Thread: Rings of Fractions - Dummit and Foote Section 7.5 - elements of Q as r.d-1

I am reading Dummit and Foote, section 7.5 Rings of Fractions. I am working through Theorem 15 on page 261 (see attachment)

I am happy with D&F's prrof of Th 15 (thanks in part to a post by Deveno) down to the following paragraph: (see attachment)

"Next note that each has a multiplicative inverse in Q: namely if d is represented by the fraction then its multiplicative inverse is . One sees that every element of Q can be written as for some and some ."

Presumably in this paragraph so that we are sure that . We can also be sure that the inverse is unique because if d is represented by with inverse where then ("cross multiply")

But when D&F write:

"One sees that every element of Q can be written as for some and some ." - what exactly do they mean by - is this just shorthand for ?

Can someone please clarify this for me?

Peter

pretty much.

yes, we are assuming e is an element of D. the fraction e/e (or d/d) acts as the unity of D^-1R (remember we are dealing with equivalence classes of RxD, not actual pairs (r,d). so there may be (and usually IS) MANY ways to represent "the same fraction". this is not much more than what we have in ordinary rational numbers:

3/4 = 6/8 = 9/12.....etc.)

this ring is usually written D^-1R, instead of R/D, to avoid thinking we are dealing of a "quotient ring" (that is, D is not necessarily an ideal of R).

the troubling inclusion of the "e" in re/de is only because 1/d may make "no sense", R may not, in fact, posses an element 1. if it does, we can use e = 1.

here is another example, which may give some additional perspective: suppose R = 2Z[x], all integer polynomials with even coefficients.

suppose we need to create a ring that contains this ring as a subring, but in which 2x⁴ has an inverse.

we can take D = 2Z[x⁴] - {0}, non-zero polynomials in x⁴ with even integer coefficients (not closed under addition, but *is* closed under multiplication).

what is our "newly-created" inverse of 2x⁴ in D^-1R? well one possible form of it is:

2/(4x⁴) (the top is in R, and the bottom is in D).

of course, for any p(x) in D, 2p(x)/(4x²p(x)) would work just as well, for example, we could use:

p(x) = 2x⁴ + 4, and in which case:

(4x⁴ + 8)/(8x⁸ + 16x⁴) is also a form of (2x⁴)^-1.

Thanks for the guidance on the issue ...

Again, the examples are most helpful

Peter