Why does $Re=S^{-1}R$?

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 $$R$$).

• What is $S^{-1}R$? What is $S^{-1}$? – YiFan Apr 25 at 22:22
• $S^{-1}R$ is the localization of $R$ w.r.t. $S$. – user437309 Apr 25 at 22:24
• 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. – Ted Apr 26 at 5:59

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 $$S$$.

• $\psi(e)$ is an invertible element of $T$, but why must it be equal to $1$? And how do you use $\psi(e)^2=\psi(e)$? – user419669 May 1 at 17:05
• @user419669 If $x$ is invertible and $x^2=x$, multiply both sides by $x^{-1}$ to get $x=1$.. – Luiz Cordeiro May 1 at 19:43

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.