\(L(0)=A_0\) is invertible, so \(L\) is full (§2.2(a)). A right monic matrix is not a unit [Coh06, Lemma 5.8.1, p. 312]. For \(N=1\), [Coh06, Thm 5.8.8, p. 317] says that \(L\) is an atom, and the subspace condition holds trivially. Let \(N{\gt}1\). By [Coh06, Thm 5.8.8], \(L\) is not an atom iff there are \(P,Q\in GL_N(k)\) and \(r,s{\gt}0\) with \(r+s=N\) such that every \(PA_jQ\) (\(j=0,\dots ,g\)) has a zero \(r\times s\) block in the upper right corner.

• Given such \(P,Q\), put \(W=Q\, \mathrm{span}(e_{r+1},\dots ,e_N)\) and \(W'=P^{-1}\mathrm{span}(e_{r+1},\dots ,e_N)\). Then \(A_jW\subseteq W'\) for all \(j\). Since \(A_0\) is injective and \(\dim W=\dim W'\), we get \(A_0W=W'\), hence \(M_iW\subseteq A_0^{-1}W'=W\), with \(0{\lt}\dim W=s{\lt}N\).

• Conversely, let \(W\) be invariant with \(0{\lt}s=\dim W{\lt}N\). Take \(Q\) whose last \(s\) columns span \(W\), and \(P=Q^{-1}A_0^{-1}\). Then \(PA_0Q=I\) and \(PA_iQ=Q^{-1}M_iQ\) maps \(\mathrm{span}(e_{r+1},\dots ,e_N)\) into itself, so the upper right \(r\times s\) block vanishes for all \(j\). \(\square \)

(1) For \(n{\lt}\omega \), \(D_n\) is not additively closed (it misses sums of \(n+1\) distinct monomials), so Case 1 applies. For \(n{\lt}\omega \), every multiset smaller than \((n+1)[1]\) involves only the ordinals \(0\) and \(1\), with at most \(n\) copies of \(1\), so its sum lies in \(D_n\); and \((n+1)[1]\) gives sums of \(n+1\) distinct monomials, which are not in \(D_n\).

1. \(D_\omega =A\) is a ring and \(D_1^{-1}\subseteq D_1\). A sum \(\alpha g+\beta h\) of two monomials with \(g\ne h\) is not a unit of \(A\) [Jak19, Thm 1.3.10], so \(D_2^{-1}\not\subseteq A\); and \((\alpha g+\beta h)^{-1}=\ell ^{-1}(\alpha g)^{-1}\) with \(\ell =1+\alpha ^{-1}\beta g^{-1}h\).

2. Case 2 needs an additively closed set and Case 3 a ring. Sums of elements of \(D_{\omega +1}\) lie in \(V_1\), which is the additive closure of \(D_{\omega +1}\) (§2.1: \(m\ell ^{-1}m'\) is a unit multiple of an inverse binomial). So the first additively closed set is \(V_1\).

3. In \(N(\sigma )\) the multisets below \(2[\omega +1]\) are those over \(\{ 0,\dots ,\omega \} \) and the \([\omega +1]+x\) with \(x\) over \(\{ 0,\dots ,\omega \} \); any multiset containing an ordinal in \((\omega +1,\sigma ]\) is larger. The corresponding products lie in \(A\cdot (A\cup D_2^{-1})\cdot A\subseteq V_1\).

4. As long as Case 1 applies, the indices are taken in increasing order; every multiset containing an ordinal born after \(\pi \) exceeds every multiset of \(N(\pi )\) [Jak19, p. 3], and \(j[\pi ]+x{\lt}(j+1)[\pi ]\). The strata over \(N(\sigma )\) lie in \(V_1\), so \(D_{j[\pi ]+x}\subseteq D_\pi ^{+j}+V_1\) (recall \(V_1\subseteq D_\pi \)). Before \(k[\pi ]\) the current set lies in \(D_\pi ^{+(k-1)}+V_1\), which by hypothesis does not contain \(D_\pi ^{+k}\); it is therefore not additively closed, so neither Case 2 nor Case 3 can occur, and \(k[\pi ]\) is eventually the least multiset with a new sum. \(\square \)

1. By (T3) the torsion submodules of \(M\) form a sublattice \(L(M)\) of the submodule lattice, so \(L(M)\) is modular.

2. For \(L\subseteq L'\) in \(L(M)\), the quotient \(L'/L\) is torsion by (T2). The interval \([L,L']\) has no interior point exactly when \(L'/L\) is \(0\) or Tor-simple: the torsion submodules of \(L'/L\) are the \(L''/L\) with \(L''\in [L,L']\), by (T2).

3. By the Schreier refinement theorem [Coh06, App. Thm A.3, p. 521], applied in the modular lattice \(L(M)\) of torsion submodules (not in the lattice of all submodules), the chains \(0\subseteq N\subseteq M\) and \((M_j)\) have refinements in \(L(M)\) whose intervals pair off into projective intervals.

4. Projective intervals are isomorphic as lattices and give isomorphic quotient modules [Coh06, p. 521]. A refinement of \((M_j)\) only repeats terms, so its nontrivial intervals are the \([M_{j-1},M_j]\).

5. Hence every nontrivial interval of the refinement of \(0\subseteq N\subseteq M\) has a Tor-simple quotient isomorphic to some \(M_j/M_{j-1}\). The part between \(0\) and \(N\) is a \(\mathcal C\)-filtration of \(N\).

6. The part between \(N\) and \(M\) is a chain \(N=L_0\subseteq \cdots \subseteq L_t=M\). The \(L_i/N\) are torsion submodules of \(M/N\) with \((L_i/N)/(L_{i-1}/N)\cong L_i/L_{i-1}\), which gives a \(\mathcal C\)-filtration of \(M/N\).

7. If \(M/N\) is Tor-simple, a \(\mathcal C\)-filtration of it with distinct terms is \(0\subset M/N\), so \(M/N\) is isomorphic to a member of \(\mathcal C\). \(\square \)

(1) \(\operatorname {im}(G_1FG_2)\subseteq \operatorname {im}(G_1F)\), so \(\operatorname {coker}(G_1F)\) is a quotient of \(\operatorname {coker}Y\); likewise \(\operatorname {coker}F^{\mathsf T}\) is a quotient of \(\operatorname {coker}(F^{\mathsf T}G_1^{\mathsf T})\), and transposition does not change Smith forms; a quotient does not increase \(\mu (t^{j-1}\, \cdot )\). (2) \(\operatorname {coker}Y\) maps onto \(\operatorname {coker}[L\ \boldsymbol \omega ]\) with kernel generated by \(m\) elements, and \(\operatorname {coker}[L\ \boldsymbol \omega ]\) is a quotient of \(\operatorname {coker}L\); over a discrete valuation ring a submodule of an \(m\)-generated module is \(m\)-generated. (3) \(\operatorname {coker}[L\ \boldsymbol \omega ]=\operatorname {coker}L/\operatorname {im}\boldsymbol \omega \) and \(t^{j-1}(\mathcal X/\mathcal X')\cong t^{j-1}\mathcal X/(t^{j-1}\mathcal X\cap \mathcal X')\) with \(t^{j-1}\mathcal X\cap \mathcal X'\) at most \(m\)-generated, so \(\mu (t^{j-1}\, \cdot )\) drops by at most \(m\); then \(\operatorname {coker}Y\twoheadrightarrow \operatorname {coker}[L\ \boldsymbol \omega ]\). (4) Since each \(\Delta _i\) is injective, \(\operatorname {coker}L\) is an iterated extension of the \(\operatorname {coker}\Delta _i\), and \(t^{e_1+\dots +e_q}\) kills it. \(\square \)

Suppose that \(H^{-1}\) has entries in \(U_\Sigma \). By Lemma 3.1 there are \(Q\in \Sigma \) and an atomic factor of \(Q\) stably associated with \(G\). In a complete factorization, \(Q=G_1\mathcal A G_2\) with \(\mathcal A,G_1,G_2\) square of the size of \(Q\) and \(\mathcal A\) an atom stably associated with that factor, hence with \(G\) [Coh06, Cor. 0.5.5]. Then \(\mu _Q(G)=\mu _Q(\mathcal A)\) by condition 1, and \(\mu _Q(\mathcal A)\le \mu _Q(Q)\) by condition 2, a contradiction. \(\square \)

