Integral Domains/Algebraic Number Theory - notation/meaning

• Jun 3rd 2011, 07:50 PM
Bernhard
Integral Domains/Algebraic Number Theory - notation/meaning
I am a math hobbyist reading Alaca and Williams "Algebraic Number Theory"

At the bottom of page 9 Alaca & Williams present Theorem 1.3.1 - see attached

The proof begins as follows:

Proof: If a/b $\in$ U(D) then a = bu for some u $\in$ U(D). etc etc (see attached)

[ U(D) is Alaca & WIlliams notation for the set of units of D. The authors do not explain the notation a/b.]

My question is what does he mean by this first statement? What is a/b? is it a divided by b? If so surely 'divides' is not a defined operation in an integral domain? Why start with such a peculiar element as a/b? Can someone clarify for me what is going on here please.

Bernhard.
• Jun 4th 2011, 12:14 AM
Deveno
without more context, i can only put forth a guess, but i would surmise that a/b is meant to be an element of the field of fractions of D (for the integral domain Z, this is the rational numbers). we can form this field from any integral domain, in the same way we form rational numbers from integers.

in such a field, b(a/b) = a, so if a/b = c is in D, then a = bc in D (this is not, strictly speaking, correct. but there is a subring of the field of fractions isomorphic to D, and this isomorphism is often treated as equality. for example, we regard the fraction 4/2 as being an integer).

i daresay you could rephrase the theorem to read:

for a,b in D*, <a> = <b> iff a = bu, for some u in U(D).

i think that in this context, you could also take a/b in D to mean: "a is divisible by b". by analogy with the rational numbers: a is divisible by b iff a/b is an integer.

as i am not familiar with this text, all of the above is an educated guess (but it does fit the situation).

there is another use of a/b, with a similar construction based on a multiplicative subset S (all non-zero) of a commutative ring R, usually written S^-1R,

(sometimes R/S), of which the field of fractions of an integral domain D is a special case (clearly, D* is multiplicatively closed in an integral domain,

by definition). in this more general setting, R may not embed into S^-1R (if s is a zero-divisor in S, so that as = 0 for some a in R, then a/1 = as/s = 0. not good.).

if R is an integral domain, we happily avoid this unfortunate situation.
• Jun 4th 2011, 05:47 PM
Bernhard
Thanks Deveno,

I will reflect on your post.

I like your suggestion to rephrase the theorem best.

I am assuming that the field of fractions you mention is the same as Alaca & Williams reference on page 3 of their book [note I wanted to upload a pdf of this but could not find an upload icon for replies!] to the field of quotients associated with and isomorphic to D. Mind you I cannot see how you would use this field in the proof of the theorem - it seems somewhat irrelevant - but maybe I am missing something.