Exercise A0.III.5.14

Proposition

Let $N$, $P$ be submodules of an $R$-module $M$. Then

• $N + P$ is a submodule of $M;$
• $N \cap P$ is a submodule of $P$ and $$\frac{N+P}{N} \cong \frac{P}{N\cap P}$$

Proof

First, observe that $N + P$ is a subgroup of $M$. Any element of $N+P$ is already an element of $M$, and since $R$-action on $N + P$ is inherited from $M$, it is sufficient to make $N + P$ into a module, provided that closure holds.

Let $n \in N$ and $p \in P$. For an arbitrary $r \in R$, consider $r(n + p) = rn + rp$. Since $N$ is a submodule, $rn \in N$. Similarly, $rp \in P$. This demonstrates that $rn + rp \in N + P$, as required.

Similarly, for $N \cap P$ we only need to check closure. Let $u \in N \cap P$ and take an arbitrary $r \in R$. Since $u \in N$, $ru \in N$ as well. Similarly, $ru \in P$. So $N \cap P$ is a submodule.

Let $K = N \cap P$ and consider the map $\varphi : (N + P) \to P/K$ defined as $n + p \mapsto p + K.$ It is clearly a surjective homomorphism. So ${\rm im}(\varphi) = P/K$ and $\ker(\varphi) = N$.

Let’s check that $\varphi$ is well-defined. Let $n_1 + p_1 = n_2 + p_2$. Note that $n_1 - n_2 = p_1 - p_2$ and since $n_1 - n_2 \in N$ and $p_1 - p_2 \in P$, we know that this difference lives in $K$.

Finally, the canonical decomposition

gives the required isomorphism.

$\blacksquare$