\(\mathcal U_r\subseteq U_{\Sigma _r}\): \(U_{\Sigma _r}\) is a ring containing \(E_1\) and \(f^{-1}\), since \(-f^{-1}\) is a corner entry of \(\tilde B_f^{-1}\). \(U_{\Sigma _r}\subseteq \mathcal U_r\): the entries of \(\tilde B_f^{-1}\) are polynomials in the entries of \(L^{-1}\) (whose entries lie in \(E_1\), because \(\ell ^{-1}\in E_1\) and \(a_i\in A\)), in \(f^{-1}\) and in the entries of the inverse pivot, which lie in \(A\). \(\square \)

\(\tilde B_M\) is full, because \(M\) is full and \(\Delta \) is invertible over \(D\). \(\Gamma _2'\subseteq U_{\Sigma _2'}\): \(U_{\Sigma _2'}\supseteq U_{\Sigma _1}=\mathcal U_1\); the block of \(\tilde B_M^{-1}\) complementary to the triangular pivot is \(B_M^{-1}\), and the \((1,1)\) block of \(B_M^{-1}\) is \(M^{-1}\) (Schur complement of \(\Delta \)). \(U_{\Sigma _2'}\subseteq \Gamma _2'\): the entries of \(\tilde B_M^{-1}\) are polynomials in those of \(B_M^{-1}\), of the inverse pivot (in \(A\)) and of elements of \(A\), and

with \(\Delta ^{-1}\in M_m(E_1)\subseteq M_m(\mathcal U_1)\) and \(R,C\) over \(A\). \(\square \)

\(U_0\) has zero trace, determinant \(-t\varepsilon \) and \(U_0^2=t\varepsilon I\), and conjugation by \(K^i\) preserves all three. The parts: every entry is divisible by \(t\) (since \(k\ge 1\)), the \((1,2)\) entry is \(-t\), and \(\det =-t^{k+1}\). \(\square \)

(1) A monomial of \(\mathcal J^r\) is \(a_0\, u_{i_1}^2a_1\cdots u_{i_r}^2a_r\) with \(a_i\in A\); each \(u_i^2\) evaluates to the central scalar \(t^{k+1}I\) (Lemma 4.1) and each \(a_i\) integrally. (2) Apply the same to \(U_i^r\), whose square is \(t^{r(k+1)}I\); \(\det \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}=3\ne 0\). \(\square \)

(1) At \(t=0\), \(\ell \mapsto (1-\beta \chi (h))I\). (2) Write \(\ell =g(1-\beta x_j^n)g^{-1}\) with \(\beta (-1)^n=1\). For \(n{\lt}0\) use \(1-\beta x_j^n=-\beta x_j^n(1-\beta ^{-1}x_j^{-n})\), a unit times a binomial of the same kind; so let \(n{\gt}0\). The roots of \(1-\beta X^n\) are simple, so \(1-\beta X^n=(1+X)c(X)\) with \(c(-1)\neq 0\); since \(x_j^\gamma \to -I\), \(c(x_j)^\gamma \) and \(g^\gamma \) are units, and the parts are those of \(u_j^\gamma =U_j\), namely \((1,k)\) (Lemma 4.1). (3) After conjugation, which changes \(\ell ^\gamma \) by a unit, write \(h=x_{j_1}^{n_1}\cdots x_{j_q}^{n_q}\) cyclically reduced. If \(q=1\), then \(\beta \chi (h)=1\) makes \(\ell \) exceptional, so \(q\ge 2\) and \(j_{i+1}\ne j_i\) for all \(i\), indices mod \(q\). Write \(N=tE_{12}\) and \(N_j=K^jNK^{-j}=t\, c_jr_j\) with \(c_j=(1,jt)^{\mathsf T}\) and \(r_j=(-jt,1)\). Then \(r_ac_b=(b-a)t\), \(N_j^2=0\), and \(\rho _t(x_j)^n=(-1)^n(I+nN_j)\) for \(n\in \mathbb Z\). Expanding the product,

over the subsets \(S=\{ i_1{\lt}\dots {\lt}i_m\} \) of \(\{ 1,\dots ,q\} \), with the factors in increasing order. For \(m\ge 1\),

which vanishes for \(m=1\). Hence \(\operatorname {tr}\rho _t(h)\) is a polynomial in \(t^2\) of degree \(q\) with leading coefficient \(\chi (h)\prod _in_i\prod _i(j_{i+1}-j_i)\ne 0\) (trace law; compare [LU69] for two parabolic generators). Since \(\det \rho _t(x_j)=1\), \(\det (I-\beta \rho _t(h))=1-\beta \operatorname {tr}\rho _t(h)+\beta ^2\) is a nonconstant polynomial, hence \(\not\equiv 0\), and \(e_\ell {\lt}\infty \). Finally \(\ell ^{\gamma _k}=\ell ^{\gamma }|_{\varepsilon =0}+O(t^k)\), and a perturbation \(O(t^k)\) of a matrix whose parts are all \({\lt}k\) does not change them (\(D+O(t^k)=D(I+O(t))\)). \(\square \)

Each chain \(C_\tau \) has at most \(q\) binomials on its diagonal, each with \(\det \not\equiv 0\) on the arc and largest part \(\le k\) (Lemma 4.3). By Lemma 3.2(4) the parts of \(L^\gamma \) are at most \(qk\), so \(\nu _{qk+1}(L^\gamma )=0\). The pivots \(x_j\) of the lift evaluate to units, so \(\tilde B_f^\gamma \) and \(B_f^\gamma \) have the same parts up to zeros. The border is one row and one column, i.e. \(m=2\) after evaluation, and Lemma 3.2(2) gives \(\nu _{qk+1}\le 0+2\). \(\square \)

Entries of \(h^\Theta \) are regular at \(t=0\) and \(t=2\), so evaluation there is a ring homomorphism. If \(\beta \ne 1\): at \(t=0\), \(h^\Theta =I\) and the determinant is \((1-\beta )^2\neq 0\). If \(\beta =1\): at \(t=2\), \(\Theta (x_j)\) is the Lipschitz quaternion \(1+2\mathbf q_j\) of norm \(5\), and \(5(1+2\mathbf q_j)^{-1}=1-2\mathbf q_j\). For a reduced word \(h\) of length \(|h|\) with \(n_-\) negative letters, \(h^{\Theta }(2)=5^{-n_-}\Pi \) with \(\Pi \) a product of \(|h|\) elements of \(\{ 1\pm 2\mathbf i,1\pm 2\mathbf j,1\pm 2\mathbf k\} \), so \(\operatorname {Norm}\Pi =5^{|h|}\). Reduce modulo \(5\) through the ring homomorphism from the Lipschitz order to \(M_2(\mathbb F_5)\) with \(\mathbf i\mapsto \operatorname {diag}(2,3)\), \(\mathbf j\mapsto \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\). The six residues have rank one, \(M_a=u_av_a^{\mathsf T}\), and \(M_aM_b=(v_a^{\mathsf T}u_b)\, u_av_b^{\mathsf T}\) vanishes exactly when \(b\) is the residue of the conjugate \(1\mp 2\mathbf q\) of \(a=1\pm 2\mathbf q\) (a check of the \(36\) products; Appendix A.2, supplement C05). A reduced word has no consecutive inverse pair, so \(\Pi \not\equiv 0\pmod5\). If \(h^\Theta (2)=I\), then \(\Pi =5^{n_-}I\), \(5^{2n_-}=5^{|h|}\), \(|h|=2n_-\ge 1\), so \(n_-\ge 1\) and \(\Pi \equiv 0\), a contradiction. Finally, for a real quaternion \(q\ne 1\), \(\det (I-q)=\operatorname {Norm}(1-q){\gt}0\). Two-term elements are \(\alpha g(1+\alpha ^{-1}\beta g^{-1}h)\) with \(\det \Theta (g)\ne 0\). \(\square \)

\(s\in \mathcal F_2\) because \(u^{-2}=u^{-1}\cdot 1\cdot u^{-1}\), so \(s^{-1}\in \mathcal U_2\). Suppose \(s^{-1}\in \mathcal U_1\). By §5.1 and Lemma 3.5, \(H_0^{-1}\) has entries in \(U_{\Sigma _1}\). We check the hypothesis of Theorem 3.4 with \(G=H=H_0\) and, for each \(Q\), \(\mu _Q=\nu _{k+1}\) along \(\gamma _k\) (Example (a)), with \(k\) chosen after \(Q\) (the case \(G=H\), \(\mu =\nu _j\) of Theorem 3.4).

