\section{Actual images from an attained rank-one nilpotent}\label{sec:even-prelim} The remainder of the article deals with $s_{2r}$. Throughout this part, its trace-zero property follows directly from cyclic alternation: for any word of length $2r$, moving the first letter to the last preserves the trace but changes the permutation sign by $(-1)^{2r-1}=-1$. Consequently \begin{equation}\label{eq:even-trace} \tr s_{2r}(A_1,\ldots,A_{2r})=0, \qquad \im s_{2r}(\Mat_n)\subseteq\Sl_n. \end{equation} The identity $s_{2n}=0$ on $\Mat_n$ implies the entire zero branch for $n\le r$. Moreover $s_{2r}(A_1,\ldots,A_{2r-1},I_n)=0$. For nonzero images the following local-to-global mechanism is essential. It is the smooth-map formulation of the nilpotent argument in \cite[Lemma~2]{Tang2026FiveSix}; the elementary degeneration is included to make the actual-value conclusion explicit. \begin{lemma}[Attained-nilpotent actual-image criterion] \label{lem:nilpotent-open} Let $F:\mathbb A^m_K\to\Sl_n(K)$ be a polynomial map whose image contains zero and is invariant under similarities and multiplication by nonzero scalars. Suppose that for some $x$ the value $R=F(x)$ is nonzero, rank one and square zero, and \(dF_x:K^m\to\Sl_n(K)\) is surjective. Then $\im F=\Sl_n(K)$. \end{lemma} \begin{proof} Both affine spaces are smooth. Surjectivity of the differential makes $F$ smooth at $x$, and smoothness is open; the restriction to a suitable neighborhood maps onto a Zariski-open subset containing $R$. In particular, the image contains an open neighborhood $U$ of $R$ in $\Sl_n$. Let $T\in\Sl_n$ be nonzero. It is nonscalar in characteristic zero. A nonscalar linear transformation has a vector $v$ for which $v,Tv$ are linearly independent: otherwise every line is invariant, forcing the transformation to be scalar. Thus after a similarity $T_{01}\ne0$. For $t\ne0$, write \begin{equation}\label{eq:nilpotent-degeneration} D(t)=\diag(t^{-1},t,1,\ldots,1),\qquad T(t)=t^2D(t)TD(t)^{-1}. \end{equation} All entries of $T(t)$ have nonnegative powers of $t$; its value at $t=0$ is $T_{01}E_{01}$. Every nonzero rank-one square-zero matrix is similar, up to a nonzero scalar, to $R$. By similarity and scaling invariance there is therefore an open neighborhood of $T_{01}E_{01}$ contained in the image. Its preimage under the affine-line morphism $t\mapsto T(t)$ contains zero and hence some nonzero $t$. Thus $T(t)$ is an actual value. Undoing the similarity and scalar $t^2$ shows that $T$ is an actual value. The zero target is already in the image. \end{proof} A standard polynomial is equivariant for simultaneous conjugation; scaling one input scales its output, and setting an input to zero gives zero. Thus every standard evaluation map satisfies the invariance assumptions of Lemma~\ref{lem:nilpotent-open}. \section{A uniform minimal-size seed}\label{sec:even-seed} The first matrix order at which $s_{2r}$ need not vanish is $n=r+1$. The following construction treats every such order with one matrix-unit formula, including the full differential required by Lemma~\ref{lem:nilpotent-open}. \begin{theorem}[Uniform even-standard seed]\label{thm:even-seed} For every $n\ge2$, \begin{equation}\label{eq:even-seed} \im s_{2n-2}(\Mat_n(K))=\Sl_n(K). \end{equation} More precisely, the evaluation constructed below has a nonzero rank-one square-zero output and surjective full differential. \end{theorem} \begin{proof} Write $p=n-2$ and index matrix units by $0,\ldots,n-1$. For $n\ge3$, use the ordered $2p+2$ inputs \begin{equation}\label{eq:even-tuple} \begin{aligned} \mathcal A=(&L=E_{00},\ X=E_{01},\\[-2pt] &S_2=E_{02},T_2=E_{20},\ldots,\\[-2pt] &S_{n-1}=E_{0,n-1},T_{n-1}=E_{n-1,0}). \end{aligned} \end{equation} The signed matrix-unit trail calculation gives \begin{equation}\label{eq:even-seed-evals} \begin{aligned} s_{2n-2}(\mathcal A)&=(p+1)!E_{01},\\ \partial_{X\gets E_{10}}s_{2n-2}(\mathcal A)&=-(p+1)!E_{10},\\ \partial_{X\gets E_{12}}s_{2n-2}(\mathcal A)&=-p!E_{12},\\ \partial_{S_2\gets E_{10}}s_{2n-2}(\mathcal A)&=-(p+1)!E_{20},\\ \partial_{S_2\gets E_{12}}s_{2n-2}(\mathcal A)&=p!