## Rings with nontrivial idempotents decompose as a direct product

Let $R$ be a ring with $1 \neq 0$. An element $e \in R$ is called idempotent if $e^2 = e$. Assume $e$ is an idempotent in $R$ and that $er = re$ for all $r \in R$. (That is, $e \in Z(R)$.) Prove that $Re$ and $R(1-e)$ are two-sided ideals in $R$ and that $R \cong Re \times R(1-e)$. Show that $e$ and $1-e$ are identities in $Re$ and $R(1-e)$ respectively.

Note first that $(1-e)^2 = 1 -2e + e = 1-e$.

First we’ll show that $Re$ is a two-sided ideal. First $0 \cdot e = 0 \in Re$, so $Re$ is nonempty. Let $re, se \in Re$. Then $re-se = (r-s)e \in Re$. Now if $t \in R$, then $tre \in Re$ and $ret = rte \in Re$. Thus $Re$ is a two-sided ideal of $R$.

Next we’ll show that $R(1-e)$ is a two-sided ideal. First $0 \cdot (1-e) = 0 \in R(1-e)$, so $R(1-e)$ is nonempty. Let $r(1-e), s(1-e) \in R(1-e)$; then $r(1-e) - s(1-e) = (r-s)(1-e) \in R(1-e)$. Now if $t \in R$, then $tr(1-e) \in R$ and $r(1-e)t = r(t-et)$ $= r(t-te)$ $= rt(1-e) \in R(1-e)$. Thus $R(1-e)$ is a two-sided ideal in $R$.

Now suppose $x \in Re \cap R(1-e)$. Then we have $re = s(1-e)$ for some $r,s \in R$. Now $re = s - se$, and right multiplying by $e$, we see that $re^2 = se - se^2$, so that $re = se - se = 0$. Thus $x = 0$, and so $Re \cap R(1-e) = 0$ is trivial.

Let $r \in R$; since $re + (1-r)e = r$, $R \subseteq Re + R(1-e)$. By the recognition theorem for direct products of groups, the map $\varphi : Re \times R(1-e) \rightarrow R$ given by $\varphi(a,b) = a+b$ is a group isomorphism. Now let $(r_1e, s_1(1-e)), (r_2e,s_2(1-e)) \in Re \times R(1-e)$. Then $\varphi((r_1e,s_1(1-e))(r_2e,s_2(1-e))) = \varphi(r_1r_2e,s_1s_2(1-e))$ $= r_1r_2e + s_1s_2(1-e)$ $= r_1r_2e^2 + s_1s_2(1-e)^2$ $= r_1er_2e + r_1s_2(e-e^2) + r_2s_1(e-e^2) + s_1(1-e)s_2(1-e)$ $= (r_1e + s_1(1-e))(r_2e + s_2(1-e))$ $\varphi(r_1e,s_1(1-e)) \cdot \varphi(r_2e,s_2(1-e))$. Thus $R \cong Re \times R(1-e)$ as rings.

Note that for all $re \in Re$, $ere = re^2 = re$ and $re^2 = re$. Thus $e \in Re$ is a multiplicative identity. Similarly, $r(1-e)^2 = r(1-e)$ and $(1-e)r(1-e) = (r - er)(1-e)$ $= r(1-e)^2$ $= r(1-e)$, so that $1-e$ is a multiplicative identity in $R(1-e)$.