The Khovanov-Seidel path ideal

$J$ is spanned by the paths of length at least one, and is homogeneous. Write $B_{\ge 1}$ for the set of basis elements of [L1] that are arrows or returns, and $P_{\ge 1}$ for the $\mathbb{Z}$-span of all paths of length at least one. Every generator of $J$ lies in $P_{\ge 1}$, and $P_{\ge 1}$ is closed under left and right multiplication by $A_m$: the product of a path of length at least one with any path is either $0$ or a concatenation of length at least one, and multiplication is bilinear; hence $J\subseteq P_{\ge 1}$ by minimality of the generated ideal. Conversely every path of length at least one is a product of arrows, hence lies in $J$ because a product of arrows belongs to $J$ and $J$ is closed under multiplication; so $P_{\ge 1}\subseteq J$, and $P_{\ge 1}=J$. Every basis element of $B_{\ge 1}$ is homogeneous by [L2], so $J$, the span of the basis elements of $B_{\ge 1}$, is homogeneous: it is the direct sum of its intersections with the homogeneous components of $A_m$.

$J^2$ is spanned by the returns, and $J^3=0$. By step 1.1 it suffices to compute products of two basis elements of $B_{\ge 1}$ and of three such elements. A concatenation of two paths of length at least one has length at least two, and by [L1] the only nonzero classes of length at least two in $A_m$ are the returns $(i\mathrm{\mid }i-1\mathrm{\mid }i)$ for $1\le i\le m$ (equal to $(i\mathrm{\mid }i+1\mathrm{\mid }i)$ only for $i<m$), each of which is a product $(i\mathrm{\mid }i-1)(i-1\mathrm{\mid }i)$ of two arrows; hence $J^2$ is spanned by the returns. A concatenation of three paths of length at least one is a path of length at least three, which vanishes in $A_m$ by [L1], so $J^3=0$.

The quotient is ${\mathbb{Z}}^{m+1}$. The assignment ${\psi }_0(e_i):=u_i$ for the standard basis vectors $u_0,\dots ,u_m$ of ${\mathbb{Z}}^{m+1}$ and ${\psi }_0(b):=0$ for every non-vertex basis element $b$ of [L1] is a unital ring homomorphism: on the multiplication table of [L1] one checks that a product of two basis elements, when nonzero, is either a vertex (and the product of the corresponding idempotents is ${\delta }_{ij}{e}_i$, matching $u_i{u}_j={\delta }_{ij}{u}_i$) or a non-vertex basis path (whose image and whose factors' product of images are both $0$ unless both factors are vertices), and paths of length at least three vanish on both sides; bilinearity extends the check to $A_m$. Since ${\psi }_0$ kills every arrow, it kills $J$, define $\psi ([x]):={\psi }_0(x)$. This is well defined: if $[x]=[y]$, then $x-y\in J$ and ${\psi }_0(x)-{\psi }_0(y)={\psi }_0(x-y)=0$. The coset formulas [L4] show that $\psi$ preserves addition and multiplication and sends $[1]$ to $1$, so it is a unital ring homomorphism with $\psi \pi ={\psi }_0$. Let $\phi :{\mathbb{Z}}^{m+1}\to A_m\mathrm{/}J$, $\phi (a_0,\dots ,a_m):={\sum }_i{a}_i[e_i]$, where $[e_i]=\pi (e_i)$; the classes $[e_i]$ are orthogonal idempotents with ${\sum }_i[e_i]=1$ because $\pi$ is a unital ring homomorphism, so $\phi$ is a unital ring homomorphism, and $\psi \phi =\mathrm{i}\mathrm{d}$ since $\psi ([e_i])=u_i$. Conversely $\phi \psi ([b])=[b]$ for every basis element $b$ of [L1]: for a vertex this is immediate, and for a non-vertex element both sides are $0$ because $b\in J$ by step 1.1; hence $\phi \psi ={\mathrm{i}\mathrm{d}}_{A_m\mathrm{/}J}$ and $A_m\mathrm{/}J\cong {\mathbb{Z}}^{m+1}$.

Conclusion. The ideal $J$ generated by the arrows is homogeneous and consists of the paths of length at least one, its square is spanned by the returns and its cube vanishes, and the quotient is ${\mathbb{Z}}^{m+1}$ on the vertex idempotents, all by steps 1.1–3.1. No choice principle is used.