Thread: Rings of Fractions - Dummit and Foote Section 7.5 - Uniqueness of Q

I am working through Theorem 15 (Rings of Fractions) of Dummit and Foote Chapter 7 (see attachment)

I am unable to completely follow the details of the argument for the uniqueness property of Q - see page 263 of D&F (see attachment) where D&F's argument is as follows:

"It remains to establish the uniqueness propoerty of Q.

Assume is an injective ring homomorphism such that is a unit in S for all .

Extend to a map by defining for all .

The map is well defined since implies , so and then

.

It is straightforward to check that is a ring homomorphism - details left as an exercise.

Finally, is injective because implies

Since is injective ths forces r and hence to be zero.

This completes the proof"


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My problems are as follows:


1. I cannot follow the argument that implies

Can someone please give an explicit and formal set of steps that establishes this.


2. in the statement of Theorem 15 on page 261 (see attachment) D&F state that Q is unique and that the ring Q is the "smallest" ring containing R in which all elements of D become units

Can someone please explain how the above proof actually establishes these facts.

Peter

suppose rd^-1 is in ker(Φ).

this means Φ(rd^-1) = φ(r)φ(d)^-1 = 0_S.

thus φ(r) = φ(r)(1_S) = φ(r)(φ(d)^-1φ(d)) = (φ(r)φ(d)^-1)φ(d) = (0_S)φ(d) = 0_S.

hence r is in ker(φ). thus (since φ is injective, r = 0_R) and thus rd^-1 = 0_Q

(the equivalence class of (0,d) is indeed the additive identity of Q = D^-1R).

**********

now, suppose T is a ring such that: T contains R as a subring, and every d in D (which is a subset of R) is a unit of T.

since T contains R as a subring, the inclusion mapping i:R→T given by i(r) = r, is clearly an injective ring homomorphism.

moreover, i(d) = d is a unit in T.

we thus get an injective ring homomorphism I:Q→T. thus I(Q) is a subring of T (ring)-isomorphic to Q.

it is common practice to "identify" isomorphic structures (just as we regard the rational number n/1 as "the integer" n).

Q is only unique, UP TO ISOMORPHISM. that is what the proof shows: if we have a ring S that embeds R and in which all elements of D are units, it has a subring isomorphic to Q.

that is, if the mapping Φ in D&F's proof is surjective, it is an isomorphism.

this is "common" in abstract algebra: since one is only concerned with classifying algebraic objects up to isomorphism type (which, being an equivalence, is "almost as good as equality") one often sees the following definitions:

a subobject B (up to isomorphism) of an object A, is an injective homomorphism from B to A.
a quotient object C (up to isomorphism) of an object A is a surjective homomorphism from A to C.

(note that these definitions only hold for "certain kinds" of objects: most notably- sets, groups, rings, modules, and vector spaces).

the basic idea is: it turns out that all the "interesting stuff" in algebra isn't really in the objects themselves, it's happening in the "structure-preserving maps" BETWEEN objects.

if you want to understand groups, you study homomorphisms. if you want to understand vector spaces, you study linear transformations. if you want to understand sets, you look at functions.

it is a bit unsettling to think of a sub-something as just "an injective homomorphism". as humans, we're more "object-oriented", and not "process-oriented". but that is where the abstraction leads, it turns out that the properties of "object A" are better described by "isomorphically preserved information" than explicit, intrinsic description.

for example, what is relevant about Z₄ is that it is a cyclic group of order 4, it really "doesn't matter" (as far as GROUP theory is concerned) if we are talking about:

{0,1,2,3} under addition mod 4
{1,i,-1,-i} under complex multiplication <---aka the roots of x₄ - 1 (see last example).
{e, (1 2 3 4), (1 3)(2 4), (1 4 3 2)} under functional composition
, under matrix multiplication
{e,x,x²,x³; x⁴ = e}


the group information that is pertinent is the same for all of these.

Thanks Deveno

Working through your post now

Peter