Every equivalence relation $x \equiv y$ compatible with the structure of a module $E$ is of the form $y-x \in M$ for some submodule $M$ of $E$.

Question

On the end of page 196 of Bourbaki’s Algebra I, it says:

Let $E$ be an $A$-module. Every equivalence relation $x \equiv y$ compatible with the structure of a module $E$ is of the form $y - x \in M$ for some submodule $M$ of $E$.

Say we start with an equivalence relation $R$ on a module $E$. For $R$ to be compatible with the module structure of $E$ means that if $x \equiv y$ and $x' \equiv y'$ then $x + x' \equiv y + y'$. How can we deduce from here that there must exist a submodule $M$ of $E$ such that $y - x \in M$ for all $x \equiv y$? Thanks.

Answer 1

Let $M := \{x \in E : x \equiv 0\}$. Clearly $0 \in M$. Moreover, if $x_1,x_2 \in M$ and $a \in A$, then $ax_1+x_2 \in M$ since $\equiv$ is compatible: $$ ax_1+x_2 \equiv a0+0 = 0. $$ Thus, $M$ is a submodule of $E$.

And, again, since $\equiv$ is compatible, for any $x_1,x_2 \in E$ we have that $$ x_1 \equiv x_2 \iff x_1-x_2 \equiv x_2-x_2=0 \iff x_1-x_2 \in M. $$