Morphism of rings and localization Let $ \varphi : A \to B $ be a morphism of rings. Why are the two following assertions equivalent:
$ 1) $ There exists a multiplicative subset $ S $ of the ring $ A $, and an ideal $ I $ of $ A $, such that the $ A $-algebra $ B $ is identified to $ S^{-1} A / IS^{-1}A $.
$ 2) $ $ \forall b \in B $ : $ b = \dfrac{ \varphi (a)}{\varphi (s)} $ such that $ a,s \in A $ and $ \varphi ( s ) \in B^{ \times } $.
Thanks a lot.
 A: (1) => (2) is clear (if not, recall the definitions of localizations and quotients).
Now assume that (2) holds. Let $S = \{a \in A : \phi(a) \in B^\times\}$ and $I = \ker(\phi)$. The universal property of localizations gives us a morphism of $A$-algebras $S^{-1} A \to B$. It vanishes on the image of  $I$. Then, the universal property of quotients gives us a morphism of $A$-algebras $S^{-1} A / I S^{-1} A \to B$. By assumotion (2) it is surjective. Now let us consider an element in the kernel. It is the residue class of a fraction $\dfrac{a}{s}$ (with $s \in S$) such that $\dfrac{\phi(a)}{\phi(s)}=0$ in $B$. If we multiply with $\phi(s) \in B^\times$, we get $\phi(a)=0$, i.e. $a \in I$. But then $\dfrac{a}{s} \in I S^{-1} A$, i.e. its residue class vanishes. $\square$
Addendum. If $\phi : A \to B$ is a homomorphism of rings, we can consider the smallest subring $B'$ of $B$ which contains the images of $\phi$ and the inverses of the invertible images of $\phi$. Hence, $A \to B$ factors as $A \to B' \to B$, where $B' \to B$ is an inclusion and $A \to B'$ is a composition of a localization and a regular quotient. Furthermore, $B' \to B$ has the property that $B'^\times = B' \cap B^\times$. Actually $B'$ is the smallest $A$-subalgebra of $B$ with this property.
