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The problem is to show that if $e$ is an idempotent in a ring $R$, then $Re=S^{-1}R$ where $S=\{1,e,e^2,e^3,\dots\}=\{1,e\}$. In fact this doesn't even seem plausible to me, because $Re$ is "smaller" than $R$ (it's a subring of $R$) whereas $S^{-1}R$ is "larger" than $R$ (it's an extension of $R$).

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  • $\begingroup$ What is $S^{-1}R$? What is $S^{-1}$? $\endgroup$ – YiFan Apr 25 at 22:22
  • $\begingroup$ $S^{-1}R$ is the localization of $R$ w.r.t. $S$. $\endgroup$ – user437309 Apr 25 at 22:24
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    $\begingroup$ The idea that $S^{-1}R$ is "larger" than $R$ is not in general correct, because $S^{-1}R$ may collapse some existing elements of $R$ together. $\endgroup$ – Ted Apr 26 at 5:59
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I assume you're considering commutative rings for localizations to make sense.

The idea is that $S^{-1}R$ is a "generalization of a field of fractions". If $e^2=e$ in a field, then $e=0$ or $e=1$, and in the latter case $e=e^{-1}$, so it is not unexpected to have something like $Re=Re^{-1}\simeq RS^{-1}$.


Formally, you can prove that $Re$ is (naturally isomorphic to) $S^{-1}R$ using the universal property of localizations: There is a ring homomorphism $\phi\colon R\to Re$ taking $S$ to units, and which is universal (initial) with this property.

To prove this, just take $\phi\colon R\to Re$ as $\phi(r)=re$. Then $\phi$ is a ring homomorphism (because $e$ is idempotent), taking each element of $S$ to the unit $e$ of $Re$.

If $T$ is any other commutative ring and $\psi\colon R\to T$ is a ring homomorphism taking elements of $S$ to units of $T$, define $\psi'\colon Re\to T$ as the restriction of $\psi$ to $Re$. Then $\psi=\psi'\circ\phi$: Indeed, $\psi(e)^2=\psi(e)$, because $\psi$ is a ring homomorphism and $e^2=e$. Since $\psi(e)$ is a unit then $\psi(e)=1$. Given arbitrary $r\in R$, we multiply both sides of the equation "$1=\psi(e)$" by $\psi(r)$ to conclude $$\psi(r)=\psi(re)=\psi'(\phi(r))$$ so $\psi$ factors through $\phi$. This factor is unique since $\phi$ is obviously surjective, and this gives the universal property of $(Re,\phi)$ as the localization of $R$ wrt $S$.

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  • $\begingroup$ $\psi(e)$ is an invertible element of $T$, but why must it be equal to $1$? And how do you use $\psi(e)^2=\psi(e)$? $\endgroup$ – user419669 May 1 at 17:05
  • $\begingroup$ @user419669 If $x$ is invertible and $x^2=x$, multiply both sides by $x^{-1}$ to get $x=1$.. $\endgroup$ – Luiz Cordeiro May 1 at 19:43
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If $e\in\{0,1\}$ then the claim is clear. Otherwise $e$ is a zero divisor, and so the canonical map $$R\ \longrightarrow\ S^{-1}R,$$ is not injective; the localized ring $S^{-1}R$ is not "larger" than $R$. In particular, because $e(1-e)=0$, the entire ideal $(1-e)R$ is mapped to zero by this map.

In fact the equivalence relation on $R\times S$ that defines $S^{-1}R$, which is in general given by $$\frac{r_1}{s_1}\sim\frac{r_2}{s_2} \qquad\iff\qquad \exists t\in S:\ t(r_1s_2-r_2s_1)=0,\tag{1}$$ greatly simplifies here. Because $S=\{1,e\}$ the right hand side of $(1)$ is equivalent to $$e(r_1s_2-r_2s_1)=0,$$ where also $s_1,s_2\in\{1,e\}$. If $s_1=s_2$ then this is in turn equivalent to $$e(r_1-r_2)=0,$$ and if $s_1\neq s_2$ then without loss of generality $s_1=e$ and $s_2=1$, and we find that $$e(r_1s_2-r_2s_1)=e(r_1-r_2e)=e(r_1-r_2+(1-e)r_2)=e(r_1-r_2),$$ because $e(1-e)=0$. So in fact in this particular case the equivalence relation $(1)$ is equivalent to

$$\frac{r_1}{s_1}\sim\frac{r_2}{s_2} \qquad\iff\qquad er_1=er_2.$$ This shows that the map $$S^{-1}R\ \longrightarrow\ eR:\ \frac{r}{s}\ \longmapsto\ er,$$ is a bijection, and it is easily verified to be a ring homomorphism.

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