\prime }}(p)\}$ for $S^{\mathrm{\prime }\mathrm{\prime }}$; regard the latter displayed polynomials also as their specified lifts in $S$. For each $p$ with $m(p)=m^{\mathrm{\prime }}(p)+1$, the unique excess vector is $p_0=p{b}_k^{m(p)-1}$. Express its image in $S^{\mathrm{\prime }\mathrm{\prime }}$ uniquely as a $B$-linear combination $p_1$ of ${\mathcal{B}}_1^{\mathrm{\prime }}$, and define $q_p=p_0-p_1\in K$. Replacing every excess vector $p_0$ by $q_p$ is a triangular change of basis, with diagonal entries one and correction terms only in the unchanged subset ${\mathcal{B}}_1^{\mathrm{\prime }}$. Hence $\{q_p\}\cup {\mathcal{B}}_1^{\mathrm{\prime }}$ is a basis of $S$. The quotient map sends $q_p$ to zero and sends ${\mathcal{B}}_1^{\mathrm{\prime }}$ isomorphically to its basis in $S^{\mathrm{\prime }\mathrm{\prime }}$, so $\{q_p\}$ is a basis of $K$. To check simultaneous compatibility, fix $I=I(m^{\mathrm{\prime }},r^{\mathrm{\prime }},k^{\mathrm{\prime }})$ in the base. The basis of $IS$ is ${\mathcal{A}}_1=\{p{b}_k^j\in {\mathcal{B}}_1:p\in I\}$ and that of $I{S}^{\mathrm{\prime }\mathrm{\prime }}$ is the analogous subset ${\mathcal{A}}_1^{\mathrm{\prime }}\subset {\mathcal{B}}_1^{\mathrm{\prime }}$. If $p\in I$, then the image of $p_0$ belongs to $I{S}^{\mathrm{\prime }\mathrm{\prime }}$; uniqueness of its basis expansion forces $p_1$ to use only ${\mathcal{A}}_1^{\mathrm{\prime }}$. Thus $q_p\in IS\cap K$. Rewriting the basis ${\mathcal{A}}_1$ by this same triangular replacement shows that $IS$ has basis ${\mathcal{A}}_1^{\mathrm{\prime }}\cup \{q_p:p\in I\}$. Intersecting with the coordinate submodule $K=B⟨q_p⟩$ gives exactly $IS\cap K=B⟨q_p:p\in I⟩$. The construction is simultaneous for all the base ideals and preserves homogeneity, since every unique expansion in a homogeneous quotient basis has the same total degree. No freeness of a general kernel of free modules is used.

Write $B=\mathbb{C}[y_1,\dots ,y_n]$ and $K=\mathbb{C}(y_1,\dots ,y_n)$. Let $R$ be a $B$-algebra, $J,I_1,\dots ,I_q$ ideals, and suppose every $R\mathrm{/}{I}_i$ is $B$-torsion-free. Assume $J$ has a $B$-basis $\mathcal{D}$ with subsets spanning every $I_i\cap J$, and $R\mathrm{/}J$ has a $B$-basis $\mathcal{C}$ with subsets spanning every $\sqrt{I_i+J}\mathrm{/}J$. Suppose that for every intersection $I$ of some of the $I_i$, $I+J$ is radical locally over $U_2$. Here, as in step 6.1, local reducedness means reducedness at every prime of $R$ above the indicated coefficient point; consequently $(I+J{)}_{B_Q}$ is radical for each coefficient prime $Q\in U_2$, since a ring is reduced if all its prime localizations are reduced. This includes $Q=(0)$, and $U_2$ contains every height-one prime of $B$. Then $R$ has a common $B$-basis for $J$ and all $I_i$. The same assertion holds in each fixed homogeneous $x$-degree when $B$ has $x$-degree zero and the given ideals and bases are homogeneous.