(E_{11}-E_{22}). \end{aligned} \end{equation} The notation $\partial_{A_j\gets V}$ denotes the directional derivative in the $j$th argument; by multilinearity it is exactly the value with that argument replaced by $V$. Here is a signed-trail derivation valid for all $n$. A nonzero matrix-unit word is precisely an oriented Euler trail on the labelled directed graph. In the unmodified tuple, each trail starts at $0$, traverses the loop $L$ and the $p$ excursions $S_jT_j$ in any order, and ends with $X$. The excursions have even length and commute in permutation parity; moving $X$ to the end crosses $2p$ labels. These $(p+1)!$ trails all have positive sign. With $X=E_{10}$ the trail instead begins with $X$, making its sign negative. With $X=E_{12}$ it starts with $X,T_2$, traverses the remaining root blocks and ends with $S_2$; there are $p!$ such trails and the word has odd inversion parity. With $S_2=E_{10}$ the trail begins with $T_2$ and traverses the root loop, the new excursion $XS_2$, and the remaining $p-1$ excursions; the resulting $(p+1)!$ trails are negative. Finally, for $S_2=E_{12}$, there is a directed triangle $X,S_2,T_2$ together with a loop and the other two-edge excursions. Trails starting at $0$ or another leaf cancel by exchanging the odd-length loop and triangle. Starting at $1$ leaves $p!$ positively signed trails, whereas starting at $2$ leaves $p!$ negatively signed trails. This proves the last formula in \eqref{eq:even-seed-evals}, including its diagonal signs. Let $\mathcal D$ be the image of the \emph{full} differential of $s_{2n-2}$ at $\mathcal A$. It lies in $\Sl_n$ by \eqref{eq:even-trace}. Equivariance supplies every commutator $[Z,E_{01}]\in\mathcal D$ (the scalar $(p+1)!$ is nonzero). Consider a change of basis $g=\diag(1,1,G)$, $G\in\mathrm{GL}_{p}(K)$, acting on the leaf coordinates. It fixes $L$ and $X$ and sends the $p$ outgoing and $p$ incoming leaf arguments to mutually contragredient linear combinations of their original arguments. The combined transformation on these $2p$ frozen slots has determinant one. The alternating multilinear map, and hence its differential image at the tuple, is therefore invariant under conjugation by $g$. The orbits of $E_{12}$ and $E_{20}$ in \eqref{eq:even-seed-evals} supply all $E_{1j}$ and $E_{i0}$ for $i,j\ge2$. The orbit of $E_{11}-E_{22}$ spans \begin{equation}\label{eq:even-seed-block} \bigl\{(\tr B)E_{11}-B: B\text{ is supported on the leaf block}\bigr\}, \end{equation} because the conjugates of a rank-one diagonal idempotent span its full matrix block: rank-one idempotents include $E_{ii}$ and $E_{ii}+E_{ji}$ for distinct leaf indices. Together with $E_{10}$ and $[Z,E_{01}]$, these directions give every off-diagonal matrix unit and every diagonal difference. Thus $\mathcal D=\Sl_n$, as required. When $n=2$, take $(E_{00},E_{01})$. Its commutator value is $E_{01}$; replacing $E_{01}$ by $E_{10}$ gives $-E_{10}$ and the conjugation direction gives $E_{00}-E_{11}$. The full differential is $\Sl_2$. In each order the output is a nonzero rank-one square-zero matrix, so Lemma~\ref{lem:nilpotent-open} proves \eqref{eq:even-seed}. \end{proof} \begin{remark}[The size-four seed in Tang's construction]\label{rem:tang-seed} At $n=4$, the tuple in \eqref{eq:even-tuple} is $(E_{00},E_{01},E_{02},E_{20},E_{03},E_{30})$. Qiyue Tang's Appendix~A uses $(4E_{11},E_{12},E_{13},E_{14},E_{31},E_{41})$ in one-based coordinates \cite[Proposition~11]{Tang2026FiveSix}. The two coincide after relabelling the indices, interchanging the fourth and fifth slots, and multiplying the first slot by $4$; alternating multilinearity accounts for the sign and scale. Thus the size-four instance agrees, up to these changes, with Tang's matrix-unit construction. \end{remark} \section{Admissible cores and an additive transfer}\label{sec:even-cores} The seed can be transferred to larger orders by a normal-map criterion that keeps track of the \emph{full differential}, rather than merely of a frozen generic kernel. Fix $r\ge1$ and write \begin{equation}\label{eq:even-rk} k=2r-1. \end{equation} For $B=(B_1,\ldots,B_k)\in\Mat_N(K)^k$ and $u\in K^N$, define \begin{equation}\label{eq:core-definitions} \begin{aligned} S&=s_k(B_1,\ldots,B_k),\\ L(P)&=s_{k+1}(B_1,\ldots,B_k,P),\\ C_i(P)&=s_k(B_1,\ldots,B_{i-1},P,B_{i+1},\ldots,B_k),\\ \Gam_{B,u}(P)&=\bigl(L(P),C_1(P)u,\ldots,C_k(P)u\bigr). \end{aligned} \end{equation} Thus $\Gam_{B,u}:\Mat_N(K)\to\Mat_N(K)\oplus(K^N)^k$ is a linear map in $P$; the frozen data $(B,u)$ vary polynomially. Call the size $N$ \emph{admissible for $r$} if there exists one such tuple with \begin{equation}\label{eq:core-admissible} \det S\ne0,\qquad \ker\Gam_{B,u}=\{0\}. \end{equation} Both conditions are polynomial nonvanishing conditions in unrestricted parameters. A witness therefore defines a nonempty Zariski-open subset of the core parameter space. \begin{lemma}[Normal-map criterion]\label{lem:core-normal} If $N$ is admissible for $r$, then \begin{equation}\label{eq:core-conclusion} \im s_{2r}(\Mat_{N+1}(K))=\Sl_{N+1}(K). \end{equation} Indeed the defining witness yields an attained nonzero rank-one square-zero value at which the full differential is surjective. At the particular border tuple used below, assuming $S$ invertible, this differential condition is equivalent to injectivity of $\Gam_{B,u}$. \end{lemma} \begin{proof} Embed the fixed matrices and the final input as \begin{equation}\label{eq:core-border} A_i=\begin{pmatrix}B_i&0\\0&0\end{pmatrix},\qquad Y=\begin{pmatrix}0&u\\0&0\end{pmatrix}. \end{equation} Only words ending in $Y$ can survive, because $YA_i=0$. Since $k$ is odd, the value of the even alternator is \begin{equation}\label{eq:core-border-value} s_{k+1}(A_1,\ldots,A_k,Y) =\begin{pmatrix}0&Su\\0&0\end{pmatrix}. \end{equation} If $u=0$, then $\Gam=L$ and its kernel contains all $B_i$ (and also the identity), contradicting injectivity. Hence $Su\ne0$. The value is nonzero, square zero and rank one. To compute the differential, pair its values with a general matrix using the nondegenerate trace form. Variations of $Y$ in the upper and lower borders give the full upper and lower border directions by left and right multiplication by $S$ and $-S$, respectively. Thus every normal is block diagonal. The differential values have trace zero, so scalar matrices are normals by \eqref{eq:even-trace}. Subtracting a scalar we may write a normal uniquely as $\diag(P,0)$. Variations of the old $N\times N$ block of $Y$ give $\tr(P L(Q))$ for arbitrary $Q$. The map $L$ is skew-adjoint for the trace pairing. To see this, collect the cyclic classes in $\tr s_{k+2}(B_1,\ldots,B_k,Q,P)$. The length $k+2$ is odd, so a cyclic shift preserves sign; moving $Q$ past the $k$ frozen letters contributes a minus sign. Consequently the trace equals $-(k+2)\tr(Q L(P))$. Interchanging $P,Q$ in the alternating trace yields \begin{equation}\label{eq:core-skew} \tr(P L(Q))=-\tr(Q L(P)). \end{equation} The old-block normal equations are therefore $L(P)=0$. Now vary the $i$th $A_i$ in the lower border by a row $c\in(K^N)^*$. For a nonzero word the unique upper-border factor $Y$ must be immediately followed by this lower-border variation. Remove $Y$ and rotate the remaining $k$ letters cyclically. If $a$ of the original old letters followed the adjacent pair, the removal has sign $(-1)^{a+1}$ and moving the varied letter to its original position has sign $(-1)^a$. The product is $-1$, and the normal pairing is exactly \begin{equation}\label{eq:core-low-normal} -c\,C_i(P)u. \end{equation} Because all rows $c$ are available, these normal equations say $C_i(P)u=0$ for each $i$. Therefore every normal modulo scalars satisfies $\Gam_{B,u}(P)=0$. Conversely, in the pure border tuple \eqref{eq:core-border}, these are all the normal equations: other source-block variations yield only upper-border outputs or zero. This also proves the asserted equivalence. If $\Gam$ is injective, all normals are scalar, so the full differential has image $\Sl_{N+1}$. Lemma~\ref{lem:nilpotent-open} now proves \eqref{eq:core-conclusion} as an \emph{attained} image equality. \end{proof} A complementary rectangular nonvanishing calculation records the invertibility available for independently varying block-diagonal inputs. \begin{lemma}[Mixed-rectangle nonvanishing]\label{lem:mixed-rect} Let $a,b\ge1$ and $m<2(a+b)$. For an $m$-variable standard polynomial, the free rectangular map $\Mat_{a\times b}\to\Mat_{a\times b}$ obtained by freezing $m-1$ block-diagonal inputs of sizes $a$ and $b$ is