If $ S \subset A^{\times}$, then $S^{-1}A \simeq\ ?$ (A question related to localization of rings)

Question

This question was asked in my quiz of commutative algebra and I was unable to solve this particular question.

So, I am posting it here.

If $S\subset A^{\times} $ (set of all units), then $S^{-1} A \simeq\ $?

This question is from localization of rings. I have gone through results and theorems but still I am not sure how should I approach this question.

I wasn't able to attempt or start this question as I am not sure how exactly I should approach this question.

Can you please outline a solution briefly?

Answer 1

Localizing at a certain set $S$ means that you force the elements in $S$ to become invertible. However, if $S\subseteq A^\times$, then we are not doing anything (as they were already invertible). This can be seen quite easily in a rigorous way. Consider the map $$ \varphi : A \rightarrow S^{-1} A, a \mapsto \frac{a}{1}.$$ This is clearly a morphism of rings, so we are left to check that it is bijective.

Let $a\in \ker (\varphi)$. Then we have that $\frac{a}{1}=0$, which implies that $a=0$ as $S$ does not contain any zerodivisors.

For surjectivity, we note that $\frac{a}{s} = \frac{s^{-1}a}{1}$, which works as $s\in S \subseteq A^\times$ (hence $s^{-1}$ is well-defined).

Answer 2

$S^{-1} A=A$. Since each $s\in S$ is a unit, we have $s^{-1}\in A$.