Exercise 2.26 (Idempotents and products): Show that $R$ can be written as a direct product of two or more (nonzero) rings iff $R$ contains a non-trivial idempotent. Show that if $e$ is an idempotent, then $R = Re\times R(1-e)$ and that $Re$ may be realized as a localization, $Re = R[e^{-1}]$.
I'm most interested in showing that $Re = R[e^{-1}]$. I assume that $R[e^{-1}]$ means localizing at the multiplicative set $U = \{ 1,e \}$ where the fact that $U$ is multiplicative follows from idempotence: $e^2=e$.
After writing it out, I can't seem to picture what $R[e^{-1}]$ looks like. I show below that $r/1=r/e$ in $R[e^{-1}]$ so does that mean $R[e^{-1}]$ have the same number of elements as $R$? If so, why wouldn't the same proof show that $R\cong Re$? What is special about $R[e^{-1}]$ that makes this work? Thanks for your help!
I will use the universal property of localization. Let $\varphi:R\to Re$ be the ring homomorphism $r\mapsto re$. (Here, we are treating $Re$ as a ring in its own right with $e\in Re$ being the identity element.) We see that $\varphi(1)=\varphi(e)=e$ and hence $\varphi$ maps $U$ to units.
By the universal property, there is a map $\psi:R[e^{-1}]\to Re$ given by the following definition (recall that $s=1,e$): $$ \psi\left(\frac{r}{s}\right) = \varphi(r)\varphi(s)^{-1} = \varphi(r)e = re^2 = re $$ It is clearly surjective. Notice that $r/1=r/e$ because $e(er-r)=er-er=0$. Therefore, to show injectivity it suffices to show that $re=r'e$ (in $Re$) implies $r/1=r'/1$ (in $R[e^{-1}]$). Suppose $re=r'e$ then $(r-r')e=0$. But this exactly means that $r/1=r'/1$. So $\psi$ is a ring isomorphism as desired.
Question: "Exercise 2.26 (Idempotents and products): Show that R can be written as a direct product of two or more (nonzero) rings iff R contains a non-trivial idempotent. Show that if e is an idempotent, then R=Re×R(1−e) and that Re may be realized as a localization, Re=R[e−1]."
Partial-answer: Assume $A\cong A_1 \oplus A_2$. It follows $e:=(1,0)\neq 1$ is a non trivial idempotent. Assume conversely that $1\neq e\in A$ is a non trivial idempotent. Let $I:=(e), J:=(e-1) \subseteq A$ be the ideal generated by $e,e-1$. It follows $I+J=(1)$ hence these ideals are coprime. Moreover $IJ=I\cap J$. Since $IJ:=A(e(e-1))=A(e^2-e)=A(0)=(0)$ is the zero ideal, we get by the CRT an isomorphism
$$ A\cong A/IJ\cong A/I \times A/J$$
where $A/I, A/J \neq 0$. Hence by the CRT we get that $A$ has a non-trivial idempotent iff $A$ may be written as a (non-trivial) direct sum of two rings.