Lemma. A partition splits the field module [fdip-001I]
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  • October 20, 2026

Let \(R\) be a commutative unital ring and choose a partition \(V=V_0\sqcup V_1\) of a finite set. Put \(H=\mathcal F_R(V)\) and \(K_i=\mathcal F_R(V_i)\) for \(i=0,1\). Restriction and extension by zero are ordinary \(R\)-linear maps \[\pi _i:H\to K_i,\qquad \iota _i:K_i\to H,\] with \((\pi _i f)(v)=f(v)\) for \(v\in V_i\), and \((\iota _i u)(v)=u(v)\) on \(V_i\), zero elsewhere. The support submodules are \(H_i=\iota _i(K_i)\subseteq H\). The support projections \(p_i=\iota _i\pi _i:H\to H\) satisfy \[\pi _i\iota _i=I_{K_i},\quad \pi _i\iota _j=0\ (i\ne j),\quad p_i^2=p_i,\quad p_ip_j=0\ (i\ne j),\quad p_0+p_1=I_H.\] Every \(f\in H\) has a unique expression \(f=f_0+f_1\) with \(f_i\in H_i\), namely \(f_i=p_if\).

Proof. 

  • October 20, 2026

Evaluation at a point proves linearity and all the displayed composition identities. A field belongs to \(H_i\) precisely when it vanishes outside \(V_i\). Thus \(H_0\cap H_1=\{0\}\). The identity \(p_0+p_1=I_H\) gives existence of the decomposition. If \(f_0+f_1=g_0+g_1\), then \(f_0-g_0=g_1-f_1\in H_0\cap H_1\), proving uniqueness.

The partition is additional data, not part of the general incidence-field definition. If one part is empty, its field module and support projection are zero. Nothing requires division by \(2\); the same construction works in characteristic \(2\) and over the zero ring. This is the direct finite-function realization of a module decomposition.

[stacks2017stacks, Section 10.10], Tag 0581 supplies the module convention.