Let
$$ S=\mathbb Q[e_1,\ldots,e_j] $$
be the symmetric-polynomial ring in elementary coordinates, and define
$$ I_d=(p_{d+1},p_{d+2},\ldots,p_{d+j}), $$
where the \(p_n\) are power sums in \(j\) variables.
The generators have different weighted degrees, so one might expect a complicated tangent cone. Instead, for every monomial order compatible with ordinary degree in the \(e_r\),
$$ \operatorname{in}(I_d) =(e_1,\ldots,e_j)^{d+1}. $$
The quotient has the brutally simple basis
$$ \left\{ e_1^{a_1}\cdots e_j^{a_j}: \sum_{r=1}^j a_r\le d \right\}. $$
A weighted complete intersection has compressed all the way down to the ordinary degree-\(d\) simplex.
Why the quotient has the right size#
Work first in
$$ R=\mathbb Q[x_1,\ldots,x_j] $$
and let
$$ J_d=(x_1^{d+1},\ldots,x_j^{d+1}). $$
The consecutive power sums lie in \(J_d\). Their only common zero is the origin: if the distinct nonzero coordinate values are \(y_s\) with multiplicities \(m_s\), the first relevant vanishing equations form a Vandermonde system in the \(y_s\), forcing every \(m_s=0\), which is impossible.
Thus the power sums form a homogeneous system of parameters and hence a regular sequence. The quotient length is
$$ \dim_\mathbb Q S/I_d =\prod_{r=1}^j\frac{d+r}{r} =\binom{d+j}{j}. $$
On the other side, the symmetric invariants of \(R/J_d\) have the monomial-symmetric basis indexed by partitions inside a \(j\times d\) rectangle, giving the same dimension. Therefore
$$ I_d=J_d\cap S. $$
The quotient is literally the symmetric part of a box-truncated polynomial ring.
Why the tangent cone is ordinary#
The elementary monomials of ordinary degree at most \(d\) are triangularly related to the monomial-symmetric basis of the \(j\times d\) box. Their residue classes are therefore linearly independent and already have the full quotient cardinality \(\binom{d+j}{j}\).
No one of them can be an initial monomial of \(I_d\). Every monomial of degree \(d+1\) or more must be. Hence
$$ \operatorname{in}(I_d) =(e_1,\ldots,e_j)^{d+1}. $$
This is independent of the chosen degree-compatible monomial order. There is no delicate Gröbner fan to navigate.
Multiplying by the determinant shifts the box exactly#
The top elementary coordinate is
$$ e_j=x_1x_2\cdots x_j. $$
For every \(L\ge0\),
$$ (I_{L+d}:e_j^L)=I_d. $$
Indeed, in the full polynomial ring,
$$ (J_{L+d}:(x_1\cdots x_j)^L)=J_d, $$
and contraction to the symmetric ring preserves the identity.
So multiplication by \(e_j^L\) embeds the degree-\(d\) quotient into the degree-\(L+d\) quotient as the exact interval of exponent vectors shifted by \(L\) in every coordinate. The determinant coordinate does not merely suggest a filtration; it implements the box translation algebraically.
The beautiful structure does not triangularize the derivative map#
This quotient geometry was sought as a route to binary-margin transversality. It does not finish that job.
Already at \((j,L,d)=(4,1,2)\), the natural conjugate-partition filtration is not preserved by the derivative map. Two source vectors satisfy
$$ \Phi(e_3e_1^3)=12e_2+30e_1^2, \qquad \Phi(e_2^2e_1^2)=28e_2+54e_1^2, $$
and the second image leaks into the next filtration layer.
So the tangent cone and colon theorem exactly describe the target complete intersection, but they do not manufacture a triangular proof of transversality. A continuation needs a new source basis or a comparison that is willing to mix the canonical layers.
Notebook references: C-0192