Normality of the Pfaffian rank ring

The Gram embedding in [F1] makes $R$ a domain. We describe its monomial degeneration explicitly. Choose positive integer weights on the finitely many variables $x_{ij}$ which realize the column-first lexicographic comparisons between all terms in all the finitely many Gram images $P_S$, $\mathrm{\mid }S\mathrm{\mid }\le m$. Such weights exist: list the variables in that lexicographic order, give successive variables weights $N^{tm-1},N^{tm-2},\dots ,1$, and take $N>2t+1$, larger than the total degree of each polynomial being compared. Thus $d_S$ is the unique term of largest weight in each image. Filter $R$ by these ambient polynomial weights. A standard product has unique term of largest weight equal to the corresponding product of the $d_S$; these monomials are distinct by [F1]. In a linear combination of standard products the terms of maximum weight therefore do not cancel. Consequently the associated graded ring is exactly $k[\mathrm{\Gamma }]$, where $\mathrm{\Gamma }$ is the additive semigroup generated by the exponent vectors of all $d_S$, for even $\mathrm{\mid }S\mathrm{\mid }\le m$. In particular this is a filtration of the entire ring, not merely a computation of the initial terms of its generators.

Write an exponent array as $b=(b_{ij})$, put ${\lambda }_j={\sum }_i{b}_{ij}$, and put $C_j(u)={\sum }_{i\le u}{b}_{ij}$, with $C_j(0)=0$. The semigroup $\mathrm{\Gamma }$ consists exactly of the integer arrays satisfying [step 1.1, algebra]

Necessity follows by adding these inequalities for individual increasing even rows. For sufficiency sort the row indices in each column increasingly, with multiplicities. The cumulative inequalities say that the $a$th entry of column $j$ is strictly less than the $a$th entry of column $j+1$ whenever the latter exists: a failure at value $u$ would give $C_{j+1}(u)>C_j(u-1)$, and the converse follows by taking $a=C_{j+1}(u)$. The inequalities on column lengths make a left-justified tableau; their pairwise equalities say that every row length is even. Each row is therefore an allowed $S$. Reading these rows from top to bottom gives a standard product and expresses $b$ as a sum of exponent vectors of $d_S$. All displayed conditions are homogeneous integral linear inequalities or equalities. Thus if $b$ is in the group $L$ generated by $\mathrm{\Gamma }$ and $qb\in \mathrm{\Gamma }$ for a positive integer $q$, then $b\in \mathrm{\Gamma }$. More precisely $\mathrm{\Gamma }=L\cap \mathcal{C}$, where $\mathcal{C}$ is the rational polyhedral cone cut out by the displayed conditions.

This proves normality of $k[\mathrm{\Gamma }]$ directly. The group algebra $k[L]$ is a Laurent polynomial ring after choosing an integral lattice basis and is integrally closed by [F8]. Each defining integral linear inequality $\mathrm{\ell }\ge 0$ gives the valuation $v_{\mathrm{\ell }}(\sum c_b{X}^b)={\min }_{c_b\mathrm{\ne }0}\mathrm{\ell }(b)$ on its fraction field (combine equal-weight terms first); its nonnegative valuation ring is integrally closed by [F8]. The equality $\mathrm{\Gamma }=L\cap \mathcal{C}$ gives [step 2.1, algebra]

Indeed for a Laurent polynomial nonnegative minimum under every $\mathrm{\ell }$ means that every exponent lies in $\mathcal{C}$. An intersection of integrally closed subrings of one field is integrally closed, by applying a monic equation in every member of the intersection. Hence the special ring of the filtration is a normal domain.

We give the lifting argument to avoid assuming normality from distinct initial terms. Let $\mathcal{B}={⨁}_{d\ge 0}{F}_{d}R{z}^d\subset R[z]$ be the Rees ring of step 1.1. It is generated by $z$ and $P_S{z}^{w(d_S)}$: expand an element in the standard basis, and use the absence of highest-weight cancellation in step 1.1. It is a finitely generated domain, hence Noetherian by [F2], with $\mathcal{B}\mathrm{/}(z)=k[\mathrm{\Gamma }]$, and $\mathcal{B}[z^{-1}]=R[z,z^{-1}]$. Retain also the ordinary grading $\mathrm{deg}(P_S)=\mathrm{\mid }S\mathrm{\mid }\mathrm{/}2$ and $\mathrm{deg}z=0$; then ${\mathcal{B}}_0=k[z]$ and all other generators have positive ordinary degree. We use the following local normality argument. For a finite-type domain $D$ over a field and a nonzero element $z$, if $(D\mathrm{/}(z){)}_{\mathfrak{p}}$ is normal, then $D_{\mathfrak{p}}$ is normal for every prime $\mathfrak{p}$ containing $z$. To check this by [F3], localize at such $\mathfrak{p}$. At a prime $Q$ containing $z$, [F4] increases depth by one from the quotient and [F6] increases dimension by one, so the $S_2$ bounds follow from $S_2$ of the quotient. Height-one primes containing $z$ have field quotients, so their local rings are regular by [F6]. Height-zero primes of the domain localize to its fraction field and are regular. For $Q$ not containing $z$ and of positive height, choose a prime $P$ minimal over $Q+(z)$ in this local ring. Applied to the corresponding primes in the original affine domain, [F5] gives $\mathrm{ht}P=\mathrm{ht}Q+1$ and $\dim (D_P\mathrm{/}Q{D}_P)=1$. If $\mathrm{ht}Q=1$, the quotient of $D_P$ by $z$ is a normal one-dimensional local domain, hence regular by $R_1$ in [F3]; [F6] makes $D_P$ regular of dimension two and then makes its localization $D_Q$ regular. If $\mathrm{ht}Q\ge 2$, the quotient has depth at least two by [F3], so $D_P$ has depth at least three by [F4]; the depth-localization inequality in [F4] $\mathrm{depth}{D}_Q\ge \mathrm{depth}{D}_P-\dim (D_P\mathrm{/}Q{D}_P)$ gives depth at least two at $Q$. Thus $R_1$ and $S_2$ hold throughout the local ring, which proves the assertion. All these applications concern Noetherian local rings by [F2], nonzero regular elements because the Rees ring is a domain, and primes of an affine domain, as required by [F3]–[F6]. Applied to $\mathcal{B}$, step 3.1 proves normality at every point of the fiber $z=0$.

For completeness, this local conclusion controls every fiber here. The normalization is finite and its nonnormal locus is closed by [F7]. The nonnormal locus of $\mathcal{B}$ is stable under both its ordinary grading action and its filtration grading action: each scaling is a ring automorphism, extends to the fraction field, and preserves integral elements and therefore the normalization quotient. This uses no assertion that normalization itself has a chosen grading. If it contained a point over $z=c\mathrm{\ne }0$, its closure under the ordinary grading action would contain the point of the section with all positive-ordinary-degree generators zero and $z=c$: the orbit map extends to scaling parameter zero. The filtration grading action then gives all points of that section with $z\mathrm{\ne }0$; closedness gives its point at $z=0$, contradicting step 4.1. By [F9], over the algebraically closed field $k$ a nonempty closed subset has a $k$-rational closed point, so this excludes the entire nonnormal locus. Hence $\mathcal{B}$ and $R[z,z^{-1}]$ are normal. Finally an element of $\mathrm{Frac}R$ integral over $R$ is integral over this Laurent extension and therefore belongs to it; its Laurent expansion has only the degree-zero coefficient, so it belongs to $R$. This proves normality of $R$. When $m=0$, $R=k$ directly; all other boundary cases are included in the filtration argument.