Exercise A0.III.6.03

Proposition

Let $R$ be a ring and let $M$ be an $R$-module. Suppose $p : M \to M$ is such that $p \circ p = p.$ Then $M \cong \im p \oplus \ker p$.

Proof

Let $f : M \to \im p \oplus \ker p,$ given as $f = (p, \id_M - p)$ where we restrict each component to its image. To see that $\im(\id_M - p) = \ker p$ note that for any $a \in M$,

$$(\id_M - p)(a) = a - p(a)$$

and

$$p(a - p(a)) = p(a) - p(p(a)) = p(a) - p(a) = 0.$$

The other inclusion is simple, since any $v \in \ker(p)$ can be written as $v - p(v)$, and this lives in $\im(\id_M - p).$

Consider also the map $g : \im p \oplus \ker p \to M,$ given as $g =\id_{\im p} + \id_{\ker p}.$ Note that it’s an $R$-module homomorphism.

To check that they are mutually inverse, we take arbitrary $m \in M$, $(a,b) \in \im p \oplus \ker p$ and check

$$g(f(m)) = g(p(m), m - p(m)) = p(m) + m - p(m) = m,$$

and

$$\begin{align*} f(g(a,b)) &= f(a + b) \\ &= (p(a + b), a + b - p(a + b)) \\ &= (p(a) + p(b), a + b -p(a) - p(b)) \\ &= (a + 0, a + b - a - 0) \\ &= (a, b). \\ \end{align*}$$

Note that the second derivation uses the fact that $a \in \im p$ (hence $p(a) = a$) and $b \in \ker p$ (hence $p(b) = 0$).

Since an $R$-module homomorphism with an inverse function is an isomorphism, this completes the proof.

$\blacksquare$