Exercise 18, § 1

a) Let A be a ring, S a subset of A, and $S'$ the smallest subset of A containing 1 and S and stable under multiplication. We say that S admits a calculus of left fractions on A if the following two (tautological when A is commutative) conditions are satisfied:

(i) For every $a\in A$ and every $s\in S$, there exist $b\in A$ and $t\in S'$ such that $ta=bs$.

(ii) For every $a\in A$ and $s\in S$ such that $as= 0$, there exists a $t\in S'$ such that $ta= 0$.

This is equivalent to saying that S admits a calculus of left fractions on the monoid obtained by endowing the set A with only multiplication (Exercise 17). Prove that on the multiplicative monoid $S^{-1}A$, there then exists a unique addition such that $S^{-1}A$, endowed with this addition and its multiplication, is a ring and that the canonical mapping $\varepsilon : A\rightarrow S^{-1}A$ (loc. cit.) is a ring homomorphism. We say that $S^{-1}A$ is the ring of left fractions of A associated with S. When A is commutative, this coincides with the ring defined in I, §8, No. 12, p. 113.

b) The set S admits a calculus of left fractions on A if and only if there exist a ring B and a ring homomorphism $u: A\rightarrow B$ such that every element of $u(S)$ is invertible in B and that for every $a\in u^{-1}$(0), there exists an $s\in S$ such that $sa= 0$.

c) Suppose that S admits a calculus of left fractions on A. The kernel of the canonical homomorphism $\varepsilon : A\rightarrow S^{-1}A$ is the set of $a\in A$ for which there exists an $s\in S$ with $sa= 0$.

Let $f$ be a homomorphism from A to a ring B such that every element of $f(S)$ is invertible in B. There exists a unique homomorphism $\overline{f}: S^{-1}A\rightarrow F$ such that $f=\overline{f}\circ s$. If $f$ is injective, then $\overline{f}$ is injective.

d) Formulate the statements analogous to a), b), c) for a calculus of right fractions; if S admits a calculus of right fractions on A, we denote the ring of right fractions of A by AS$^{-1}$. In this case, we have (AS$^{-1}$)$^o= S^{-1}(A^o)$. If S admits a calculus of left and right fractions on A, then the rings $S^{-1}A$ and AS$^{-1}$ are canonically isomorphic.

e) Let A be a nonzero ring without zero divisors. Then A admits a field of left fractions (I, §9, p. 177, Exercise 15) if and only if the set S of nonzero elements of A admits a calculus of left fractions on A, and in this case, $S^{-1}A$ is the field of left fractions of A.