Choose one lift $c\in R$ of every vector of $\mathcal{C}$. These lifts together with $\mathcal{D}$ are already a $B$-basis of $R$: projecting a putative relation to $R\mathrm{/}J$ kills its lift coefficients, and subtracting the lifted expansion of an arbitrary element leaves an element of $J$. For a lift $c$, let $\alpha =\{i:\bar{c}\in \sqrt{I_i+J}\mathrm{/}J\}$ and put $I={\bigcap }_{i\in \alpha }{I}_i$, with the empty intersection interpreted as $R$. The elementary identity $\sqrt{I+J}={\bigcap }_{i\in \alpha }\sqrt{I_i+J}$ holds because the zero set of an intersection of ideals is the union of their zero sets. It puts $c$ in $\sqrt{I+J}$. At every $Q\in U_2$, the local radical hypothesis therefore gives $c\in I_{B_Q}+J_{B_Q}$. Hence there is a correction $u_Q\in J_{B_Q}$ with $c+u_Q\in I_{B_Q}$. The class of $u_Q$ in $(J\mathrm{/}(I\cap J){)}_{B_Q}$ is unique: any two corrections differ by $I_{B_Q}\cap J_{B_Q}=(I\cap J{)}_{B_Q}$, since localization is exact. Intersections of the subset-spanned modules $I_i\cap J$ are spanned by the intersections of their subsets of $\mathcal{D}$, so $J\mathrm{/}(I\cap J)$ is free with basis the complementary subset ${\mathcal{D}}_c\subseteq \mathcal{D}$. Expand the unique correction class in that basis. The solution at every $Q$ maps to the unique solution at $Q=(0)$, so its finitely many nonzero coefficients are the same rational functions $a_d\in K$, independent of $Q$, and belong to $B_Q$ for every $Q\in U_2$. In particular they belong to every height-one localization of $B$. A reduced denominator with an irreducible factor would give a pole at that factor's height-one prime, so unique factorization implies $a_d\in B$. Thus $u={\sum }_{d\in {\mathcal{D}}_c}{a}_{d}d$ is a finite global sum in $J$ and $c^{\ast }=c+u$ is a globally defined element of $R$. Its generic class belongs to $I_K$. For $i\in \alpha$, torsion-freeness of $R\mathrm{/}{I}_i$ injects it into its generic localization; hence $c^{\ast }\in I_i$ globally. The set $\mathcal{B}=\mathcal{D}\cup \{c^{\ast }:c\in \mathcal{C}\}$ remains a $B$-basis because each $c^{\ast }\equiv c(\mathrm{m}\mathrm{o}\mathrm{d}J)$. Fix $i$. At $Q=(0)$ the local radical hypothesis gives $(\sqrt{I_i+J}{)}_K=(I_i+J{)}_K$. The corrected lifts associated to the subset $\mathcal{C}\cap (\sqrt{I_i+J}\mathrm{/}J)$ consequently span the image $(I_i+J{)}_K\mathrm{/}{J}_K$, and the given subset $\mathcal{D}\cap I_i$ spans $(I_i\cap J{)}_K$. The exact sequence $0\to (I_i\cap J{)}_K\to (I_i{)}_K\to (I_i+J{)}_K\mathrm{/}{J}_K\to 0$ shows that $\mathcal{B}\cap I_i$ spans $(I_i{)}_K$. Every element of $I_i$ has its unique finite expansion in the global basis $\mathcal{B}$; after passing to $K$, its coefficients outside this subset vanish, so they already vanish in $B$. Therefore $\mathcal{B}\cap I_i$ spans $I_i$ over $B$, simultaneously for every $i$. If a fixed $x$-degree is being constructed, choose homogeneous lifts and take the homogeneous component of every local correction in that degree. Since $B$ has $x$-degree zero, the uniqueness and coefficient-extension argument use only basis vectors of that degree, and the corrected lifts remain homogeneous. This proves the claimed gluing assertion and its degreewise version.

First, $R(n,l)\mathrm{/}{J}_l$ is $R([n]\setminus \{1\},[l])\otimes C[y_1]$, since on its components $x_1=0$ and none of the f(i) equals 1. A common ideal basis for R(n−1,l), shifted by θ, is therefore a common basis here. The restriction of Y(m,r,k) for $r>0$ is Y(m−1,r−1,k) in the remaining indices, directly from the cardinality definition; $r=0$ gives only trivial ideals.

For the second induction stage put $J_s=I(1,1,s)$, $R=R(n,l)\mathrm{/}{J}_{t-1}$ and $J=J_t\mathrm{/}{J}_{t-1}$, with $1\le t\le l$. Assume the stipulated homogeneous common bases for $R(n,l)\mathrm{/}{J}_t$ and $R(n,l-1)$. The colon identity $J_{t-1}:(a_t)=J_t$ follows componentwise in the reduced arrangement: on a component of $Y(1,1,t-1)$ one has $x_1=0$ and $1\mathrm{\notin }f([t-1])$, and $a_t=x_{f(t)}$ vanishes identically precisely when $f(t)=1$. Taking the colon removes just these components and leaves $Y(1,1,t)$. Thus multiplication by $a_t$ identifies $(a_t)\subset R$ with $R(n,l)\mathrm{/}{J}_t$, so $(a_t)$ has a homogeneous free $B$-basis $B_1$. We also have a homogeneous free basis $B_2$ of $R\mathrm{/}\sqrt{(a_t)}$: transpose the first and $t$-th auxiliary pairs, so $a_t=0$ becomes $a_1=0$, and project away the first pair. The resulting arrangement is the reduced preimage of $Y_{n,L}(1,1,t-1)$ inside the special arrangement $Z^{\mathrm{\prime }}(n,1)$, with $L=[l]\setminus \{1\}$ increasingly ordered. Indeed the old first $t-1$ auxiliary indices become exactly the first $t-1$ labels of $L$, and their avoidance of the distinguished main index $1$ is unchanged. The radical-extension and quotient-basis conclusions of step 39.1 therefore give $B_2$. Choose global homogeneous lifts of $B_2$ to $R$. By step 22.1, $J_{t-1}+(a_t)$ is radical locally above $U_2$; equivalently $(a_t)$ and $\sqrt{(a_t)}$ have equal localizations there. Hence $B_1$ together with those lifts is a basis of $R$ after every $B$-localization in $U_2$, by the exact sequence through $(a_t)$. All the vectors are global elements and $R$ is $B$-torsion-free, since it is a reduced arrangement ring. The coefficient-extension argument of step 6.1 now makes this a global homogeneous free $B$-basis of $R$. This establishes freeness of $R$ before any assertion about the freeness of $J$.

