Is the ring homomorphism from a ring A to its localization ring $S^{-1}A$ surjective?

Question

Given the ring $A$ and the multiplicative subset $S\subset A$, we define the ring homomorphism: $$\phi_S: A \rightarrow S^{-1} A, \,\,\ \phi_S(a)=\frac {a}{1}.$$ It is easy proven that this map is injective. I was wondering if this map is also surjective in general, something that I do not find in books.

We assume that $0 \notin S.$ Let $c\in S^{-1} A.$ Then it must be expressed as $c=\frac {b}{1},\,\, b\in A$. Thus $\exists b \in A$ such that $\phi_S(b)=\frac {b}{1}=c.$

Does this prove the surjectivity ? It seems too simple to be true.

Thanks.

Answer 1

First of all, the map is not in general injective.

Let $A=\mathbb Z/6\mathbb Z,$ and $S=\{2,4\}.$ Then, $3/1=0/1.$

Furthermore, it is not in general surjective either. When $A=\mathbb Z,$ $S=A\setminus \{0\},$ there are no $a$ such that $a/1=1/2.$

P.S.: Note that, when $a/s,a'/s'\in S^{-1}A,$ $a/s=a'/s'$ if and only if for some $t\in S,$ $t(as'-sa')=0.$

In particular, when $s=s'=1,$ this means $a/1=a'/1$ if and only if for some $t\in S,$ $t(a-a')=0.$ So, if $S$ contains a zero-divisor, one can find $a\neq a'$ so that $a/1=a'/1.$

Answer 2

It is easy proven that this map is injective.

I wonder how exactly you've proven that? Because in general it is not. $\phi_S$ is injective if and only if $S$ does not contain zero divisors.

Then it must be expressed as $c=\frac {b}{1},\,\, b\in A$.

Not every element is of the form $\frac{b}{1}$. Consider $A=\mathbb{Z}$ and $S=\mathbb{Z}\backslash\{0\}$. You can easily verify that $S^{-1}A\simeq\mathbb{Q}$ (in fact you can define $\mathbb{Q}$ like that) and $\phi_S:\mathbb{Z}\to\mathbb{Q}$ is given by $\phi_s(z)=\frac{z}{1}$. Is that surjective? Is $\frac{1}{2}=\frac{z}{1}$ for some integer $z\in\mathbb{Z}$?