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The subobject lattice of an abelian category is modular

Statement

For every object $A$ in an abelian category, the lattice of subobjects of $A$ is modular.

Facts & Assumptions

Given: An object $A$ in an abelian category.

[L1]

A modular lattice is one satisfying $x \leq z \Rightarrow x \lor (y \land z) = (x \lor y) \land z$ (Modular lattice).

[L2]

The subobjects of $A$ form a lattice (The subobjects of an object in an abelian category form a lattice).

[L3]

Quotienting by the kernel identifies a morphism with its image (First isomorphism theorem in an abelian category).

[L4]

Quotienting nested subobjects satisfies the third isomorphism theorem (Third isomorphism theorem in an abelian category).

[L5]

In an abelian category, a morphism that is both monic and epic is an isomorphism (An abelian category is balanced).

[L6]

The join of two subobjects is the image of the induced map from their biproduct (The join of two subobjects in an abelian category).

[L7]

The meet of two subobjects is represented by their pullback (The meet of two subobjects is their pullback).

Proof

technique · direct

1.1construct

Consider any lattice in which comparable complements of a fixed element coincide in every interval. Let $X \leq Z$ and put $U : = X \lor (Y \land Z), V : = (X \lor Y) \land Z .$ Then $U \leq V$ by monotonicity.

1.2L4construct

Fix an interval $[B_{1}, B_{2}]$ of subobjects of $A$ , and write $Q : = B_{2} / B_{1}$ with quotient map $q : B_{2} \to Q$ . If $D$ is any intermediate subobject, let $q_{D} : B_{2} \to B_{2} / D$ be its quotient map. Because $B_{1} \leq D$ , the composite $q_{D}$ kills $B_{1}$ , so it factors through $q$ as a map ${\overline{q}}_{D} : Q \to B_{2} / D$ . Define $Φ (D)$ to be the kernel subobject of ${\overline{q}}_{D}$ .

2.1L1step 1.1algebra

Both $U$ and $V$ are complements of $Y$ in the interval $[Y \land Z, X \lor Y]$ : one has $U \lor Y = X \lor Y = V \lor Y$ , and $Y \land Z \leq U \leq Z$ gives $U \land Y = Y \land Z$ , while $V \leq Z$ and $Y \leq X \lor Y$ give $V \land Y = Y \land Z$ . By step 1.1, the interval hypothesis forces $U = V$ . Therefore the modular-law identity of [L1] holds.

2.2L3L4step 1.2construct

Conversely, if $S \leq Q$ with quotient map $t_{S} : Q \to Q / S$ , define $Ψ (S)$ to be the kernel subobject of $t_{S} q$ in $B_{2}$ . Then $B_{1} \leq Ψ (S) \leq B_{2}$ . If $D$ lies in the interval, the equality $q_{D} = {\overline{q}}_{D} q$ gives $Ψ (Φ (D)) = D$ . If $S \leq Q$ , the morphism $t_{S} q$ is epic, so its image is all of $Q / S$ ; by [L3], $B_{2} / Ψ (S) ≅ Q / S .$ Under this identification, ${\overline{q}}_{Ψ (S)}$ is a cokernel of $S$ , so $Φ (Ψ (S)) = S$ . Thus $Φ$ and $Ψ$ are inverse bijections between $[B_{1}, B_{2}]$ and the subobject lattice of $Q$ .

3.1step 1.2step 2.2

If $D \leq E$ in the interval, then $q_{E}$ factors through $q_{D}$ , so ${\overline{q}}_{E}$ kills $Φ (D)$ . Hence $Φ (D) \leq Φ (E)$ . The same argument applied to $Ψ$ shows that $Φ$ is an order isomorphism from $[B_{1}, B_{2}]$ onto $Sub (Q)$ .

4.1construct

By step 3.1, it is enough to prove the comparable-complements property in $Sub (Q)$ . Let $E \leq Q$ , and let $U \leq V$ be two complements of $E$ . Write $t : Q \to Q / E$ for the quotient map.

5.1L5L6L7step 4.1construct

Because $U \land E = 0$ , the pullback description [L7] makes the kernel of the restricted map $t ∣_{U} : U \to Q / E$ trivial. Because $U \lor E = Q$ , the canonical map $[i_{U}, i_{E}] : U \oplus E \to Q$ has image $Q$ by [L6], so composing with $t$ shows that $t ∣_{U}$ is epic. By [L5], $t ∣_{U}$ is an isomorphism. The same argument shows that $t ∣_{V}$ is an isomorphism. Let $i : U \to V$ represent the comparison $U \leq V$ . Then $t ∣_{U} = (t ∣_{V}) i$ , so $i$ is an isomorphism. Therefore $U$ and $V$ represent the same subobject of $Q$ .

6.1L1L2step 2.1step 3.1step 5.1∎

Step 5.1 proves that every interval in the subobject lattice has the comparable-complements property, and step 2.1 shows that this property implies the modular law of [L1]. With [L2], this proves that the subobject lattice of $A$ is modular.

Depends on

Used by

A subobject lattice of an abelian category need not be distributive Counterexample
The subobject lattice of a cyclic group of order twelve Example
The published subgroup modular law is the instance Remark
Zassenhaus butterfly lemma in an abelian category Theorem

Dependency tree · two levels

23 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, Proposition 73 and Corollary 72 (standard reference, not scraped)