Domain question

  • #1

Homework Statement



Let c and d be two non-zero elements of a domain D. If a and b are integers s.t gcd(a,b)=1, a>0, b > 0.

If we know c^a=d^a and c^b=d^b,

does it follow that c=d?

Homework Equations





The Attempt at a Solution



I'm thinking divinding the two might be pointless ..but subtracting might be a good idea.
c^a-c^b=d^a-d^b.

Somehow have to use the fact that we are working in a domain? The hint is to consider the Quot(D) and work in it.
 

Answers and Replies

  • #2
actually...maybe I'm wrong, now I'm thinking to consider

(c^a)/(d^a)=(c^b)/(d^b) (given since non-zero) ...

I'm thinking if one messes around with this ...I'll get the answer. Does this seem on the right track?
 
  • #3
I'm getting then

(c^a)*(d^b)-(c^b)*(d^a)=0

I sbustituted in c^b=d^b to get

(c^a)*(d^b)-(d^b)*(d^a)=0

but that gets me back to where I started...
 
  • #4
22,129
3,297
Hi Metric Space! :smile:

Do you know Bezout's theorem? The one that says that if gcd(a,b)=1, then there exist integers k and l such that ka+lb=1.

I suggest you use this.
 
  • #5
interesting...I'll give it a shot -- I know the result but not the name...until now
 
  • #6
22,129
3,297
Well, actually, Bezout's theorem says that for each a and b, there exists k and l such that

[tex]ka+lb=gcd(a,b)[/tex]

I just applied it here with gcd(a,b)=1.

It's one of the more important theorems in abstract algebra and number theory. It's very useful!
 
  • #7
care to give another hint? I plugged that into one of my equations and just got a big mess
 
  • #8
22,129
3,297
Complete this:

[tex]c=c^{ak+lb}=...[/tex]
 
  • #9
ah...very useful hint

I solved for a before a in ak+lb=1...and that's what made a mess ...this hint 'solved' it - thanks!
 

Related Threads on Domain question

  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
4
Views
1K
Replies
0
Views
460
  • Last Post
Replies
0
Views
868
  • Last Post
Replies
16
Views
4K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
0
Views
3K
Replies
8
Views
15K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
Top