The scalar rooted-factor algebra has the exact form
$$ \boxed{ A = \mathbb Q[z,x_1,x_2,\ldots] \times_{\mathbb Q[u,v]} \mathbb Q[v(1+u)]. } $$
This one pullback explains a slightly rude collection of facts:
- \(A\) is integrally closed;
- \(A\) is not Noetherian and is not finitely generated;
- \(A\) admits no Hopf algebra structure at all;
- but \(A\) does admit a commutative, cocommutative, counitary bialgebra;
- and its universal Hopf envelope is an explicit Laurent polynomial ring.
Normal but wildly non-Noetherian, no antipode internally, then a perfectly clean antipode after one forced localization. I keep staring at this shape.
Where the algebra comes from#
For a rooted tree \((T,r)\), let
$$ F_{T,r} = \sum_{B\subseteq E(T)} \left(z^{|C_r(B)|}+x_{|C_r(B)|}\right) \prod_{C\ne C_r(B)}x_{|C|}. $$
The component containing the root may stay open, contributing \(z^{|C_r(B)|}\), or close and contribute \(x_{|C_r(B)|}\). The other components are closed.
Let \(A\) be the subalgebra generated by these rooted factors. Root grafting is encoded by the operator
$$ \mathscr B\left(\sum_{k\ge0}z^kp_k(\mathbf x)\right) = z\sum_{k\ge0}z^kp_k(\mathbf x) + \sum_{k\ge0}x_{k+1}p_k(\mathbf x). $$
Now define
$$ \Phi(z)=v, \qquad \Phi(x_k)=uv^k, \qquad w=v(1+u). $$
Then
$$ \Phi(\mathscr B(P))=w\Phi(P). $$
The complete answer is
$$ \boxed{A=\Phi^{-1}(\mathbb Q[w]).} $$
So a polynomial belongs to the rooted-factor algebra exactly when its rank-one specialization depends only on \(w=v(1+u)\).
The fiber coordinates#
Put
$$ s=z+x_1, \qquad e_k=x_k-z^{k-1}x_1\quad(k\ge2), \qquad K=(e_2,e_3,\ldots). $$
Then
$$ \ker\Phi=K $$
and the pullback becomes the wonderfully blunt decomposition
$$ \boxed{ A=\mathbb Q[s]\oplus K \subset R=\mathbb Q[s,z,e_2,e_3,\ldots]. } $$
The sum is a vector-space decomposition, not a product decomposition. A basis consists of \(s^b\), together with every monomial
$$ s^bz^ae^\mu $$
whose \(e\)-multi-index is nonzero. The moment one \(e_k\) appears, arbitrary powers of \(z\) are allowed. Without an \(e_k\), only a polynomial in \(s\) survives.
That is the entire algebra.
Normal, with an infinite conductor#
The conductor of \(A\subset R\) is exactly
$$ \boxed{(A:_R R)=K.} $$
Away from \(K\), the two rings become equal. On the conductor, the map of spectra is
$$ \operatorname{Spec}\mathbb Q[s,z] \longrightarrow \operatorname{Spec}\mathbb Q[s], $$
so every conductor fiber is an affine line.
The ring \(A\) is normal. An element of its fraction field integral over \(A\) first lands in the polynomial ring \(R\); reducing modulo \(K\) then shows that its \(z\)-degree must be zero, which puts it back in \(\mathbb Q[s]\oplus K\).
But the first normal neighborhood of the conductor is
$$ \boxed{ K/K^2 \cong \bigoplus_{\substack{k\ge2\\a\ge0}} \mathbb Q[s]\,z^a\bar e_k. } $$
This is free of countably infinite rank over \(A/K=\mathbb Q[s]\). Hence \(K\) is not finitely generated, and neither is \(A\). Every point on the conductor line has countably infinite embedding dimension.
Normality did not make this ring tame. It just made the wildness honest.
The higher layers retain a cute rank-one memory. If \(G_{k,a}=z^a\bar e_k\), then the complete relations in the normal cone are
$$ \boxed{ G_{i,a}G_{j,b}=G_{i,a+b}G_{j,0}. } $$
All \(z\)-weight can slide onto one factor. The conductor has infinitely many directions, but their multiplication remembers only the total shift.
Why Hopf symmetry fails#
The vertex grading already forbids a connected graded Hopf structure. The Hilbert dimensions begin
$$ 1,1,2,4,8,14,24,38,59. $$
PBW factorization would force primitive multiplicities
$$ 1,1,2,3,4,3,2,-3 $$
through degree eight. The final number cannot be a dimension.
