ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-28

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The subobject lattice of a two-dimensional vector space over F_2 is the diamond M_3

Example

Let $V = F_{2}^{2}$ . Its subspaces are $0$ , the three lines through the origin, and $V$ itself. So the subobject lattice of $V$ in $F_{2} - M o d$ is exactly the diamond $M_{3}$ .

Facts & Assumptions

Given: The vector space $V = F_{2}^{2}$ .

[L1]

Subobjects form a lattice in every abelian category (The subobjects of an object in an abelian category form a lattice).

[L2]

The A-page counterexample identifies the same diamond pattern as non-distributive (A subobject lattice of an abelian category need not be distributive).

Verification

technique · direct

1.1L1algebra

Over $F_{2}$ , the nonzero vectors of $V$ are $(1, 0)$ , $(0, 1)$ , and $(1, 1)$ , and each spans a distinct one-dimensional subspace. Any two distinct lines meet only in $0$ , and because they are not equal each pair spans all of $V$ . So the subspace lattice has exactly the five elements $0$ , the three lines, and $V$ .

2.1L2step 1.1∎

This is the same five-element diamond described in [L2]. Hence the subobject lattice of a very familiar module can already be modular without being distributive.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Daniel Murfet, Abelian Categories, Section 4.2 (standard reference, not scraped)