Stanley–Reisner rings and simplicial complexes
Encode nonfaces by squarefree monomials and recover face counts through a graded Hilbert series.
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The question for today
What you’ll understand
- Build the squarefree nonface ideal of a finite simplicial complex.
- Derive its Hilbert series by monomial supports.
- State field and homological hypotheses behind Cohen–Macaulay h-positivity.
A simplicial complex specifies which subsets of vertices may occur together. Its Stanley–Reisner ring turns forbidden subsets into monomials that vanish. This translation preserves enough combinatorial information to recover every face count, while opening access to dimension, depth and homology. Fix a field k, a finite vertex set {1,…,n}, and a nonempty complex Δ containing the empty face and every listed vertex. Give each variable degree one; both the field and this standard grading are part of the construction.
Support determines which monomials survive
The Stanley–Reisner ideal is generated by the squarefree product of variables in every nonface; minimal nonfaces suffice. The quotient k[Δ] retains a monomial exactly when its support is a face. Powers themselves are not forbidden: if vertex i is present, all positive powers of xi survive. Thus this ring is not the finite-dimensional face vector space whose basis is only squarefree faces. It is an infinite-dimensional graded algebra with a finite combinatorial support rule.
For a face F, monomials with support exactly F have positive exponents on its vertices and exponent zero elsewhere. Their generating series is (t/(1−t)) to the size of F, with the empty face contributing one. Summing over faces gives the Hilbert series. This counts an explicit monomial basis and works over every field, so the resulting face numbers and Hilbert coefficients are characteristic-independent even though depth and homology can later depend on characteristic.
Dimension and the h-polynomial
Write d for the maximum face size, so the complex has dimension d−1 and its face ring has Krull dimension d. Put f−1=1 for the empty face and fi−1 for the number of i-element faces. Clearing the denominator of the support formula defines the h-polynomial of degree at most d. This change of coordinates between f and h is purely algebraic and does not require purity, shellability or Cohen–Macaulayness. The resulting h-coefficients need not be nonnegative for an arbitrary complex.
When k is infinite and k[Δ] is Cohen–Macaulay, a linear system of parameters is a regular sequence. Quotienting by those d linear forms produces a finite-dimensional graded algebra whose Hilbert series is the h-polynomial. Hence every h-coefficient is a nonnegative integer. These hypotheses explain the positivity, rather than allowing one to infer it from the rational-form Hilbert series alone. Over a finite field, one may extend scalars to an infinite field without changing the relevant Hilbert numbers.
Homology tests depth through every link
For a face F, its link consists of faces G disjoint from F such that their union remains a face. Reisner’s criterion says that a finite complex is Cohen–Macaulay over k exactly when every link, including the link of the empty face, has vanishing reduced k-homology below its own dimension. This is stated as a theorem, not proved from the elementary support count. The empty face test matters: inspecting only vertex links can miss a disconnected complex.
Homology over k can depend on characteristic because integral torsion behaves differently after changing coefficients. Consequently the face ring may be Cohen–Macaulay over one field and fail over another, although its monomial Hilbert series stays the same. Even under Cohen–Macaulayness, h-positivity alone does not imply unimodality, log-concavity or real-rootedness. Each stronger property needs its own assumptions or structure. Keeping these distinctions prevents an attractive translation between algebra and combinatorics from becoming an unjustified chain of implications.
Work it through
The boundary of a triangle
Let Δ contain every proper subset of {1,2,3}, but not the full triple. Compute its face ring, Hilbert series and h-polynomial.
The only minimal nonface is {1,2,3}, so the ideal is (x₁x₂x₃). All three vertices and all three edges are present; the maximum face size is d=2.
The support count gives 1+3t/(1−t)+3t²/(1−t)². Putting these terms over the common denominator yields (1+t+t²)/(1−t)².
Alternatively the nonzero homogeneous element x₁x₂x₃ is a degree-three nonzerodivisor in the polynomial ring. Its quotient has series (1−t³)/(1−t)³, equal to the same expression. The first graded dimensions are 1,3,6,9.
The ring is k[x₁,x₂,x₃]/(x₁x₂x₃), of dimension two, with h-vector (1,1,1). Its degree-three dimension is nine rather than the polynomial ring’s ten.
The living notebook
Explore the model
Interactive lab · probability
Faces versus all subsets
Pause, reset, then set p=0.5 and n=3. Predict masses 1,3,3,1 divided by eight. These count subsets by cardinality; removing the size-three outcome leaves seven allowed faces of the triangle boundary, with probability 7/8 under uniform independent vertex selection. The lab still shows all eight subsets by rank. It does not compute monomial powers, a Hilbert series, links or Cohen–Macaulayness.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forFace counts are finite subset data; the graded ring also counts arbitrarily high powers with those allowed supports.
Pause & reason
Check your understanding
Your turn
Try it yourself
Take a complex consisting of two disjoint edges on four vertices. Find its h-vector and explain what it tells you about Cohen–Macaulayness.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
There are four vertices, two edges and d=2. Use the f-to-h formula before drawing a topological conclusion.
Compare with the solution
The numerator is (1−t)²+4t(1−t)+2t²=1+2t−t², so h=(1,2,−1). Its negative final coefficient rules out Cohen–Macaulayness, since a Cohen–Macaulay standard graded face ring has nonnegative h-coefficients after extending to an infinite field if needed. Topologically, the complex is disconnected, so its reduced H₀ is nonzero below dimension one; the empty-face instance of Reisner’s criterion also detects the failure over every field.
Keep the idea
Take it with you
- Squarefree nonface generators kill monomials by support, not by degree.
- Face sizes determine the Hilbert series and h-polynomial in every characteristic.
- Cohen–Macaulay h-positivity requires depth or link-homology hypotheses.
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