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I want to show that if $S$ consists of units then $S^{-1}R \cong R$.

Can you tell me if my proof is correct?

Since $S$ consists of units, $S$ is zero-divisor free and hence $f: R \to S^{-1}R$, $r \mapsto \frac{r}{1}$ is injective. So we have an isomorphism $h: R \to f(R)$.

Now we construct an isomorphism $S^{-1}R \to f(R)$: For this we pick any $s_0$ in $S$ and denote by $u \in R$ its inverse, i.e. $s_0 u = us_0 = 1$. Then the map $g_u : S^{-1}R \to f(R)$, $\frac{r}{s} \mapsto \frac{ru}{1}$ is an isomorphism. It is injective since if $\frac{ru}{1} = \frac{r^\prime u}{1}$ then $ru = r^\prime u$ and since $u$ is a unit, $r = r^\prime$. And it is surjective since for $\frac{r}{1}$ in $f(R)$, $g(\frac{rs}{s}) = \frac{rsu}{1} = \frac{r}{1}$. Hence $R \cong f(A) \cong S^{-1}R$.

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    Every element in $S^{-1}R$ has the form $\frac{r}{s}$, which can be viewed as the image of $r/s$ (since $s$ is a unit) by $f$. So $f$ is already surjective and hence an isomorphism.2012-06-13
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    Sure, but maybe using the universal property would be cleaner. There is a ring map $S^{-1}R \to R$, such that $R\to S^{-1}R \to R$ is the identity, since $S$ consists of units. On the other hand, the composition $S^{-1}R \to R \to S^{-1}R$ is the unique map such that $R\to S^{-1}R \to R \to S^{-1}R$ is the localization map, i.e., the identity.2012-06-13
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    When you check injectivity, you must start with $\frac{ru}{1}=\frac{r'u'}{1}$.2012-06-13
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    @JoeJohnson126 But my map $g_u$ is defined to map to $\frac{ru}{1}$.2012-06-13
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    Thank you Justin and Zhe Chen, you are both awesome! I didn't think of either things you pointed out. Now that you have pointed it out, both seems obvious. I'll post the two proofs as an answer.2012-06-13
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    Embrace the universal property! If you have a homomorphism $\alpha\colon R \to T$ under which the image of $S$ consists of units (this is always true, I guess) then, well, there is a unique homomorphism $\beta\colon R \to T$ such that $\beta \circ \mathrm{id}_R = \alpha$. Just by writing this down you see that $\beta = \alpha$ is forced upon you.2012-06-13
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    @MattN. Sorry, missed that bit. I thought you were mapping each $\frac{r}{s}$ to $\frac{ru}{1}$ where $su=1$, not a fixed $u$. Disregard my earlier comment.2012-06-13

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The natural injection is the isomorphism, but I think you don't have to work so hard.

Just note that for any $(r,s)$ in $S^{-1}R$, we have $(r,s)=(rs^{-1},1)$.

So, $S^{-1}R=Im(R)\cong R$.