# The relations of (ab, bc, ca)

My question about “rank-two freeness” leads to a relation module, an exact resolution, and three ways to check it.

2026-10-17 · Study note · worked proof

> An edited record of my questions about I=(ab,bc,ca). The complete proofs were developed with assistance; attribution and the unfinished independent reconstruction are kept explicit.

## My initial question

I proposed I=(ab,bc,ca) and wondered whether it might be “free of rank two”. I also asked how the discussion had become an exact chain, and why a proposed list really generated the whole kernel.

## The correction and its source

The ideal is not free; its relation module is free of rank two. The full kernel proof below was supplied with assistance. In the related textbook example I asked why one relation among relations did not let us delete a generator: its coefficients were not units, so that deletion was invalid.

## What the saved record establishes

I connected the inability to delete a relation with continuing the resolution. This is a recorded question and local reformulation, not evidence that I independently reconstructed every exactness and minimality argument. Rebuilding that proof without the supplied argument remains a useful review task.

## A generated ideal need not be free

$$
S=k[a,b,c],\qquad I=(ab,bc,ca),\qquad R=S/I.
$$

The three displayed monomials generate I as an S-module, but they are not a free basis: polynomial relations exist. A nonzero ideal in a domain has rank one after tensoring with the fraction field. If it is free, it must be principal. This ideal is not principal: a common generator would divide ab, bc, and ca, whose greatest common divisor is 1, yet I is proper.

The rank-two free module appears instead as the module of relations among the three generators. This separates the ideal, its chosen generating module, and the kernel of that presentation.

## Generate the whole kernel

$$
A_1=\begin{pmatrix}ab&bc&ca\end{pmatrix},\qquad A_2=\begin{pmatrix}c&0\\0&a\\-b&-b\end{pmatrix}.
$$

The columns (c,0,−b) and (0,a,−b) are relations. To prove they generate every relation, take abp+bcq+car = 0. Since c divides the last two summands and is coprime to ab, p = cu. Similarly q = av. Substitution and cancellation in the domain give r = −b(u+v).

$$
(p,q,r)=u(c,0,-b)+v(0,a,-b).
$$

If this combination is zero, its first two coordinates give cu = 0 and av = 0, hence u = v = 0. Thus A₂ is injective, and its image is exactly the kernel of A₁.

Checking A₁A₂ = 0 proves only that the columns are relations. The argument with an arbitrary (p,q,r) is what establishes exactness.

## Keep the degree shifts visible

$$
0\longrightarrow S(-3)^2\xrightarrow{A_2}S(-2)^3\xrightarrow{A_1}S\longrightarrow R\longrightarrow0.
$$

Give each variable degree one. The generators of I have degree two, and each relation has total degree three. All matrix entries between the free modules lie in the homogeneous maximal ideal, so this is a minimal graded resolution.

$$
\beta_{0,0}(R)=1,\quad\beta_{1,2}(R)=3,\quad\beta_{2,3}(R)=2.
$$

Removing the last S gives the resolution of I itself, with Betti ranks (3,2). The quotient R has Betti ranks (1,3,2). The two rank lists refer to different modules.

## The Hilbert series and the complex agree

$$
\begin{aligned}H_R(t)&=\frac{1-3t^2+2t^3}{(1-t)^3}\\&=\frac{1+2t}{1-t}\\&=1+3t+3t^2+3t^3+\cdots.\end{aligned}
$$

One can count this directly: in positive degree, every surviving monomial is a pure power of exactly one of a,b,c. There are three in every positive degree. The alternating sum from the resolution gives the same series.

The Stanley–Reisner complex consists of three isolated vertices. In multidegree abc its reduced zeroth homology has dimension two. Hochster’s formula therefore recovers β₂,₃(R) = 2. In a two-vertex degree, two isolated vertices give one relation-generator contribution, producing the three β₁,₂ entries.

This links an exact kernel proof, a generating function, and topology through the same ideal. It is useful preparation for the characteristic-dependent project: one must know exactly which module, degree, and homology group a Betti number measures.

## Sources

- [Miller & Perkinson — Eight Lectures on Monomial Ideals, Lecture I](https://sites.math.duke.edu/~ezra/Queens/cocoa.pdf)