$\phi(X)$ is essential and $\psi \circ \phi$ is injective then $\psi$ is injective.

Question

Let $R$ be a commutative ring, and $M$ an $R$-module. An $R$-submodule $N$ of $M$ is said to be essential if for any $R$-submodule $U$, $U \cap N = \{ 0 \}$ then $U = \{ 0 \}$.

My problem is the following:

Let $\phi:X \rightarrow Y, \psi:Y \rightarrow Z$ be homomorphisms of $R$-modules. Suppose that $\phi(X)$ is essential and $\psi \circ \phi$ is injective. Prove that $\psi$ is injective.

My attempt: Assume that we have $\psi(y_1)=\psi(y_2)$ for some $y_1,y_2 \in Y$. Note that $\phi(X)$ is essential, by statement in here there exist $r_1,r_2 \in R\setminus\{0\}$ such that $r_1y_1=\phi(x_1), r_2y_2=\phi(x_2)$ for some $x_1,x_2 \in X$. It follows that, $$\psi \circ \phi(r_2x_1)=r_2\psi \circ \phi(x_1)=r_2\psi(r_1y_1)=r_1r_2\psi(y_1)=r_1r_2\psi(y_2)=r_1\psi \circ \phi(x_2)=\psi \circ \phi(r_1x_2).$$ So the injectivity of $\psi \circ \phi$ implies that $r_2x_1=r_1x_2$. Thus, $$r_1r_2y_1=r_2\phi(x_1)=r_1\phi(x_2)=r_1r_2y_2.$$ We conclude that $y_1=y_2$(?) i.e. $\psi$ is injective.

It seems a fake proof because $r_1r_2$ may not a unit of $R$ in general. Does this approach help me closer the actual proof? Thanks for looking at it.

By hypothesis, $\ker(\psi)\cap \phi(X) = {0}$, so you can conclude that $\ker(\psi) = {0}$. — Prahlad Vaidyanathan, Oct 16 '15 at 12:27
@PrahladVaidyanathan wow! I like your one-line proof! Is there any way to find a proof with my posted approach? — user, Oct 16 '15 at 13:27

score 1 · Accepted Answer · answered Oct 16 '15 at 12:33

1

Hint:

$$\ker(\psi\circ\phi)=\phi^{-1}(\ker\psi)=\phi^{-1}\bigl(\ker\psi\cap\phi(X)\bigr).$$

answered Oct 16 '15 at 12:33

Bernard

175,478

$\phi(X)$ is essential and $\psi \circ \phi$ is injective then $\psi$ is injective.

1 Answers1