The localisation of a ring over the non-zero divisors is the quotient field of this ring

Question

Let $A$ a ring and let $S(A)=\left\{a\in A\; |\; a\; \text{non-zero divisor}\right\} $. This is a multiplicative system, so the localisation $A_{S(A)}$ is well-defined. Actually, I am trying to show that $A_{S(A)}=\text{Quot} \left(A\right)$. Of course the localisation is contained in the quotient but I cannot prove that a zero divisor is invertible in $A_{S(A)}$. I have seen this in an algebraic geometry course but I cannot justify it.

Answer 1

What you are trying to prove is false. You certainly have an injective morphism from $Frac(A)$ to $A_{S(A)}$ but theses two objects are not always isomorphic. The localisation $A_{S(A)}$ you defined has a name and is called the total ring of fraction. You might be interested by its Wikipedia page.

Here are two counter examples of your assumption :

• You can prove that the total ring of fraction of a product of two rings is the product of the total rings of fraction of the rings. Hence the product $\mathbb{Z}\times \mathbb{Z}$ have $\mathbb{Q}^2$ as total ring of fraction which is not a field and so not isomrphic to the field of fractions.

• In the ring $\mathcal{H}(U_1 \cup U_2)$ of holomorphic function on two disjoint open sets of $\mathbb{C}$ the invertible elements are the functions that do not vanishes. But this ring has non invertible zero-divisors, for example $z\mapsto z~~if~z\in U_1~~or~~0~~if~~ z\in U_2$. Note that for holomorphic functions on a connected open set, this cannot happen : the ring is an integral domain.