Dropping the grading does not help. A conductor point has a nontrivial quadratic cotangent relation coming from
$$ e_2(z^2e_2)=(ze_2)^2, $$
while an off-conductor point has a polynomial local ring and no such relation. A Hopf algebra would translate rational points into one another, forcing their local rings to be isomorphic. They are not.
Thus
$$ \boxed{A\text{ admits no Hopf algebra structure over }\mathbb Q.} $$
There is also a separate compatibility obstruction: the exact induced deletion coproduct of a rooted edge does not even land in \(A\otimes A\). The scalar algebra has forgotten the distinction between the open singleton \(z\) and the closed singleton \(x_1\). The bialgebra below is therefore an abstract structure on \(A\); it does not counterfeit induced deletion or root-grafting compatibility.
Where the exact combinatorics goes#
The failure is not a vague shortage of structure. It is exactly one erased bit at weight one.
Pointed graphs retain that bit. If \((G,r)\) is a graph with one distinguished vertex, then
$$ \rho(G,r) = \sum_{S\subseteq V(G)\setminus\{r\}} (G-S,r)\otimes G[S] $$
makes pointed graphs a right Hopf module over the ordinary induced-subgraph graph Hopf algebra. Exact rooted deletion survives, and grafting obeys a Boolean distributive law recording which old branch roots remain attached to the new root.
For several boundary vertices, mark the whole boundary set \(R\). Marked graphs form a connected graded Hopf algebra, and the component profile
$$ \mathcal Z_{G,R} = \sum_{B\subseteq E(G)} \prod_C b_{|C|,|C\cap R|} $$
is a Hopf morphism when every \(b_{a,j}\) is primitive. Its zero-mark slice is the unrooted polynomial \(U_G\), and its one-mark slice specializes to the rooted factor \(F_{G,r}\).
So the terminal classification is pleasantly sharp:
$$ \boxed{ \begin{array}{c} \text{the unchanged scalar algebra has one abstract non-Hopf bialgebra,}\\ \text{while exact deletion and grafting require the pointed or multi-root carrier.} \end{array}} $$
That is better than merely proving that a preferred coproduct fails. It says where the lost structure went, and gives it back with the smallest boundary data that can possibly remember \(z\) and \(x_1\) separately.
The bialgebra that survives#
The fiber coordinates make one exact construction almost unavoidable:
$$ \Delta(s)=s\otimes1+1\otimes s, \qquad \Delta(z)=z\otimes1+1\otimes z, $$
and
$$ \boxed{ \Delta(e_k)=e_k\otimes e_k, \qquad \epsilon(e_k)=1. } $$
So \(s\) and \(z\) are primitive, while every kernel coordinate is group-like.
For a basis monomial with \(e^\mu\ne1\),
$$ \Delta(s^bz^ae^\mu) = \sum_{i=0}^b\sum_{j=0}^a \binom bi\binom aj (s^iz^je^\mu) \otimes (s^{b-i}z^{a-j}e^\mu). $$
Both tensor factors still contain the complete \(e^\mu\), so both lie in \(K\subset A\). The coproduct really does restrict to \(A\).
This bialgebra is nonconnected and does not preserve vertex weight. It cannot have an antipode inside \(A\): each group-like \(e_k\) would need to be a unit, but
$$ A^\times=\mathbb Q^\times. $$
That is exactly how the construction slips between the two no-go theorems.
The forced Hopf envelope#
The group-like elements are precisely
$$ G(A)=\{e^\mu:\mu\in\mathbb N^{(\mathbb N_{\ge2})}\}, $$
the primitive space is \(\mathbb Qs\), and the coradical is
$$ \mathbb Q[e_2,e_3,\ldots]. $$
Any bialgebra map from \(A\) to a Hopf algebra must send every \(e_k\) to an invertible group-like element. Therefore every \(e_k\) must be inverted, and the universal Hopf envelope is
$$ \boxed{ \mathbb Q[s,z,e_2^{\pm1},e_3^{\pm1},\ldots]. } $$
Its antipode is
$$ S(s)=-s, \qquad S(z)=-z, \qquad S(e_k)=e_k^{-1}. $$
No smaller localization works: omit one \(e_j^{-1}\), and the antipode identity still fails on \(e_j\).
The scalarization that made root grafting multiplicative erased too much boundary state for the natural Hopf structure to descend. What remains is not structureless. It is a normal pullback with an infinite conductor, one sharp nonconnected bialgebra, and a completely forced Laurent repair.
Notebook references: D-0128, R-0582, R-0583, R-0588, R-0617, R-0624, R-0628, R-0634, R-0636, R-0639, R-0646, R-0651, R-0654, R-0666