• \(H_0\) is a full atom over \(P\). It is \(H_{3,2}\), an atom of \(\mathbb C\langle u,v,w\rangle =P\) by Lemma 5.7 below (the case \(n=3\), \(p=2\)).

• Target. By Lemma 4.2(2) with \(r=1\), \(\nu _{k+1}(H_0^{\gamma _k})=4\) for every \(k\).

• Denominators. For \(Q=Q_s\) or \(x_j\), \(\nu _{k+1}\le 2\). For \(Q=\tilde B_f\), \(0\ne f\in V_1=\mathcal F_1\), choose \(k{\gt}e(f)\); Lemma 4.4 with \(q=1\) gives \(\nu _{k+1}(\tilde B_f^{\gamma _k})\le 2\).

Since \(4{\gt}2\), Theorem 3.4 gives \(H_0^{-1}\notin M_2(U_{\Sigma _1})\), a contradiction. \(\square \)

Figure 4 shows this comparison on computed examples, together with the corresponding comparisons in the proofs of Theorems 5.2, 7.2 and 7.3.

For \(r=1\): the element \(S_3=u^{-1}+v^{-1}+w^{-1}\in V_1=\mathcal F_1\) has \(S_3^{-1}\notin E_1=\mathcal U_0\). This is Theorem 3.4 with \(\mu \) the corank at \(\chi \) (Example (b), a homomorphism \(P\to M_1(\mathbb C)\); the case \(G=H\), \(\mu =\operatorname {corank}\) of Theorem 3.4), with target \(H_{3,1}=\operatorname {diag}(v,w)+u\mathbf1\mathbf1^{\mathsf T}\) (an atom, Lemma 5.7(1) with \(n=3\), \(p=1\)): at \(\chi \), \(H_{3,1}\) vanishes and has corank \(2\), while every \(Q\in \Sigma _0\) has corank \(\le 1\) there (the pivot \(T\) of \(Q_s\) evaluates at \(\chi \) to an invertible triangular matrix, since \(x_j\mapsto -1\), so \(Q_s^\chi \) is equivalent to \(I\oplus s^\chi \); and \(x_j^\chi =-1\)). For \(r=2\): Theorem 5.1, since \(s\in \mathcal F_2\). (The case \(r=1\) also follows from [HS15, Thm 4.6] by two pivots; see §8.8.)

For \(r\ge 3\): \(\varphi _r\in \mathcal F_r\), so \(\varphi _r^{-1}\in \mathcal U_r\). Suppose \(\varphi _r^{-1}\in \mathcal U_{r-1}\). By Lemma 5.5 and Lemma 3.5, \(Q_r^{-1}\) has entries in \(U_{\Sigma _{r-1}}\). Apply Theorem 3.4 with \(G=H=Q_r\) (a full atom, Lemma 5.7(2)) and \(\mu _Q=\nu _j\) along \(\gamma _k\), \(j=(r-1)k+1\): given \(Q\in \Sigma _{r-1}\), choose \(k{\gt}\max (e(Q),r)\). Then \(\nu _j(Q_r^{\gamma _k})\ge 3\) by Lemma 5.6, while \(\nu _j(Q^{\gamma _k})\le 2\) by Lemma 4.4 (with \(q=r-1\)) or by the remark after it for \(Q\in \Sigma _0\). Contradiction. \(\square \)

\(g_r\in \mathcal F_{2r}\) (\(u^{-2r}\) is a product of \(2r\) inverses of \(u\)), so \(g_r^{-1}\in \mathcal U_{2r}\). Suppose \(g_r^{-1}\in \mathcal U_r\). By §5.1 and Lemma 3.5, \(H_{3,2r}^{-1}\) has entries in \(U_{\Sigma _r}\). Apply Theorem 3.4 with \(G=H=H_{3,2r}\) (an atom, Lemma 5.7(1)) and \(\mu _Q=\nu _j\) along \(\gamma _k\), \(j=rk+1\): for \(Q\in \Sigma _r\) choose \(k{\gt}e(Q)\). By Lemma 4.2(2) all four parts of \(H_{3,2r}^{\gamma _k}\) equal \(r(k+1)\ge rk+1\), so \(\nu _j=4\), while \(\nu _j(Q^{\gamma _k})\le 2\) by Lemma 4.4 with \(q=r\), or by the remark after it for \(Q\in \Sigma _0\). Contradiction. \(\square \)

\(L\) is triangular with letters (units of \(\Lambda \)) on the diagonal, and the inverse of the block of chain \(\tau \) has \((1,r)\) entry \(\ell _{\tau ,1}^{-1}\cdots \ell _{\tau ,r}^{-1}\); block elimination gives the first two claims. The block inverse of \(Q_r\) uses only \(L^{-1}\) (entries are products of inverse letters, in \(E_1\)), \(\varphi _r^{-1}\) and the entries of \(\boldsymbol \alpha ,\boldsymbol \omega \), which lie in \(A\). \(\square \)

A chain block is equivalent to \(I\oplus \ell _{\tau ,r}^\gamma \cdots \ell _{\tau ,1}^\gamma \) (the \(-1\)’s are unit pivots). For \(r=3\) the product is \(U^3=t^{k+1}U\), with parts \((k+2,2k+1)\). For \(r\ge 4\) write \(U_i=t(\mathbf a_i\mathbf b_i-t^{k-1}E_{21})\) with \(\mathbf a_i=(1,it)^{\mathsf T}\), \(\mathbf b_i=(it,-1)\), so \(\mathbf b_a\mathbf a_b=(a-b)t\). If the letters of the product have indices \(i_1,\dots ,i_r\) with \(i_j\ne i_{j+1}\), its leading term is \(t^r\cdot t^{r-1}\prod _j(i_j-i_{j+1})\, \mathbf a_{i_1}\mathbf b_{i_r}\), whose \((1,2)\) entry has valuation exactly \(2r-1\); corrections have valuation \(\ge r+k-1\ge 2r\). So the first invariant factor is \(t^{2r-1}\) and, as \(\det U_i=-t^{k+1}\), the second is \(t^{r(k+1)-(2r-1)}\). Now \(rk-r+1\ge (r-1)k+1\iff k\ge r\) and \(2r-1{\lt}(r-1)k+1\); for \(r=3\), \(2k+1\ge 2k+1{\gt}k+2\). Five chains give \(\nu (L^\gamma )=5\), and Lemma 3.2(3) with \(m=2\) gives \(\ge 3\). \(\square \)

Proof of (1). Let \(R=[\mathbf1\mid I_{n-1}]\in M_{(n-1)\times n}(\mathbb C)\), with columns \(r_1=\mathbf1\) and \(r_i=e_{i-1}\) for \(i\ge 2\), and let \(\mathsf U=\operatorname {diag}(u_1,\dots ,u_n)\). Then \(H_{n,p}=R\, \mathsf U^pR^{\mathsf T}\). The matrix \(H_{n,p}\) is full, since \(H_{n,p}(1,\dots ,1)=I+\mathbf1\mathbf1^{\mathsf T}\) has determinant \(n\). It is not invertible, since it vanishes at \(0\) (§2.2(a)). Put \(N=np-1\).

The pencil. For \(p=1\) let \(\mathbf L=H_{n,1}=R\, \mathsf UR^{\mathsf T}\). For \(p\ge 2\) take the vector \(v=(v_0,v_1,\dots ,v_{p-1})\) with \(v_0\in \mathbb C^{n-1}\) and \(v_k\in \mathbb C^n\), and let \(\mathbf L\) be the \(N\times N\) block matrix with

and zero blocks elsewhere. The block \(T\) on the indices \(1,\dots ,p-1\) is \(I-\mathsf N\), where \(\mathsf N\) is block strictly upper triangular with \(\mathsf U\) on the block superdiagonal. So \(T^{-1}=\sum _{m{\lt}p-1}\mathsf N^m\) is polynomial, and \((T^{-1})_{1,p-1}=\mathsf U^{p-2}\). The Schur complement of \(T\) is \(-R\, \mathsf U\, \mathsf U^{p-2}(-\mathsf UR^{\mathsf T})=H_{n,p}\). By §2.2(b),(c) it suffices to prove that \(\mathbf L\) is an atom.