Let $L=[l]\setminus \{t\}$ in increasing order, $Z^{\mathrm{\prime }}=Z^{\mathrm{\prime }}(n,t)$ and $Z^{\mathrm{\prime }\mathrm{\prime }}=Z^{\mathrm{\prime }}(n-1,t)$, and write $K=I(Z^{\mathrm{\prime }\mathrm{\prime }})\mathrm{/}I(Z^{\mathrm{\prime }})\subset \mathcal{O}(Z^{\mathrm{\prime }})$. Separate the components of $Z^{\mathrm{\prime }}$ as $Z^{\mathrm{\prime }}=Z^{\mathrm{\prime }\mathrm{\prime }}\cup Z_1$, where $Z_1$ is the union with $f(t)=n$, $n\mathrm{\notin }f([t-1])$ and $x_n=0$. There is no common component. The cyclic main-index shift $\theta$ gives $Z_0=\theta Z_1$, and $Y(1,1,t-1)=Y(1,1,t)\cup Z_0$, again with no common component: its new components have $f(t)=1$, $1\mathrm{\notin }f([t-1])$ and $x_1=0$. Denote the shifted ring $\mathcal{O}(\theta Z^{\mathrm{\prime }})$ by $\widehat{S}$ and the shifted ideal by $\widehat{K}=\theta K$. Restriction to $Z_0$ embeds both $J$ and $\widehat{K}$ into $R_0=\mathcal{O}(Z_0)$. For example a function of $J$ vanishing also on $Z_0$ vanishes on the reduced union $Y(1,1,t)\cup Z_0$ and is zero in $R$; the proof for $\widehat{K}$ uses the other reduced union. This is injectivity of restriction, not freeness of an arbitrary kernel.

The needed equality is the equality of scheme-theoretic intersections, locally above $U_2$, $\theta Z^{\mathrm{\prime }\mathrm{\prime }}\cap Z_0=Y(1,1,t)\cap Z_0,$ or equivalently $I(\theta Z^{\mathrm{\prime }\mathrm{\prime }})+I(Z_0)=J_t+I(Z_0)$ in the localized ambient arrangement ring. Neither intersection is assumed reduced.

The equation is empty away from a two-$y$ collision involving main index $1$: a point of $Z_0$ has an admissible assignment sending $t$ to $1$, while membership in either other arrangement requires an assignment sending $t$ away from $1$. When $y_1$ differs from every other $y_j$ these cannot describe the same point. At the unique collision $N=\{1,q\}$, discard the auxiliary pairs whose $y$-values are outside $N$; they have fixed assignments and give only an affine factor. If the $t$-th pair is outside this factor the intersection is again empty. Otherwise relabel $N$ as $\{1,2\}$, keep the remaining auxiliary labels in increasing order, and let the old label $t$ become $s$. In this two-coordinate ring the three arrangements have exactly the following ideals.

Put $I_0=(x_1,a_s-x_1,b_s-y_1,a_i-x_2,b_i-y_2(i<s)),$ $I_{\ast }=(x_2,a_s-x_2,b_s-y_2,a_i-x_1,b_i-y_1(i<s)),$ $J_{\ast }=(x_1,a_s-x_2,b_s-y_2,a_i-x_2,b_i-y_2(i<s))\mathrm{.}$ These are respectively the reduced ideals of $Z_0$, $\theta Z^{\mathrm{\prime }\mathrm{\prime }}$ and $Y(1,1,s)$ in the local two-coordinate factor. For the first, eliminate the displayed fixed auxiliary pairs. Its quotient before imposing $x_1=0$ is precisely the smaller polygraph ring with the first $s$ labels omitted, because substitution and forgetting those pairs are inverse on all remaining generators. Imposing $x_1=0$ gives its reduced quotient by the coordinate-translation retraction of step 22.1. This proves radicality and the asserted ideal, including the collision points. The second follows by exchanging the two main labels; the third is exactly the supplied two-coordinate equation $I(1,1,s)$ from step 19.1. Thus no reduced ideal has been guessed from its points alone.

Direct addition gives $I_0+I_{\ast }=I_0+J_{\ast }=(x_1,x_2,y_1-y_2,a_i,b_i-y_1(i\le s))\mathrm{.}$ For instance the two prescriptions for $b_s$ force $y_1=y_2$, the prescriptions for $a_s$ together with $x_1$ force $x_2=0$, and the remaining fixed pairs then give exactly the displayed generators. These computations are in the actual quotient ring, so give equality even when the sums have nilpotents. Restoring the affine factors proves step 46.1 scheme-theoretically at every localization above $U_2$.