Proof of $M_1 + (M_2 \cap M_3) = M_2 \cap (M_1 + M_3)$ if $M_1 \subset M_2$ in abelian category

Question

Let $\mathcal{A}$ be an abelian category, $M_1,M_2$ and $M_3$ three objects with $M_1 \subset M_2$. The sum and intersection of two sub-object is defined as usual in abelian category. I want to prove : If $M_1 \subset M_2$ then,$$M_1 + (M_2 \cap M_3) = M_2 \cap (M_1 + M_3).$$ I think this is true as we have this result on modules. It's easy to prove that: $$M_1 + (M_2 \cap M_3) \subset M_2 \cap (M_1 + M_3)$$ because:

• $M_1 \subset M_2$ and $M_1 \subset M_1 + M_3$ so $M_1 \subset M_2 \cap (M_1 + M_3)$
• $M_2 \cap M_3 \subset M_2$ and $M_2 \cap M_3 \subset M_3 \subset M_1 + M_3$ so $M_2 \cap M_3 \subset M_2 \cap (M_1 + M_3)$

But I don't have any idea for the converse. I just obtain ugly big diagramm... It's frustrating when you see the very easy proof in modules:
Let $x_2 = x_1+ x_3 \in M_2 \cap (M_1 +M_3)\ $ (with $x_1 \in M_1,\ x_2 \in M_2,\ x_3 \in M_3$). Then $x_3 = x_2 - x_1$ and since $x_1 \in M_1 \subset M_2$, $x_3 \in M_2$. We obtain $x_2 = x_1 + x_3 \in M_1 + (M_2 \cap M_3)$

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Do you have any ideas? I think this is not an easy question, any help was welcome. - M.T.

Answer 1

The answer is "yes".

If $L$ is a lattice, then it's called Modular precisely when we get this kind of associativity between $\land$ and $\lor$:

If $a \leq b$ in $L$, then for each $x$, $a \lor (x \land b) = (a \lor x) \land b$

These lattices are called "modular" because the lattice of subobjects of a module (with $+$ for $\lor$ and $\cap$ for $\land$) is the primary example. In this language, modularity says that whenever $M_1 \subseteq M_2$ and $N$ are submodules of $M$, $M_1 + (N \cap M_2) = (M_1 + N) \cap M_2$, which is your condition. As a brief aside, you shouldn't be discouraged! The direction you've proven ($a \lor (x \land b) \leq (a \lor x) \land b$) is true in every lattice, and the other direction (which you're struggling with) is always the difficult direction to check.

The benefit of this language is that it makes it easier to google for your problem. Indeed, if we try "abelian category subobjects modular lattice" we find a paper Abelian Categories by Daniel Murfet. Here proposition 73 says that for any abelian category $\mathcal{A}$ and for any object $A \in \mathcal{A}$, the lattice of subobjects $\text{Sub } A$ is modular!

The proof is a bit long to show here in full, but the sketch is this.

One can show that a lattice is modular if and only if every interval $I = [x,y] = \{M \mid x \leq M \leq y\}$, for every $N \in I$, if $N$ has two complements $M_1 \leq M_2$, then actually $M_1 = M_2$. This is pure lattice theory. Next, specific to abelian categories, we prove that if $M \subseteq N$ is a subobject, then $\text{Sub } M \cong [0,M]$ and $\text{Sub }N/M \cong [M,1]$. These are sometimes called the "lattice isomorphism theorems", and they're true for modules. So it's reassuring that they're also true for objects in some abelian category. Lastly, we use this lattice isomorphism theorem to reduce the lattice theoretic equivalence to a special case, which we can then just... check.

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There is actually another, proof-by-sledgehammer, approach to this theorem which perhaps you'll like more.

By the Embedding Theorem, we know that $\mathcal{A}$ is a full, faithful, exact subcategory of some $R\text{-mod}$. If $F : \mathcal{A} \to R\text{-mod}$ is the embedding, this tells us that $\text{Sub } X$ as computed in $\mathcal{A}$ is a sublattice of $\text{Sub } FX$ as computed in $R\text{-mod}$. But we know already that subobject lattices of modules are modular (again, that's where the name comes from!), and it's not hard to show that sublattices of modular lattices are modular!

You can check this directly, or for fun you can use another high power theorem that every nonmodular lattice contains "$N_5$" as a sublattice. So if $\text{Sub } X$ were nonmodular, it would contain an $N_5$, but then since $\text{Sub } X$ is a sublattice of $\text{Sub } FX$, we see that this would also have to contain an $N_5$, contradicting modularity of $\text{Sub } FX$.

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I hope this helps ^_^

Answer 2

You can adjust the given argument to use generalized elements. So, suppose $M$ is the containing object of $M_1,M_2,M_3$, and suppose you have a "test object" $U$ and a morphism $m : U \to M$ such that $m$ factors through $M_2 \cap (M_1 + M_3)$. That implies that $m$ factors both through $M_2$ (call the factor $m_2$) and through $M_1 + M_3$ (call the factor $s$). Now, since we have a canonical epimorphism $M_1 \oplus M_3 \to M_1 + M_3$, there exists some "refined test object" $V$ and an epimorphism $\pi : V \to U$, along with $m_1 \in \operatorname{Hom}(V, M_1)$ and $m_3 \in \operatorname{Hom}(V, M_3)$ such that $m_1 + m_3 = s \circ \pi$. (For example, you can take $V$ to be the pullback of $s$ and this epimorphism and $\pi$ to be the projection from the pullback to $U$.) Now by your argument, $m_1 + m_3 = m_2 \circ \pi$ implies both sides factor through $M_1 + (M_2 \cap M_3)$. However, since $\pi$ is an epimorphism, that implies that $m_2 = s$ also both factor through $M_1 + (M_2 \cap M_3)$. This in turn implies that $m$ factors through $M_1 + (M_2 \cap M_3)$.

Since this is true for any $U$ and $m$, we can now conclude that $M_2 \cap (M_1 + M_3) \le M_1 + (M_2 \cap M_3)$ in the subobject lattice of $M$. (For example, take $U = M_2 \cap (M_1 + M_3)$ and $m$ the inclusion morphism.)