Does $(M_1 / K) \cap (M_2 / K) = (M_1 \cap M_2) / K$?

Question

Let $M$ be an $R$-module, $M_1$, $M_2$, $K$ be three submodules such that both $K \subseteq M_1$ and $K \subseteq M_2$. The question is: $$ \frac{M_1}{K} \cap \frac{M_2}{K} = \frac{M_1 \cap M_2}{K} \,? $$ (Where $M_i / K$ is the quotient module.) I have tried to prove this and have managed to prove the $\supseteq$ inclusion, but I couldn’t prove the other direction. So I wonder whether maybe it is not true, and there is a counterexample.

Answer 1

The tricky part of this is whether $M_1/K\cap M_2/K\subseteq (M_1\cap M_2)/K$ (the reverse inclusion should be entirely straightforward). Suppose you have an element $x\in M/K$ which is in both $M_1/K$ and $M_2/K$, meaning we have $x=a+K=b+K$ for some $a\in M_1$, $b\in M_2$. You want to write $x=c+K$, for some $c\in M_1\cap M_2$. Since $a+K=b+K$, $a-b\in K$. But this means $a=b+(a-b)\in M_2$, since $b\in M_2$ and $a-b\in K\subseteq M_2$. Thus $a\in M_1\cap M_2$, and you can just take $c=a$.

Answer 2

Short version: The lattice of submodules of $M/K$ is isomorphic to the lattice of submodules of $M$ that contain $K$, via $N \mapsto N/K$. Intersections are mimima in these lattices, and are therefore preserved by the isomorphism.

---

Long version: For every module $M$ its submodules form a partially ordered set under inclusion, which we shall denote by $\mathcal{L}(M)$. Every homomorphism of modules $f \colon M \to M'$ induces order-preserving maps $$ f_* \colon \mathcal{L}(M) \longrightarrow \mathcal{L}(M') \,, \quad f^* \colon \mathcal{L}(M') \longrightarrow \mathcal{L}(M) $$ by assigning to every submodule of $M$ its image under $f$, and to every submodule of $M'$ its preimage under $f$. If $f$ is injective, then we have $f^*(f_*(N)) = N$ for every submodule $N$ of $M$. If $f$ is surjective, then we have $f_*(f^*(N')) = N'$ for every submodule $N'$ of $M'$.

Let now $K$ be a submodule of $M$ and let $π$ be the canonical quotient map from $M$ to $M/K$, which is surjective. For every submodule $N$ of $K$ and every submodule $N'$ of $M / K$ we have $$ π^*( π_*(N) ) = N + K \,, \quad π_*( π^*(N') ) = N' \,. $$ It follows that the two order-preserving maps $π_*$ and $π^*$ induce mutually inverse isomorphisms of partially ordered sets between $\mathcal{L}(M)_{⊇ K} = \{ N ∈ \mathcal{L}(M) \mid N ⊇ K \}$ and $\mathcal{L}(M / K)$, given in one direction by $N \mapsto N/K$. (In other words: submodules of $M/K$ are the same as submodules in $M$ containing $K$, and this identification is compatible with inclusion of submodules.)

For two submodules $N_1$ and $N_2$ of $M$, their intersection $N_1 ∩ N_2$ is precisely their minimum in $\mathcal{L}(M)$. If both $N_1$ and $N_2$ contain $K$, then $N_1 ∩ N_2$ is again contained in $\mathcal{L}(M)_{⊇ K}$, and consequently also the minimum of $N_1$ and $N_2$ in $\mathcal{L}(M)_{⊇ K}$. Under the chosen isomorphism $\mathcal{L}(M)_{⊋ K} ≅ \mathcal{L}(M/K)$ we see that $(N_1 ∩ N_2) / K$ is the miminum of $N_1 / K$ and $N_2 / K$ in $\mathcal{L}(M / K)$. But this minimum is given by the intersection $(N_1 / K) ∩ (N_2 / K)$. Therefore, $(N_1 ∩ N_2) / K = (N_1 / K) ∩ (N_2 / K)$.