Translation. Write \(\mathbf L=\mathbf L_0+\sum _iu_i\mathbf L_i\). Substitute \(u_i=1+z_i\), an affine automorphism (§2.2(d)). The new pencil has constant term \(K=\mathbf L(1,\dots ,1)\) and the same coefficients \(\mathbf L_i\). Its determinant is \(\det K=\det T(1)\cdot \det (RR^{\mathsf T})=n\ne 0\).

Rank-\(p\) factorization of the coefficients. For \(c=(c_1,\dots ,c_p)\in \mathbb C^p\) let \(\iota _i c\) be the vector with block \(0\) equal to \(r_ic_1\), block \(k\) equal to \(-e_ic_{k+1}\) (\(1\le k\le p-1\)). Let

Then \(\mathbf L_i=\iota _i\varkappa _i\). The map \(\iota _i\) is injective, because its \(p\) components land in different blocks and \(r_i,e_i\neq 0\). The map \(\varkappa _i\) is onto, for the same reason. For \(p=1\), \(\iota _i=r_i\) and \(\varkappa _i=r_i^{\mathsf T}\).

Monic. We have \(\sum _i\operatorname {im}\mathbf L_i=\sum _i\operatorname {im}\iota _i=\bigl(\sum _i\mathrm{span}(r_i)\bigr)\oplus \bigoplus _{k\ge 1}\bigl(\sum _i\mathrm{span}(e_i)\bigr)=\mathbb C^N\), because \(R\) has rank \(n-1\). Next, \(\bigcap _i\ker \mathbf L_i=\bigcap _i\ker \varkappa _i\): this forces \(v_k=0\) for \(k\ge 1\), and \(R^{\mathsf T}v_0=0\), so \(v_0=0\). Hence the row and the column of coefficients have rank \(N\).

The transfer matrices. Let \(\vartheta _i=K^{-1}\iota _i\), injective. To compute \(v=\vartheta _ic\), solve \(Kv=\iota _ic\):

• blocks \(1\le k\le p-2\) give \(v_{k+1}=v_k+e_ic_{k+1}\);

• block \(p-1\) gives \(v_{p-1}=R^{\mathsf T}v_0-e_ic_p\);

• block \(0\) gives \(Rv_1=r_ic_1\).

Hence \(RR^{\mathsf T}v_0=r_i\sigma \) with \(\sigma =\mathbf1^{\mathsf T}c\), and

The matrix \(G\) is the orthogonal projector onto \(\operatorname {im}R^{\mathsf T}=(\ker R)^\perp \). Since \(\ker R=\langle \mathbf n\rangle \) with \(\mathbf n=(1,-1,\dots ,-1)\), we have \(G=I-\mathbf n\mathbf n^{\mathsf T}/n\). Also \(r_j^{\mathsf T}v_0=e_j^{\mathsf T}Ge_i\, \sigma \). Therefore

where \((\mathsf N_p)_{km}=1\) for \(m{\gt}k\) and \(0\) otherwise. In particular \(g_{ji}=\pm 1/n\neq 0\) for \(i\ne j\). (For \(p=1\), \(\mathsf N_1=0\) and (5.2) reads \(r_j^{\mathsf T}(RR^{\mathsf T})^{-1}r_i=g_{ji}\).)

Cyclicity. Let \(M=a\mathbf1\mathbf1^{\mathsf T}-\mathsf N_p\) with \(a\in \mathbb C\). Then \(\mathbf1\) is a cyclic vector for \(M\), and \(\mathbf1^{\mathsf T}\) is cyclic for \(M\) acting on row vectors from the right.

• Since \(Mx=-\mathsf N_px+a(\mathbf1^{\mathsf T}x)\mathbf1\), induction gives \(\mathrm{span}\{ M^m\mathbf1\} _{m{\lt}q}=\mathrm{span}\{ \mathsf N_p^m\mathbf1\} _{m{\lt}q}\) for every \(q\).

• The vector \(\mathsf N_p^m\mathbf1\) vanishes below position \(p-m\) and has entry \(1\) there. So these vectors form a triangular basis.

• The row case is the same: \(\mathbf1^{\mathsf T}\mathsf N_p^m\) vanishes at positions \(\le m\) and equals \(1\) at position \(m+1\).

No invariant subspace. Let \(A_i=K^{-1}\mathbf L_i=\vartheta _i\varkappa _i\), and let \(W\neq 0\) be invariant under all \(A_i\). Put \(M_i=\varkappa _i\vartheta _i=g_{ii}\mathbf1\mathbf1^{\mathsf T}-\mathsf N_p\). Then \(A_i^m\vartheta _i=\vartheta _iM_i^m\), and \(A_j\vartheta _i=g_{ji}\vartheta _j\mathbf1\mathbf1^{\mathsf T}\) for \(j\ne i\).

1. Pick \(w\in W\setminus 0\). By the monic condition some \(c=\varkappa _iw\ne 0\), and \(A_iw=\vartheta _ic\in W\setminus 0\).

2. Since \(c\neq 0\) and \(\mathbf1^{\mathsf T}\) is cyclic for \(M_i\), there is \(m\) with \(\beta =\mathbf1^{\mathsf T}M_i^mc\neq 0\). Fix \(j\neq i\) (possible since \(n\ge 2\)). Then \(W\ni A_jA_i^m\vartheta _ic=A_j\vartheta _iM_i^mc=g_{ji}\beta \, \vartheta _j\mathbf1\), so \(\vartheta _j\mathbf1\in W\).

3. Then \(A_j^m\vartheta _j\mathbf1=\vartheta _jM_j^m\mathbf1\in W\) for all \(m\). Since \(\mathbf1\) is cyclic for \(M_j\), \(\operatorname {im}\vartheta _j\subseteq W\).

4. For every \(l\neq j\), \(\vartheta _je_1\in W\) by step 3, and \(A_l\vartheta _je_1=g_{lj}\vartheta _l\mathbf1\mathbf1^{\mathsf T}e_1 =g_{lj}\vartheta _l\mathbf1\in W\). So \(\vartheta _l\mathbf1\in W\), and by step 3 \(\operatorname {im}\vartheta _l\subseteq W\).

Hence \(W\supseteq \sum _l\operatorname {im}\vartheta _l=K^{-1}\sum _l\operatorname {im}\iota _l=\mathbb C^N\). By Lemma 2.1 the translated pencil is an atom; so are \(\mathbf L\) (§2.2(d)) and \(H_{n,p}\) (§2.2(c)). \(\square \)

The proof of Theorem 5.1 applies with three changes.

• The arc sends every letter to the exact arc, \(u_i\mapsto U_{i-1}(t,t^k)\) for \(i=1,\dots ,N\); Lemma 4.1, the trace law and Lemmas 4.3–4.4 hold verbatim for distinct indices. (Sending the extra letters to \(-I\) does not work: \(1-[x_N,x_1]\) then evaluates to \(0\).)

• The target is \(H_{n,2}\), with §5.1 read with \(u_2\) in place of \(v\). Its \(2(n-1)\) parts all equal \(k+1\), since \(\det (\mathbf1\mathbf1^{\mathsf T}+I_{n-1})=n\), against the bound \(2\) of Lemma 4.4.

• \(H_{n,2}\) is an atom of \(P_N=\mathbb C\langle x_1,\dots ,x_N\rangle \). The translated pencil of the proof of Lemma 5.7(1) is a pencil over \(P_N\) whose coefficients at \(x_{n+1},\dots ,x_N\) are zero. Zero coefficients change neither the ranks of the row and the column of coefficients nor the common invariant subspaces of the \(M_i\). So Lemma 2.1 applies over \(P_N\), and §2.2(b)–(d) give the claim as in Lemma 5.7(1). \(\square \)

Fix \(d\ge 2\). \(\hat F_d\) is a full \(d\times d\) matrix over \(A\) (proof of (6.1)), so \(\rho _d\in A_d\). By (6.2), \(\sigma \in E_1\) and \(\sigma \ne 0\), so \(\rho _d=\sigma ^{-1}\in E_2\).

