Proving D is an Integral Domain: A Prime Number Case

[Driessen12]

Homework Statement

Let p be a prime number, and let D = {m/n| m,n are integers such that p does not divide n}
Verify that D is an integral domain and find Q(D)

Homework Equations

i am unsure where to use the fact that a prime number divides n in this proof. I know how to check that D is an integral domain, but I am unclear where this fact fits in.

The Attempt at a Solution

I must show that addition is commutative associative, has an identity and inverse.
multaplication is commutative and associative, and has an identity.
and show that D is distributive.
n cannot be p, 0, or any multiple of p, but I'm not sure how that plays in the proof, any help would be much appreciated.

 

[Dick]

Most of those properties, like associativity and commutativity, you don't have to prove because you know they are true for the rational numbers. The important ones to prove are closure and inverses. Think about the example p=2. That means D is all fractions with an odd denominator. Can you show that's closed under addition?

 

[Driessen12]

[a,b] + [c,d] is defined as [ad+bc, bd]. For b, d odd integers bd will again be odd, so yes this would be closed under addition

 

[Dick]

[a,b] + [c,d] is defined as [ad+bc, bd]. For b, d odd integers bd will again be odd, so yes this would be closed under addition

Sure, now keep going. What other properties do you really need to prove? Once you're done with p=2 try and do it for general p.

 

[Driessen12]

ok i think i got it, for [m,n] and [a,b] for m,n,a,b in the integers where p does not divide n or b we define addition by
[mb + an, nb] and since p does not divide n or b p cannot divide nb so D is closed under addition.
for [m,n][a,b] we get [ma,nb] and again we know p cannot divide nb so D is closed under multiplication.
Associativity and commutativity work so i just have to show distributive, identities, additive inverse, and 1 not equal to zero which is certainly true.

 

[Dick]

ok i think i got it, for [m,n] and [a,b] for m,n,a,b in the integers where p does not divide n or b we define addition by
[mb + an, nb] and since p does not divide n or b p cannot divide nb so D is closed under addition.
for [m,n][a,b] we get [ma,nb] and again we know p cannot divide nb so D is closed under multiplication.
Associativity and commutativity work so i just have to show distributive, identities, additive inverse, and 1 not equal to zero which is certainly true.

That's the basic idea alright. The distributive property 'just works' too. Aside from closure, you need to show inverses and identities are in D and there are no zero divisors.

 

[Driessen12]

what would Q(D) look like?

 

[Dick]

what would Q(D) look like?

You'll need to define what Q(D) means. I don't know that notation.

 

[Driessen12]

the field of fractions of D or field of quotients.

 

[Dick]

the field of fractions of D or field of quotients.

You can think of D and Q(D) as subsets of the rationals. Let's go back to p=2. Can you name some rationals that are in Q(D)? Can you think of any that aren't?

 

[MikeMcCrack]

I have the same problem, but I am wondering why n can't be zero. zero is not prime, and the conditions says nothing about whether n is zero or not.

 

[Deveno]

because if n = 0, m/n is not a rational number.

 

[MikeMcCrack]

can i assume that m/n has to be rational?

 

[Deveno]

if m,n are integers and n ≠ 0, then m/n is rational. if n = 0, m/n has no meaning (or infinitely many, depending on your viewpoint).

as such, the answer is "yes".

 

[MikeMcCrack]

what is Q(D) going to look like. i know if D={a+ib | a,b in Z} it would be Q(D) iso Q but don't know how to apply it to this problem.

 

[Deveno]

try playing around with what you get for p = 3, and comparing the results to p = 2.

for example, 1/2 and 2/5 are in D₃.

what can you say about the denominators of rational numbers in D₃?

are all the integers in D₃?

you should convince yourself that the addition and multiplication in D_p is the same as the ordinary addition and multiplication of rational numbers. why does that guarantee we have a ring (provided we have closure)?

do you know how to test if a subset of a ring is a subring?

once you know that, the only additional property to verify is the zero-divisors property.

what would it mean for [m,n][m',n'] = [0,k] for some k?