Rational Garnir annihilator lines present the specified Specht generator

Put $\kappa =0$ in the residue formulas below. Work first in the affine content algebra $R_{\mathbb{Z}}$ before the cyclotomic quotient. For each row of $\lambda$, the successive residue segment has a one-dimensional integral module on which its word idempotent is $1$ and its dots and internal crossings are $0$. The KLR relations hold: successive colours are distinct, their quadratic at zero dots is $0$, and a possible three-strand repeated-endpoint pattern occurs only for $e=2$, where its double-edge braid correction also vanishes at zero dots. Induce the tensor product of these segment modules, and shift its generator $m_0$ to degree $\mathrm{deg}(T_0)$, where $T_0$ is row reading. Denote the result by $M_{\mathbb{Z}}$.

Fix a Garnir node $A=(a,b)$ with $(a+1,b)\in \lambda$. Its belt consists of the row-$a$ suffix $X$ beginning at column $b$ and row-$(a+1)$ prefix $Y$ ending at column $b$. The entries in these nodes in $T_0$ form one consecutive interval $J=\{u,\dots ,v\}$: the suffix ends at the last position of row $a$, immediately followed by the prefix at the beginning of row $a+1$. Let $G_A$ agree with $T_0$ outside the belt and fill $Y$ first, then $X$, increasingly with $J$. It is row-strict and nonstandard. This $G_A$ is the nonstandard belt filling used for the rational-kernel presentation; it is distinct from the maximal STANDARD Garnir tableau in The domination order on tableaux, Garnir tableaux and bricks.

Let $P$ be the Young row subgroup. Choose reduced expressions compatible with the unique decomposition $w=dp$, with $d$ a minimal representative for $S_d\mathrm{/}P$ and $p\in P$, and $\mathrm{\ell }(w)=\mathrm{\ell }(d)+\mathrm{\ell }(p)$. F1 then makes the inducing bimodule free over the row algebra on the minimal-representative words. Tensoring the one-dimensional row modules shows that $M_{\mathbb{Z}}$ has the integral basis $m_T={\psi }_{w_T}{m}_0$ indexed by all row-strict tableaux $T$. Its cyclic presentation is exactly the initial weight, dot and horizontal-crossing relations. In particular all generator-action matrices on this finite basis have integer entries. Put $M_{\mathbb{Q}}=\mathbb{Q}\otimes M_{\mathbb{Z}}$; scalar extension gives $M_K=K\otimes M_{\mathbb{Q}}$. By F4, there is a surjective graded map ${\pi }_K:M_K\to S_K(\lambda )$ carrying $m_0$ exactly to $z_K$.

Among ALL row-strict tableaux agreeing with $T_0$ outside this belt, $G_A$ is the only nonstandard one. For columns $c<b$, the upper entry is an unchanged number less than $u$, hence less than every changed lower entry. For columns $c>b$, the lower entry, if present, is unchanged and greater than $v$, hence greater than every changed upper entry. Comparisons with the row above or the row below have the same strict bounds, and every other column comparison is unchanged. Thus only column $b$ can fail. It fails precisely when the first upper-belt entry exceeds the last lower-belt entry. Since each of the two rows is increasing, this is equivalent to every upper-belt entry exceeding every lower-belt entry, and hence to the unique filling $G_A$.

On $J$, the residue word $i(G_A)$ is one successive-residue segment. Along $Y$ the residues increase by $1$; its final node $(a+1,b)$ has residue $\kappa +b-a-1$. The first node of $X$, $(a,b)$, has residue $\kappa +b-a$, and the rest again increase by $1$. Consequently $i(G_A{)}_r\mathrm{\ne }i(G_A{)}_{r+1}$ for every $u\le r<v$, including $e=2$. This distinctness, rather than any braid normalization, is essential.

Crossing straightening gives a filtration useful below: a word with at most $N$ crossings applied to $m_0$ is a combination of $m_U$ with $\mathrm{\ell }(w_U)\le N$. Move dots right, reduce nonreduced words by quadratic and Coxeter moves, compare reduced words, and split off the right row subgroup as in step 2.1. Positive dot powers kill $m_0$; a nontrivial row factor ends in an internal crossing and also kills $m_0$. Every correction has fewer crossings. In particular if $s_{r}T$ is row-strict and $\mathrm{\ell }(w_{s_{r}T})=\mathrm{\ell }(w_T)+1$, then ${\psi }_r{m}_T=m_{s_{r}T}$ plus terms of strictly smaller crossing length, with leading coefficient exactly $1$. If $s_{r}T$ is not row-strict, its possible leading term is killed by the right row factor and only shorter terms remain.