Suppose \(\rho _d\in W_{d-1}\). Since \(E_1\subseteq W_1\subseteq W_{d-1}\) and \(T_d^{-1}\) has entries in \(E_1\), (6.2) gives \(\hat F_d^{-1}\in M_d(W_{d-1})=M_d(U_{\Sigma ^{(d-1)}})\). Apply Theorem 3.4 with \(G=H=\hat F_d\) (a full atom, Lemma 6.2(c)), \(\Sigma =\Sigma ^{(d-1)}\) and, for every \(Q\), \(\mu _Q\) the corank at \(\Theta \) (Example (b) with \(K=\mathbb C(t)\); the case \(G=H\), \(\mu =\operatorname {corank}\) of Theorem 3.4): \(\operatorname {corank}\hat F_d^\Theta =2d{\gt}2(d-1)\ge \operatorname {corank}Q^\Theta \) for every \(Q\in \Sigma ^{(d-1)}\) (Lemma 6.2(a),(b)). This is a contradiction, so \(\rho _d\notin W_{d-1}\). (Figure 3(b) shows both sides of this comparison for \(d\le 6\).)

The consequences: \(A_{d-1}\subseteq W_{d-1}\), so \(\rho _d\in (A_d\cap E_2)\setminus A_{d-1}\) and \(\rho _d\in B_d\setminus B_{d-1}\). For \(d=2\), \(W_1\cap E_2=\mathcal U_\infty \) (§2.1), so \(\rho _2\in E_2\setminus \mathcal U_\infty \). Finally \(\bigcup _bB_b=E_2\): a nonzero \(e\in E_1\) has a representation \(e=\gamma -\mathbf uM^{-1}\mathbf v\) with \(M\) full over \(A\) [Coh06, Thm 7.1.2(d)], with \(\Sigma \) the full matrices over \(A\), since the \(\Sigma \)-rational closure of \(A\) is a subfield of \(D\) containing \(A\), hence all of \(D\); then \(N=\begin{pmatrix} M & \mathbf v \\ \mathbf u & \gamma \end{pmatrix}\) is full over \(A\) of some order \(n\) and \(e^{-1}=(N^{-1})_{nn}\in A_n\subseteq W_n\); \(E_2\) is generated by \(E_1\) and such inverses, and the \(W_b\) increase. \(\square \)

The strictness of the chain \(A_b\) alone does not need Lemma 4.5: with \(\Sigma _A^{(b)}=\Sigma _0\cup \{ \)lifts of full \(a\times a\) matrices over \(A\), \(a\le b\} \) one has \(A_b=U_{\Sigma _A^{(b)}}\) (as in Lemma 3.5), and the lifts carry only the pivots \(x_j\), which are invertible at \(\Theta \), so their corank is at most \(2a\) directly. Lemma 4.5, whose proof amounts to the freeness of the group generated by \(1+2\mathbf i,1+2\mathbf j,1+2\mathbf k\) modulo scalars, is needed only for \(W_b\), that is, for \(\mathcal U_\infty \subsetneq E_2\) and for \(B_{d-1}\subsetneq B_d\). For the witnesses \(f_k\) the character \(\chi \) already suffices (Remark 7.11); it does not detect \(\rho _d\), since \(\hat F_d^\chi =8\mathsf C_d\) is invertible.

(a) For \(Q_s\) (\(s\in A\)) the pivot block is triangular with \(\Theta (x_j)\) on the diagonal, invertible, so \(\operatorname {corank}Q_s^\Theta =\operatorname {corank}s^\Theta \le 2\), also when \(s^\Theta =0\) (e.g. \(s=c\)). \(\Theta (x_j)\) has corank \(0\). In \(\tilde B_G^\Theta \) the triangular pivot evaluates to an invertible matrix (\(\Theta (x_j)\) on its diagonal), and its Schur complement is \(B_G^\Theta \), since \(\Theta \) is a ring homomorphism on \(A\). In \(B_G^\Theta \) the block \(L^\Theta \) is block upper triangular with the \(2\times 2\) blocks \(\ell ^\Theta \) on the diagonal, invertible by Lemma 4.5. Taking Schur complements does not change the corank, so \(\operatorname {corank}\tilde B_G^\Theta \) is the corank of a \(2a\times 2a\) matrix over \(\mathbb C(t)\), at most \(2a\le 2b\). (No element of \(R_{\rm bin}\) is evaluated at \(\Theta \); only the polynomial matrix \(\tilde B_G\) is.)

1. \(q_\lambda ^\Theta =-t^2(1+(\lambda -1)-\lambda )I=0\) and \(c^\Theta =0\).

2. We prove a slightly more general statement.

Claim. Let \(C,D\in M_d(\mathbb C)\) satisfy:

• (H1) \(C\) and \(D\) have no common invariant subspace other than \(0\) and \(\mathbb C^d\);

• (H2) \(D\) is invertible;

• (H3) \(E:=D-I-C^2\) is invertible.

Then \(F=(\xi ^2-\eta ^2)I+(\eta ^2-\zeta ^2)D+cC\), with \(c=\xi \eta +\eta \xi \), is a full atom of \(P=\mathbb C\langle \xi ,\eta ,\zeta \rangle \).

Proof. Let \(\mathbf L=A_0+\xi A_\xi +\eta A_\eta +\zeta A_\zeta \) be the \(4d\times 4d\) pencil:

Its lower right block is \(I_{3d}\), and the Schur complement is \((\xi I+\eta C)\xi +(\xi C+\eta (D-I))\eta -\zeta D\zeta =\xi ^2I+cC+\eta ^2(D-I)-\zeta ^2D=F\). So \(\mathbf L\) is stably associated with \(F\) (§2.2(b)). \(F\) is full, because \(F(1,0,0)=I\), and not invertible, because \(F(0)=0\).

Monic. Write vectors as \((v_0,v_1,v_2,v_3)\) in blocks of size \(d\). For a row \(f\):

If all three vanish, then \(f=0\). For a column \(v\):

If all three vanish, then \(v_0=0\), \(v_1=-Cv_2\), \(Ev_2=0\) and \(Dv_3=0\), so \(v=0\) by (H2), (H3).

Normalization. Substitute \(\xi =1+\xi '\) (§2.2(d)). The constant term becomes

By Lemma 2.1 it suffices to show that \(M_a=K^{-1}A_a\) (\(a=\xi ,\eta ,\zeta \)) have no common invariant subspace \(W\neq 0\) other than \(\mathbb C^{4d}\). Use the coordinates \(z=(v_0,\, v_1+Cv_2,\, v_2,\, v_3)\), an invertible change. A direct computation gives

The polynomials \(\mathsf P=2M_\xi -M_\xi ^2\) and \(\mathsf N=M_\xi ^2-M_\xi \) in \(M_\xi \) act by \(\mathsf Pz=(z_0,z_1,0,0)\) and \(\mathsf Nz=(z_1,0,0,0)\). Let \(B_k\) be the \(k\)-th block subspace.

1. \(W\) meets \(B_0\). Let \(0\ne z\in W\).

   • If \(z_1\ne 0\), take \(\mathsf Nz\).

   • If \(z_1=0\ne z_0\), take \(\mathsf Pz\).

   • If \(z_0=z_1=0\ne z_2\), take \(\mathsf NM_\eta z=(Ez_2,0,0,0)\), nonzero by (H3).

   • Otherwise take \(\mathsf NM_\zeta z=(-Dz_3,0,0,0)\), nonzero by (H2).

2. \(B_0\subseteq W\). Let \(W_0=\{ a:(a,0,0,0)\in W\} \ne 0\). On \(B_0\) we have \(\mathsf PM_\eta (a,0,0,0)=(Ca,0,0,0)\) and \(\mathsf NM_\zeta ^2(a,0,0,0)=\mathsf N(Da,Da,0,0)=(Da,0,0,0)\). So \(W_0\) is invariant under \(C\) and \(D\), and \(W_0=\mathbb C^d\) by (H1).

3. The other blocks. Let \(a\in \mathbb C^d\).

   • \(M_\zeta (a,0,0,0)=(0,0,0,-a)\) gives \(B_3\subseteq W\).

   • \(M_\eta (a,0,0,0)=(Ca,0,-a,0)\), minus \((Ca,0,0,0)\in B_0\), gives \(B_2\subseteq W\).

   • \(M_\eta (0,0,a,0)=(Ea,Ea,0,0)\), minus \((Ea,0,0,0)\), gives \((0,Ea,0,0)\in W\). By (H3), \(B_1\subseteq W\).