generically invertible as the frozen blocks vary independently. \end{lemma} \begin{proof} Choose integers $e,f\ge0$ with $e+f=m-1$, $e<2a$ and $f<2b$; they exist since $m<2(a+b)$. Specialize the first $e$ frozen labels to the $a$ block and the remaining $f$ to the $b$ block. The only surviving words put the left labels before the free rectangle and the right labels after it, so the map is \begin{equation}\label{eq:mixed-rect} X\longmapsto(-1)^f s_e(A)X s_f(B),\qquad s_0=I. \end{equation} Both factors admit invertible evaluations. Indeed, for degree below the Amitsur--Levitzki threshold a standard polynomial is a nonidentity. Expand a nonzero evaluation in the clock-and-shift homogeneous basis of the corresponding matrix algebra; by multilinearity one homogeneous tuple yields a nonzero scalar multiple of an invertible basis element. Thus the rectangular map is invertible at one specialization, and its determinant is a nonzero polynomial. \end{proof} \begin{theorem}[Additivity of admissible core sizes]\label{thm:core-add} For a fixed $r\ge1$, if $a,b\ge r$ are admissible core sizes, then $a+b$ is admissible for $r$. \end{theorem} \begin{proof} Choose witnesses $(B^{(a)},u_a)$ and $(B^{(b)},u_b)$ for the two sizes. For one parameter $t$ set \begin{equation}\label{eq:core-add-tuple} B_i(t)=\diag(B_i^{(a)},tB_i^{(b)}),\qquad u=\binom{u_a}{u_b}. \end{equation} For every $t\ne0$, the diagonal endpoint blocks are $S_a$ and $t^kS_b$ and hence are invertible. The old-block $L$ and $C_i$ operators in the second block are respectively scaled by $t^k$ and $t^{k-1}$, so their individual admissibility kernels remain zero. On the two off-diagonal rectangular spaces, the map $L_t$ in \eqref{eq:core-definitions} is polynomial in $t$. At $t=0$ the top-right rectangular map is $X\mapsto S_aX$, because only words with the free variable last survive. The bottom-left rectangular map is $X\mapsto -XS_a$, because only words with the free variable first survive and $k$ is odd. Both are invertible. Therefore their determinants are nonzero polynomials in the \emph{same} scalar parameter. As $K$ is infinite, select one common $t\in K^\times$ avoiding their zero sets. At this fixed $t$, if $\Gam_{B(t),u}(P)=0$, the two mixed $L_t$ equations force the off-diagonal blocks of $P$ to be zero. The remaining $C_i(P)$ are block diagonal, and the normal-map equations reduce to those for the two original cores, with nonzero scalar factors on the second block. Both diagonal blocks of $P$ vanish by the original injectivity hypotheses. Thus $\Gam_{B(t),u}$ is injective, and $\det s_k(B(t))\ne0$. This constructs a single admissible witness of size $a+b$. \end{proof} \begin{corollary}[An infinite congruence family]\label{cor:congruence} For every $r,q\ge1$, \begin{equation}\label{eq:congruence} \im s_{2r}(\Mat_{qr+1}(K))=\Sl_{qr+1}(K). \end{equation} \end{corollary} \begin{proof} In Theorem~\ref{thm:even-seed}, set $n=r+1$, place the exit variable last, and remove its sink coordinate. The old $r\times r$ core has \begin{equation}\label{eq:core-star} B=(E_{00},E_{01},E_{10},\ldots,E_{0,r-1},E_{r-1,0}), \qquad u=e_0, \end{equation} with the obvious singleton interpretation when $r=1$. Counting closed trails through the root loop and the $r-1$ two-edge excursions gives \begin{equation}\label{eq:core-star-endpoint} S=s_{2r-1}(B) =\diag(r!,(r-1)!,\ldots,(r-1)!), \end{equation} which is invertible in characteristic zero. The full differential in Theorem~\ref{thm:even-seed} is onto $\Sl_{r+1}$; the converse statement in Lemma~\ref{lem:core-normal} shows that this old core is admissible. For $r=1$ it follows also directly from $S=1$ and $C_1(P)u=P$. Add $q$ copies of the admissible size $r$ with Theorem~\ref{thm:core-add}, then apply Lemma~\ref{lem:core-normal} in size $qr+1$. \end{proof} \section{Exact core certificates in degrees eight and ten}\label{sec:even-low} For each $r$, the uniform construction supplies an admissible core of size $r$. By additive transfer, to obtain all sizes $N\ge r$, it suffices to establish admissibility for the finite range \begin{equation}\label{eq:core-finite-reduction} r,r+1,\ldots,2r-1. \end{equation} Indeed every $N\ge r$ has the form $N=qr+s$, $q\ge1$ and $0\le s