Let ${\delta }_A=\mathrm{deg}({\psi }_{w_{G_A}}e(i^{\lambda }))+\mathrm{deg}(T_0)$ and let $T_A^{\mathbb{Q}}$ be the span of those integral basis vectors $m_T$ for which $T$ agrees with $T_0$ outside the belt, has weight $i(G_A)$, and has homogeneous degree ${\delta }_A$. It includes $m_{G_A}$. The use of the entire weight-degree slice avoids any brick-classification assumption. Write $T_A^K=K\otimes T_A^{\mathbb{Q}}$.

The vector ${\pi }_K(m_{G_A})$ is a combination of images of standard basis vectors in $T_A^K$. Indeed F3 gives standard $U$ with $w_U\le w_{G_A}$. The permutation $w_{G_A}$ belongs to the parabolic subgroup generated by $s_u,\dots ,s_{v-1}$, since it permutes only $J$. Every Bruhat-smaller permutation belongs to this same subgroup: choose a reduced word for $w_{G_A}$ in these generators, and use the defining subword characterization of strong Bruhat order. Hence $U$ agrees with $T_0$ outside the belt. Weight idempotents discard different words, and homogeneity discards different degrees, leaving exactly the standard vectors in the slice of step 3.2. By step 2.2, every other vector of the slice is standard; by F3 its image is part of the standard Specht basis and is linearly independent. Therefore $\ker ({\pi }_K{\mathrm{\mid }}_{T_A^K})$ is precisely one-dimensional, including the case where the image of $m_{G_A}$ is zero.

For every $u\le r<v$, the image ${\psi }_r{T}_A^K$ in $M_K$ is spanned by standard row-strict belt tableaux. To see the support assertion, choose reduced words for all belt permutations using only $s_u,\dots ,s_{v-1}$. Straightening a crossing supported in this interval never introduces a crossing outside it; factoring a right row permutation also uses only this interval. Thus every resulting row-strict basis tableau agrees outside the belt. The image has fixed weight $s_{r}i(G_A)$, which differs from $i(G_A)$ by step 2.3. Its support cannot include the sole nonstandard belt tableau $G_A$. Hence the entire image has standard support by step 2.2, and ${\pi }_K$ is injective on its span by F3.

If $x\in \ker ({\pi }_K{\mathrm{\mid }}_{T_A^K})$, then ${\pi }_K({\psi }_{r}x)={\psi }_r{\pi }_K(x)=0$. The injectivity in step 4.2 implies ${\psi }_{r}x=0$ in $M_K$ for every inside-belt crossing. Thus this one-dimensional Specht-kernel line lies in $B_A^K:={\bigcap }_{r=u}^{v-1}\ker ({\psi }_r:T_A^K\to M_K)$.

The $m_{G_A}$ coefficient map is injective on $B_A^K$. Suppose a nonzero $x\in B_A^K$ has that coefficient zero. Choose a tableau $T$ with nonzero coefficient in $x$ whose crossing length $N=\mathrm{\ell }(w_T)$ is maximal. It is standard by step 2.2. Record membership in $X$ or $Y$ for the successive numbers $u,\dots ,v$ as a binary word. Unless this word is all $Y$ then all $X$, it has an adjacent occurrence $XY$; that excluded word is exactly $G_A$. Choose such an occurrence at $r,r+1$. Swapping them preserves row strictness because they lie in different rows and are consecutive numbers; it increases permutation length by $1$, because $r$ is in the earlier row. Its weight becomes $s_{r}i(G_A)\mathrm{\ne }i(G_A)$, so it cannot be $G_A$ and is also standard by step 2.2. By step 3.1, the coefficient of $m_{s_{r}T}$ in ${\psi }_{r}x$ at length $N+1$ is precisely the nonzero coefficient of $m_T$ in $x$. A lower-length input cannot reach length $N+1$; a different equal-length input has a different possible leading tableau because $s_r$ is injective; and its correction terms have smaller length. Therefore ${\psi }_{r}x\mathrm{\ne }0$, contradicting $x\in B_A^K$.

Consequently ${\dim }_K{B}_A^K\le 1$, while step 5.1 gives a nonzero line inside it. Thus $B_A^K=\ker ({\pi }_K{\mathrm{\mid }}_{T_A^K})$. Define $B_A^{\mathbb{Q}}$ using the same intersection of crossing kernels over $\mathbb{Q}$. These are kernels of a single finite stacked matrix with integer entries on the explicitly selected integral row-strict bases. Row reduction over $\mathbb{Q}$ commutes with extension to $K$, so $B_A^K=K\otimes B_A^{\mathbb{Q}}$. It follows that $B_A^{\mathbb{Q}}$ is one-dimensional and has a unique rational vector $h_A$ with coefficient $1$ on $m_{G_A}$. Its scalar extension is exactly the Specht-kernel vector with that normalization. This is an explicit rational linear-system construction of the relation, not an assumption that a chosen vector belongs to a rational form.