So \(W=\mathbb C^{4d}\). By Lemma 2.1 and §2.2(c),(d), \(F\) is an atom. \(\square \)

Let \(D_x\) be as in Lemma 7.5. For \(k=3\): \(s\in D_\pi ^{+3}\) and \(s^{-1}\notin \mathcal U_1=D_x\) (Theorem 5.1). For \(k\ge 4\): \(f_k\in D_\pi ^{+k}\) and \(f_k^{-1}\notin \Gamma _{k-2}'\supseteq D_x\) (Theorem 7.3); for \(k=4\) also \(\psi \in D_\pi ^{+4}\) and \(\psi ^{-1}\notin \Gamma _2'\) (Theorem 7.2). In each case the witness lies in no stratum born before \(b_{kXX}\), since its inverse would then be in \(D_x\); so its complexity is \(\operatorname {cp}=b_{kXX}\), \(D_{b_{kXX}}^{-1}\not\subseteq D_x\), and the Case 3 at which the pointer reaches \(b_{kXX}\) selects it. That Case 3 exists: otherwise every Case 3 would select an earlier stratum, and the union of the series, which is \(D\) [Jak19, Def. 1.1.4], would lie in \(\Gamma _{k-2}'\), against \(f_k^{-1}\notin \Gamma _{k-2}'\) (or \(s^{-1}\notin \mathcal U_1\)). The ring reached after the step, \(D_x[(D_\pi ^{+k}\setminus 0)^{-1}]\), still lies in \(E_2\), since \(D_\pi ^{+k}\subseteq E_1\). \(\square \)

\(u^{-2}=u^{-1}\cdot 1\cdot u^{-1}\) and \(yu^{-2}y^{-1}\) has outer coefficients \(y,y^{-1}\in A\), so \(\psi \in \mathcal F_2\). By Lemma 7.8, \(F_\psi \) is full, so \(H_\psi =\Delta _\psi (I+\Delta _\psi ^{-1}\mathbf1u^2\mathbf1^{\mathsf T})\) is full, and since \(I+\mathbf a\mathbf b^{\mathsf T}\) is invertible iff \(1+\mathbf b^{\mathsf T}\mathbf a\) is, \(u^2\psi \ne 0\); hence \(\psi ^{-1}\in \mathcal U_2\). Also \(yu^{-2}y^{-1}=(yuy^{-1})^{-1}(yuy^{-1})^{-1}\) with \(yuy^{-1}=1+yxy^{-1}\) a binomial, so \(\psi \in D_\pi ^{+4}\).

Suppose \(\psi ^{-1}\in \Gamma _2'\). Since \(H_\psi =\Delta _\psi +\mathbf1u^2\mathbf1^{\mathsf T}\) and \(\psi =u^{-2}+\mathbf1^{\mathsf T}\Delta _\psi ^{-1}\mathbf1\), Woodbury gives \(H_\psi ^{-1}=\Delta _\psi ^{-1}-\Delta _\psi ^{-1}\mathbf1\psi ^{-1}\mathbf1^{\mathsf T}\Delta _\psi ^{-1}\) with \(\Delta _\psi ^{-1}\in M_3(E_1)\), and \(F_\psi ^{-1}=\operatorname {diag}(1,1,y^{-1})H_\psi ^{-1}\in M_3(\Gamma _2')=M_3(U_{\Sigma _2'})\) (Lemma 3.6). Apply Theorem 3.4 with \(G=H=F_\psi \) and \(\mu _Q=\nu _{k+1}\) along \(\gamma _k\) (the case \(G=H\), \(\mu =\nu _j\) of Theorem 3.4). Given \(Q\in \Sigma _2'\), choose \(k\) larger than \(e(Q)\): then \(\nu _{k+1}(Q^{\gamma _k})\le 2\) for \(Q=Q_s\) or \(x_j\) and for the borders of \(\mathcal U_1\) (Lemma 4.4 with \(q=1\)), and \(\le 4\) for \(Q=\tilde B_M\) (Lemma 7.7). Since \(\nu _{k+1}(F_\psi ^{\gamma _k})=6\) for every \(k\) (Lemma 7.8), this contradicts Theorem 3.4. Hence \(\psi ^{-1}\notin \Gamma _2'\).

\(D'\subseteq \Gamma _2\subseteq \Gamma _2'\) by Lemma 7.6, so \(\psi ^{-1}\in \mathcal U_2\setminus D'\). Each element of \(D_\pi \) lies in \(\mathcal F_2\), so \(D'\subseteq \mathcal U_2\); and \(s=u^{-2}+v^{-2}+w^{-2}\in D_\pi ^{+3}\) gives \(s^{-1}\in D'\setminus \mathcal U_1\) by Theorem 5.1. Finally \(\psi ^{-1}\in \mathcal U_2\subseteq \mathcal U_\infty \) and \(\psi ^{-1}\notin \Gamma _2\), while \(\rho _2\in \Gamma _2\) (\(\hat F_2\) is a full \(2\times 2\) matrix over \(A\)) and \(\rho _2\notin \mathcal U_\infty \) (Theorem 6.1). Both rings lie in \(E_2\): \(\mathcal U_\infty \subseteq E_2\) (§2.1), and for a full \(2\times 2\) matrix over \(A\) a pivot (after a permutation if needed) leaves a Schur complement in \(E_1\), so the entries of its inverse lie in \(E_2\). \(\square \)

Each summand of \(f_k\) is the inverse of the square of \(w\) or of a binomial \(y^iuy^{-i}=1+y^ixy^{-i}\), so \(f_k\in D_\pi ^{+k}\cap \mathcal F_2\), and \(f_k\ne 0\) (Lemma 7.9(a)). Suppose \(f_k^{-1}\in \Gamma _{k-2}'\). Woodbury gives \(F_k^{-1}=\operatorname {diag}(y^{-b})\bigl(\Delta _k^{-1}-\Delta _k^{-1}\mathbf1 f_k^{-1}\mathbf1^{\mathsf T}\Delta _k^{-1}\bigr)\) with \(\Delta _k^{-1}\) over \(E_1\), so \(F_k^{-1}\) has entries in \(\Gamma _{k-2}'=U_{\Sigma '_{k-2}}\) (Lemma 3.6 for \(a=k-2\)). Given \(Q\in \Sigma _{k-2}'\), choose \(\kappa {\gt}e(Q)\); then \(\nu _{\kappa +1}(Q^{\gamma _\kappa })\le 2(k-2)\) (Lemma 7.7, Lemma 4.4), while \(\nu _{\kappa +1}(G_k^{\gamma _\kappa })=2(k-1)\) (Lemma 7.9(d)). Theorem 3.4, with \(G=G_k\), \(H=F_k\) and \(\mu _Q=\nu _{\kappa +1}\) along \(\gamma _\kappa \), gives a contradiction. \(\square \)

Write \(f=d_1^{-1}+\dots +d_{n'}^{-1}+r\) with \(n'\le n\) pure summands and \(r\in V_1\) (§2.3; \(V_1\) is closed under sums). If \(n'\le 2\), then \(f=d_1^{-1}+d_2^{-1}+r\), \(d_1^{-1}+r\) or \(r\), and

with \(d_1rd_2+d_1+d_2,\ 1+d_1r\in V_1\), so \(f^{-1}\in \mathcal U_1\subseteq \Gamma _{k-2}'\). If \(n'\ge 3\), then \(k\ge 4\). Put \(Y=d_1rd_2+d_1+d_2\in V_1\) and

The Schur complement of the diagonal block is \(Y+d_1\bigl(\sum _{i\ge 3}d_i^{-1}\bigr)d_2=d_1fd_2\ne 0\), so \(M\) is full and \(f^{-1}=d_2(M^{-1})_{11}d_1\), with \(n'-1\le k-2\). \(\square \)

