Let R be a unitary commutative ring such that 1 noteqaulto 0

Let R be a unitary commutative ring such that 1 noteqaulto 0 and S R he closed under multiplication (i.e. x, y S, xy S) and contain 1. We define the relation E on R Times S by (a, s)E(b,t) if and only if there exists x S such that xat = xbs. Show that E is an equivalence relation. Let R_s denote the set (R Times S)/E (it is the set of E-classes). If (a, s) R/Times S, we denote by (a, s) R_S the E-class of (a, s). Show that the map ((a, s), (b, t)) rightarrow (ab, st) is well defined. We denote this map *. Show that the map ((a, s), (b, t)) rightarrow (at + bs,st) is well defined. We denote this map Show that (R_S, *) is a unitary commutative ring. Show that the map a rightarrow (a, 1) is a unitary ring homomorphism from psi: R rightarrow R_S. Show that if S contains 0 then R_s is the trivial ring. Show that psi is not injective if and only if S contains a zero-divisor. Show that R \\ s {0} is closed under multipication if and only if R is an integral domain. Assume that R is an integral domain. Show that R(R\\(0)) is a field. Show that Z_(z\\{0}) is isomorphic (as a unitary ring) to Q.

Solution

R be an unitary commutative ring such that 1 0 and S is a subset of R. S is closed under multiplication and contain 1. Define a relation E on R x S by (a, s)E(b, t) if and only if there exists x S such that xat = xbs.

1. To prove that E is an equivalence relation.

E is Reflexive:

As xas = xas for any x S, we have (a, s)E(a, s) for all (a, s) R x S. This implies that E is reflexive.

E is Symmetric:

Let (a, s)E(b, t) is true i.e. there exists x S such that xat = xbs.

Now, xat = xbs can be written as xbs = xat, i.e. (b, t)E(a, s)

Since, (a, s)E(b, t) implies (b, t)E(a, s) for all (a, s), (b, t) R x S, E is symmetric.

E is Transitive:

Let (a, s)E(b, t) and (b, t)E(c, u) is true, i.e. there exists x S such that xat = xbs; there exists y S such that ybu = yct

xat = xbs ----------------- (1)

ybu = yct ----------------- (2)

Multiply both sides of equation by yu to get, xatyu = xybus (As R is a commutative ring, multiplications are commutative)

Or, xatyu = x(ybu)s

Or, xauty = x(yct)s                 from (2), we get ybu = yct

Or, xau(ty) = xcs(ty)

Or, xau = xcs

This implies that (a, s)E(c, u) is true.

So, (a, s)E(b, t) and (b, t)E(c, u) implies that (a, s)E(c, u) for all (a, s), (b, t), (c, u) R x S

Therefore, E is transitive.

Since, E is reflexive, symmetric and transitive, E is an equivalence relation on R x S. (Proved)