Every other row-strict belt tableau has strictly smaller crossing length than $G_A$. Outside the belt there are no inversions involving belt entries because the belt interval is consecutive in row reading and its values form $J$. Inside, inversions are exactly upper/lower pairs with the upper entry larger. The tableau $G_A$ has all $\mathrm{\mid }X\mathrm{\mid }\mathrm{\mid }Y\mathrm{\mid }$ such inversions, whereas every other filling loses at least one. Hence $h_A=m_{G_A}+{\sum }_{U:\mathrm{\ell }(w_U)<\mathrm{\ell }(w_{G_A})}{c}_U{m}_U$ with rational $c_U$, and it is homogeneous of degree ${\delta }_A$ by its definition in the slice.

Put $S_{\mathbb{Q}}^{\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{v}}=M_{\mathbb{Q}}\mathrm{/}{\sum }_A{R}_{\mathbb{Q}}{h}_A$. We prove standard spanning locally. For any nonstandard row-strict $T$, choose a vertical inversion at $(a,b),(a+1,b)$. All entries in its upper suffix $X$ exceed all entries in its lower prefix $Y$. Its Garnir factor $G_A$ therefore satisfies $T=w{G}_A$ with $w=w_T{w}_{G_A}^{-1}$ and $\mathrm{\ell }(w)=\mathrm{\ell }(w_T)-\mathrm{\ell }(w_{G_A})$: reordering the contiguous belt's one-line list from $X,Y$ to $Y,X$ removes precisely its $\mathrm{\mid }X\mathrm{\mid }\mathrm{\mid }Y\mathrm{\mid }$ cross inversions, while preserving within-part and outside comparisons. Reduced-word straightening and the relation $h_A=0$ replace $m_T$ by words with fewer than $\mathrm{\ell }(w_T)$ crossings, by step 8.1. The filtration of step 3.1 expresses these as row-strict vectors of strictly smaller length. Induction proves that standard vectors span the quotient.

Scalar extension of the rational quotient presentation gives $S_K^{\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{v}}=K\otimes S_{\mathbb{Q}}^{\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{v}}$. Every $h_A$ lies in $\ker {\pi }_K$ by step 7.1, so ${\pi }_K$ induces a surjection $S_K^{\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{v}}\to S_K(\lambda )$ preserving the generator and grading. The source is spanned by the standard vectors by step 9.1, and their images are the linearly independent standard Specht basis by F3. Hence this map is an isomorphism. In particular the rational module is finite-dimensional, has its standard basis, and its generator maps exactly to $z_K$.

The rational module factors through the cyclotomic quotient. Each specified cyclotomic relation acts as zero after extension to $K$, because the module in step 10.1 is the specified cyclotomic Hecke cell Specht module. The matrix of that action on the rational standard basis has rational entries; faithful field extension forces it to be zero already over $\mathbb{Q}$. Denote this rational cyclotomic module by $S_{\mathbb{Q}}(\lambda )$, with generator $z_{\mathbb{Q}}$ the class of $m_0$. This identifies the required rational initial cyclic generator, not merely the unpointed module's isomorphism class.

If an integer cyclic quotient is desired for F5, take the saturated generator annihilator $I_{\mathbb{Z}}=\{a\in A_{\mathbb{Z}}:a{z}_{\mathbb{Q}}=0\}$ in the finite integer cyclotomic algebra of The integral form of the cyclotomic Hecke-KLR block. It is a homogeneous left ideal, and $A_{\mathbb{Z}}\mathrm{/}{I}_{\mathbb{Z}}$ identifies with the integral orbit $A_{\mathbb{Z}}{z}_{\mathbb{Q}}\subset S_{\mathbb{Q}}(\lambda )$. Its rational span is the entire rational module by cyclicity, so $\mathbb{Q}\otimes (A_{\mathbb{Z}}\mathrm{/}{I}_{\mathbb{Z}})\cong S_{\mathbb{Q}}(\lambda )$ with the specified generator. Clearing denominators in each $h_A$ is also possible, but no unsaturated integer quotient or integral Specht-freeness theorem is needed.

Apply F5 to this pointed rational form and the supplied absolutely simple normalized head. Its radical-basechange proof gives a rational normalized head with the image of $z_{\mathbb{Q}}$; absolute Schur makes every quotient map to the fixed head differ only by a scalar. Its integral-algebra orbit is finitely generated and torsion-free, with rational span that head, so its rank is exactly the head's $K$-dimension. This establishes the precise requested initial-generator line and lattice statement over arbitrary characteristic-zero $K$. No three-brick identity, unproved R-matrix formula, arbitrary-vector rational form, or change of the specified generator enters the argument.