By Lemma 2.2, every stratum born before \(k[\pi ]\) is contained in \(A\), in \(V_1\), or in \(D_\pi ^{+n}+V_1\) with \(n\le k-1\). By §2.3, at the Case 3 where the pointer reaches \(b_{kXX}\) the current ring contains \(D_\gamma ^{-1}\) for every earlier stratum \(\gamma \), whether \(\gamma \) was a stopping step (its inverses were adjoined there) or not (the pointer passed it at a Case 3 where its inverses already lay in the ring); and between two Case 3 steps Cases 1 and 2 only add sums and products. So \(D_x\) is generated by \(A\) and the inverses of the nonzero elements of all earlier strata. Conversely, for \(n\le k-1\) the elements of \(D_\pi ^{+n}+V_1\) are the sums indexed by the multiset \(n[\pi ]+[\sigma ]{\lt}k[\pi ]\); since Case 1 selects the least multiset with a new sum (Lemma 2.2(5)), they already lie in the set built before \(b_{kXX}\), so the inverses of its nonzero elements lie in \(D_x\). These inverses lie in \(E_1\), in \(\mathcal U_1\), or in \(\Gamma _{k-2}'\) by Lemma 7.4; and \(D_x\) contains \(A^{-1}\) and \(V_1^{-1}\), since \(D_\omega =A\) and \(D_\sigma =V_1\) are earlier strata. This gives the first two claims, and the third because \(\Gamma _1'=\mathcal U_1\). Finally \(\mathcal U_1\subseteq E_2\) and \(D_\pi ^{+n}+V_1\subseteq E_1\), so the inverses of its nonzero elements lie in \(E_2\). \(\square \)

(1) \(H_3=\Delta +\mathbf1d_1\mathbf1^{\mathsf T}\) with \(\Delta =\operatorname {diag}(d_2,d_3)\), and \(e=d_1^{-1}+\mathbf1^{\mathsf T}\Delta ^{-1}\mathbf1\); Woodbury gives the formula, and \(H_3\) is full iff \(1+d_1(d_2^{-1}+d_3^{-1})=d_1e\ne 0\). (2) With two pure summands and one in \(V_1\), \(e=p'^{-1}+q'^{-1}+r\) and \(e^{-1}=q'(p'rq'+p'+q')^{-1}p'\) with \(p'rq'+p'+q'\in V_1\) (Lemma 7.4); with one pure summand, \((d^{-1}+r)^{-1}=(1+dr)^{-1}d\) with \(1+dr\in V_1\); with none, \(e\in V_1\). \(\square \)

Over \(R_j=\mathbb C[[t]]/t^j\), a square matrix \(N\) of size \(n\) satisfies \(\nu _j(N)=n-\mu (\operatorname {im}N)\), with \(\mu \) the minimal number of generators. Let \(n=2a+2m\) be the size of \(B_M^{\gamma }\) and \(\mathrm{pr}\) the projection onto the last \(2m\) coordinates. Then \(\mathrm{pr}(\operatorname {im}B_M)=\operatorname {im}[-C\ \Delta ]\supseteq \operatorname {im}\Delta \). An image of a module needs no more generators, and neither does a submodule, viewing \(R_j\)-modules as \(\mathbb C[[t]]\)-modules with the same \(\mu \) (\(R_j=\mathbb C[[t]]/t^j\) is not a principal ideal domain, but \(\mathbb C[[t]]\) is), so \(\mu (\operatorname {im}B_M)\ge \mu (\operatorname {im}\Delta )\) and \(\nu _j(B_M)\le 2a+2m-\mu (\operatorname {im}\Delta )=2a+\nu _j(\Delta )=2a+\sum _s\nu _j(\ell _s)\). For \(k{\gt}e(M)\), every \(\ell _s^{\gamma _k}\) has largest part at most \(k\) (Lemma 4.3: units, exceptional binomials with parts \((1,k)\), the others with parts \({\lt}k\)), so \(\nu _{k+1}(\ell _s^{\gamma _k})=0\). The pivots of the lift are units on the arc. \(\square \)

(b) \(u^2,v^2,w^2\mapsto t^{k+1}I\) (Lemma 4.1), central; \(\det (I_3+J_3)=4\); and \(\det (U-I)=1-\operatorname {tr}U+\det U=1-t^{k+1}\) for \(U=v^{\gamma _k}\), a unit of \(\mathbb C[[t]]\).

1. Let \(Q=Q_0+uQ_u+vQ_v+wQ_w\) be the \(9\times 9\) pencil with nonzero entries (row, column, coefficient)

• \(Q_0\): \((i,i,1)\) for \(i=4,\dots ,9\);

• \(Q_u\): \((1,4,1)\), \((2,4,1)\), \((3,4,1)\), \((3,7,-1)\), \((4,1,-1)\), \((4,2,-1)\), \((4,3,1)\), \((4,9,1)\), \((7,3,-1)\), \((8,7,1)\);

• \(Q_v\): \((1,5,1)\), \((3,8,-1)\), \((5,1,-1)\), \((9,3,1)\);

• \(Q_w\): \((2,6,1)\), \((6,2,-1)\).

Its block \(Q_{22}\) (rows and columns \(4\)–\(9\)) is \(I+N\) with \(N^2=0\), so \(Q_{22}^{-1}=I-N\) is polynomial, the Schur complement of \(Q_{22}\) is exactly \(+F_\psi \), and \(Q\sim F_\psi \oplus I_6\) over \(P\). The pencil is epic (the stacked coefficient matrices have rank \(9\) both ways). At \(u=v=w=2\), \(\det Q(2,2,2)=256\), and \(K_i=16\, Q(2,2,2)^{-1}Q_i\) (\(i=u,v,w\)) are integer matrices. With \(\mathbf r=(0,0,-1,0,0,0,-2,4,1)^{\mathsf T}\) and \(\boldsymbol \ell =(2,0,-2,0,2,0,0,3,0)\),

and, writing \(K_\omega =K_{\omega _1}K_{\omega _2}\cdots \) for a word \(\omega \) (left to right),

with \(\mathcal R=(\varnothing ,u,uu,vu,wu,uuu,vuu,uvu,vvu)\) and \(\mathcal O=(\varnothing ,u,w,uu,uv,wu,ww,uuu,uuv)\). By (7.1) the rank-one matrix \(\mathbf r\boldsymbol \ell \) is a polynomial without constant term in \(K_v\), so it lies in the algebra generated by the \(K_i\); by (7.2) the vectors \(K_\omega \mathbf r\) and the covectors \(\boldsymbol \ell K_{\omega '}\) form bases, so the products \(K_\omega (\mathbf r\boldsymbol \ell )K_{\omega '}\) span \(M_9\). The translated pencil is monic with invertible constant term \(Q(2,2,2)\), and its normalized coefficients generate \(M_9\), so it is an atom by Lemma 2.1; the shift is an automorphism (§2.2(d)) and \(Q\sim F_\psi \oplus I_6\) (§2.2(b),(c)), so \(F_\psi \) is an atom. \(Q(2,2,2)\) is invertible, so \(Q\), and with it \(F_\psi \), is full (§2.2(a),(b)). \(\square \)

(a) Grade \(\Lambda _y\) by the exponent of \(w\) and put \(T=\sum _{i=0}^my^iu^{-2}y^{-i}\), so that \(f'=wf_kw=1+wTw\) has components in degrees \(0\) and \(2\) only. Let \(f'=gh\). The extreme components multiply, and the lowest ones give \(g_ah_c=1\), so \(g_a\) is a unit, i.e. a scalar times a group element; replacing \(g,h\) by \(gg_a^{-1},g_ah\) we may assume \(g_a=h_c=1\) in degree \(0\). The widths of \(g\) and \(h\) add up to \(2\). If one of them is \(0\), that factor is \(1\). Otherwise \(g=1+g_1\), \(h=1+h_1\) with \(g_1+h_1=0\) and \(g_1h_1=wTw\), so \(-g_1^2=wTw\). For a bi-invariant order of the free group the leading word of a product is the product of the leading words, so the leading word of \(wTw\), which is one of the distinct reduced words \(W_i=wy^iu^{-2}y^{-i}w\), would be a square. But \(W_i\) is cyclically reduced, so a square root of it would be cyclically reduced and \(W_i=hh\) as words; the first letter of \(W_i\) is \(w\) and the letter in position \(i+3\) is \(u^{-1}\). Hence \(f'\), which is not a unit, is an atom, and so is \(f_k=w^{-1}f'w^{-1}\).

1. Over \(\Lambda _y\), \(\Delta _k\) is invertible and \(H_k=\Delta _k(I+\mathbf a\mathbf b^{\mathsf T})\) with \(\mathbf a=\Delta _k^{-1}\mathbf1\), \(\mathbf b^{\mathsf T}=u^2\mathbf1^{\mathsf T}\); \(I+\mathbf a\mathbf b^{\mathsf T}\) is stably associated with \(1+\mathbf b^{\mathsf T}\mathbf a=u^2f_k\). Since \(\operatorname {diag}(1,y,\dots ,y^m)\) and \(u^2\) are units of \(\Lambda _y\), \(F_k\) is stably associated with \(f_k\ne 0\), hence full, and stable association preserves atoms [Coh06, Cor. 0.5.5, Thm 3.2.3 and p. 195].

