Let $R$ be a commutative ring, let $S\subset R$ and let $T$ be the smallest multiplicatively closed subset of $R$ containing $S$. Then $R_S\cong R_T$

96 Views Asked by At

If we let $\pi_T: R\rightarrow R_T$ be the canonical projection, then since by localizing at T, we have adjoined the inverses of all elements of T which itself contains S, $\pi_T(s)\in U(R_T)$ for all $s\in S$. Hence by the universal property we have a unique homomorphism $\theta :R_S\rightarrow R_T$ such that $\pi_T=\theta\circ\pi_S $ where $\pi_S:R\rightarrow R_S$ is the canonical projection.

Now I want to apply the universal property starting from $\pi_S$ but I am no sure how $R_S$ inverts $T$. If we can show that for each $x\in T$ there exists $y\in T$ such that $xy\in S$, then the inverse of $\pi_S(x)$ is $y/xy$ which is defined to be the image of $yt_{xy}$ in $R_S$. To show that such $y$ exists I assume that I need to somwhow use the fact that $T$ is as small as possible but I am not sure how.

All hints are appreciated

1

There are 1 best solutions below

1
On BEST ANSWER

$\pi_S : R \to R_S$ inverts $T$, since $\{r \in R : \pi_S(r) \text{ is invertible}\}$ is a multiplicative subset of $R$ which contains $S$, and hence also $T$.