Localization of Ring Exists

Theorem
Let $A$ be a commutative ring with unity.

Let $S \subseteq A$ be a multiplicatively closed subset with $0 \notin S$.

Then there exists a pait $(A_S,\iota)$ satisfying the definition of the localisation of $A$ at $S$.

Proof
Define a relation $\sim$ on the Cartesian product $A \times S$ by:


 * $(a,s) \sim (b,t) \Leftrightarrow \exists\: u \in S \text{ s.t. } atu = bsu$

Lemma 1
Let $a/s$ be the equivalence class of $(a,s)$ in $(A\times S)/\sim$.

Write $A_S$ for $(A\times S)/\sim$, and for $a/s,b/t \in A_S$ define:


 * $\displaystyle \frac as + \frac bt = \frac{at + bs}{st}$


 * $\displaystyle \frac as \cdot \frac bt = \frac{ab}{st}$

Lemma 2
Now define $\iota : A \to A_S$ by $\iota(a) = a/1$.

We must show that $(A_S,\iota)$ satisfy the universal property for localisation.

Let $B$ be a ring and $g : A \to B$ such that $g(S) \subseteq B^\times$.

Suppose that $h : A_S \to B$ is a ring homomorphism with $h \circ \iota = g$.

Then we must have $h(a/1) = g(a)$, and $h(1/s)\cdot h(s/1) = 1$.

Therefore $h(1/s)g(s) = 1$, so $h(1/s) = g(s)^{-1}$.

Therefore $h(a/s) = h(a/1)\cdot h(1/s) = g(a)g(s)^{-1}$.

So if such $h$ exists it must equal $g(a)g(s)^{-1}$, so is unique.

Therefore, to conclude the proof we pull out:

Also See

 * If $A$ is an integral domain and $S = A \backslash \{0\}$ then the localisation of $A$ at $S$ is precisely the field of fractions of $A$.