2. If an atomic factor were stably associated with \(u\), then at any matrix point \(X\) where \(u(X)\) is singular, that factor, and hence \(F_k(X)\), would be singular (evaluate the stable association and the factorization). Take \(n=m+1\), \(u\mapsto U=\operatorname {diag}(0,1,\dots ,1)\), \(w\mapsto I\) and \(y\mapsto Y\) the cyclic permutation matrix, so that the evaluations of \(u^2\) and of \(y^iu^2y^{-i}\) are \(I-e_ie_i^{\mathsf T}\) (\(i=0,\dots ,m\)). A vector \(\mathbf v\) in the kernel of \(H_k\) gives, with \(s=(I-e_0e_0^{\mathsf T})\sum \mathbf v\), \(\mathbf v_w=-s\), \(\mathbf v_i=-s+c_ie_i\) with \(s\perp e_0,\dots ,e_m\) and \((m+2)s=\sum c_ie_i\); since \(e_0,\dots ,e_m\) is a basis, \(s=0\), so \(\sum c_ie_i=0\), \(c_i=0\) and \(\mathbf v=0\) (no division is needed). So \(F_k\) is invertible at a point where \(u\) is singular. For \(w\) a scalar point suffices: at \(u=y=1\), \(w=0\), \(F_k=H_k=\operatorname {diag}(0,1,\dots ,1)+J_{k-1}\), and subtracting its first row from the others leaves the rows \(e_1^{\mathsf T},\dots ,e_m^{\mathsf T}\), so \(\det F_k=1\).

3. Let \(F_k=G_1\cdots G_s\) be a complete factorization over \(P\). Over a weakly finite ring a product of square matrices is invertible only if every factor is: if \(XY\) is invertible, \(X\) has the right inverse \(Y(XY)^{-1}\), so \(X\) is invertible, and then so is \(Y\). By (b), \(F_k\) is not invertible over \(\Lambda _y\), so some \(G_{i_0}\) is not. If another \(G_{i_1}\) were not invertible over \(\Lambda _y\), say \(i_1{\gt}i_0\), then \(F_k=XY\) with \(X=G_1\cdots G_{i_0}\) and \(Y=G_{i_0+1}\cdots G_s\) full and both non-invertible over \(\Lambda _y\), against (b). So each \(G_i\), \(i\ne i_0\), is invertible over \(\Lambda _y=U_{\{ u,y,w\} }\). By Lemma 3.1 with \(\Sigma =\{ u,y,w\} \) (atoms of \(P\), being of degree one), it is stably associated with \(u\), \(y\) or \(w\), and by (c) with \(y\). On the other hand, \(G_{i_0}\) is not stably associated with \(y\), since it would then be invertible over \(\Lambda _y\) (converse part of Lemma 3.1). On the arc, \(\det y^{\gamma _\kappa }=1-t^{\kappa +1}\) (as in Lemma 7.8(b)), so \(y^{\gamma _\kappa }\), and by stable association every \(G_i^{\gamma _\kappa }\) with \(i\ne i_0\), is invertible over \(\mathbb C[[t]]\). Hence \(\nu _{\kappa +1}(G_{i_0}^{\gamma _\kappa })=\nu _{\kappa +1}(F_k^{\gamma _\kappa })\). Each \(y^iu^2y^{-i}\) evaluates to \(t^{\kappa +1}I\) (Lemma 4.1: \(u^2\mapsto t^{\kappa +1}I\) is central), so \(F_k^{\gamma _\kappa }=t^{\kappa +1}\bigl((I_{k-1}+J_{k-1})\otimes I_2\bigr)\operatorname {diag}(Y^b)\) with \(\det (I+J)=k\ne 0\), and all \(2(k-1)\) parts equal \(\kappa +1\). \(\square \)

(a) Let \(d\ge 2\), \(\Delta =\operatorname {diag}(w,yuy^{-1},\dots ,y^{d-1}uy^{-(d-1)})\), \(H=\Delta +u J_d\) and \(L=H\operatorname {diag}(1,y,\dots ,y^{d-1})\in M_d(P)\), whose entries are \(uy^b+\delta \, y^bu\) (\(w\) in place of \(y^0u\) for the first index). This is the construction of §7.6 with \(u,w\) in place of \(u^2,w^2\), and Lemma 7.9(a)–(c) and the first part of (d) hold for \(q_d\) and \(L\) with the same proofs. In (a), \(wq_d=1+wT\) with \(T=\sum _{i=0}^{d-1}y^iu^{-1}y^{-i}\) has components in degrees \(0\) and \(1\) only, so one of two factors has width \(0\), hence is a unit, and the argument with squares is not needed. In (b), \(1+u\mathbf1^{\mathsf T}\Delta ^{-1}\mathbf1=uq_d\). In (c), with \(n=d\), the matrix \(U\) is idempotent, so \(u\) and \(u^2\) have the same evaluation, and the two points give the same matrices as there. Each summand of \(q_d\) is an inverse binomial (\(w^{-1}=(1+z)^{-1}\), \(y^iu^{-1}y^{-i}=(1+y^ixy^{-i})^{-1}\)), so \(q_d\in V_1\), \(q_d\ne 0\) by (a), and \(e_d\in \mathcal U_1\subseteq \mathcal U_r\). Woodbury gives \(e_d=u-u\mathbf1^{\mathsf T}H^{-1}\mathbf1u\in A_d\), since \(H\) is a full \(d\times d\) matrix over \(A\). If \(e_d\in A_{d-1}\), then \(H^{-1}=\Delta ^{-1}-\Delta ^{-1}\mathbf1e_d\mathbf1^{\mathsf T}\Delta ^{-1}\), with \(\Delta ^{-1}\) over \(E_1=A_1\), and \(L^{-1}=\operatorname {diag}(y^{-b})H^{-1}\) have entries in \(A_{d-1}\). Every entry of \(L\) vanishes at \(\chi \), and the atomic factors of \(L\) other than the exceptional one, \(G\), are stably associated with \(y\), which is invertible at \(\chi \); so \(\operatorname {corank}G^\chi =d\), while the matrices of \(\Sigma _A^{(d-1)}\) have corank at most \(d-1\) at \(\chi \) (Remark 7.11(i)). Theorem 3.4, with \(G\), \(H=L\) and \(\mu \) the corank at \(\chi \), gives a contradiction. Finally \(A_d\cap \mathcal U_0=A_d\cap E_1=E_1\), since \(E_1=A_1\subseteq A_d\). (b) \(s\) for \(n=3\) (Theorem 5.1, \(\Gamma _1'=\mathcal U_1\)) and \(f_n\) for \(n\ge 4\) (Theorem 7.3). (c) Expanding the coefficients into monomials, a term of an element of \(\mathcal F_2\) with two inverse binomials is \(m_0\ell _1^{-1}m_1\ell _2^{-1}m_2=(\ell _1'^{-1}m_0m_1)(\ell _2^{-1}m_2)\) with \(\ell _1'=m_0\ell _1m_0^{-1}\), a product of two elements of \(D_{\omega +1}\) (Lemma 2.2(2)); terms with fewer inverses lie in \(V_1\subseteq D_\pi \). With \(D_\pi \subseteq \mathcal F_2\) this gives \(\mathcal F_2=\bigcup _nD_\pi ^{+n}\). By Lemma 7.5, \(\mathcal D_k\subseteq \mathcal D_{k+1}\) and \(\mathcal D_k\subseteq \mathcal U_2\), and \(\bigcup _k\mathcal D_k\) contains \(E_1\) and the inverse of every nonzero element of \(\mathcal F_2\), so it is \(\mathcal U_2\). Finally \(s\in D_\pi ^{+3}\) gives \(s^{-1}\in \mathcal D_4\setminus \mathcal D_3\) (Theorem 5.1), and for \(k\ge 4\), \(f_k\in D_\pi ^{+k}\) gives \(f_k^{-1}\in \mathcal D_{k+1}\setminus \mathcal D_k\), since \(\mathcal D_k\subseteq \Gamma _{k-2}'\) (Lemma 7.5 and Theorem 7.3). \(\square \)