Exercise A0.III.5.15

Proposition

Let $R \in \mathbf{CRing}$ and let $I, J$ be ideals of $R$. Then

$$I \cdot (R/J) \cong (I + J)/J$$

as $R$-modules.

Proof

First, let’s recall what $I \cdot (R/J)$ stands for. It doesn’t seem that Aluffi covers this in the chapter (or maybe I’m not paying attention). Regardless, I found an explanation here.

To take a product of an ideal $I \subseteq R$ and an $R$-module $M$ we take the set of all finite sums of the form $i_km_k$ where $i_k \in I$ and $m_k \in M$. In symbols,

$$I \cdot M = \left\{\sum_{k=0}^{n \in \mathbb{N}}i_km_k \;\middle|\; i_k \in I, m_k \in M\right\}.$$

It is clear that this is a submodule of $M$.

We have $I \cdot (R/J)$ which consists of finite sums of terms like $i_k(r_k + J) = i_kr_k + J$. But is this well-defined? Well, if $r_a + J = r_b + J$ then $r_a - r_b \in J$ and so

$$ir_a + J = ir_a + i(r_b - r_a) + J = ir_a + ir_b - ir_a + J = ir_b + J.$$

Since $I$ is an ideal, $i_kr_k \in I$, so any element of $I \cdot (R/J)$ can be understood as $i + J$ with $i \in I$. Now let $M=I \cdot (R/J)$, $K=I \cap J$ and consider the map $\varphi : M \to I/K$ given by $i + J \mapsto i + K$.

Is it well-defined? If $i_1 + J = i_2 + J$ then $i_1 - i_2 \in J$ and hence $i_1 - i_2 \in K$. It is manifestly a surjective module homomorphism.

To see that it is injective, note that if $i_1 + K = i_2 + K$ then $i_1 - i_2 \in K$ and hence $i_1 - i_2 \in J$, which shows that $i_1 + J = i_2 + J$.

Thus we know that $I \cdot (R/J)$ is isomorphic to $I/(I \cap J)$. We know from 5.14 that $I/(I \cap J)$ is isomorphic to $(I + J)/J$. This completes the proof.

$\blacksquare$