GCD of a & b in Ring R: Unique or Not?

[rourky]

1. I am just looking for a defn, I can't find it on the net:
Given a ring R, a, b elements of R, (a?) gcd(a,b) is defined to be?






3. I am guessing that if d/a and d/b and for any other e such that e/a and e/b we have e/d, then d is (a?) gcd of a and b. Is this correct, and is d THE unique gcd? My guess is based mostly on the definition of a ("a", may not be unique) gcd(B), B any subset of R.

 

[NateTG]

http://mathworld.wolfram.com/GreatestCommonDivisor.html

I'm not sure that gcd is well-defined for rings. For example, the real numbers are a ring and it's very difficult to come up with anything sensible for, say, the greatest common divisor of $\pi$ and $e$.

Perhaps there's something missing?

 

[matt grime]

GCDs exist in Euclidean domains (i.e. places where you can do the euclidean algorithm) for example, but not rings in general. GCDs, when they exist, are never unique: they are only defined up to multiplication by a unit.

If you have a PID (euclidean domains are a subset of PIDs), then gcd(a,b) is the/a generator of the ideal (a,b) (which exists since ideals are principal).

 

[NateTG]

GCDs exist in Euclidean domains (i.e. places where you can do the euclidean algorithm) for example, but not rings in general. GCDs, when they exist, are never unique: they are only defined up to multiplication by a unit.

If you have a PID (euclidean domains are a subset of PIDs), then gcd(a,b) is the/a generator of the ideal (a,b) (which exists since ideals are principal).

Well, it depends on your definition of GCD. For any ring which has unique prime decompositions, so for any elements of the ring, $a$ and $b$, we have:
$$a=\prod p_i^{a_i}$$
and
$$b=\prod p_i^{b_i}$$
The GCD can be defined as:
$$\rm{GCD}(a,b)=\prod p_i^{\rm{min}(a_i,b_i)}$$

 

[matt grime]

I don't see how anything 'depends on your definition of GCD'.

 

[NateTG]

I don't see how anything 'depends on your definition of GCD'.

Well $\rm{GCD}(a,b)=\prod p_i^{\rm{min}(a_i,b_i)}$ gives a unique GCD. While defining the GCD as a generator of the of the ideal $<a,b>$ may not.

 

[matt grime]

No, it doesn't give any uniqueness. If p is a prime, so is up for any unit u. Prime decomposition is only unique up to re-ordering and units. Your GCD is only defined up to associates, just like any other GCD. The only reason we think GCD is unique in Z, is because we have the > relation, and think that 'greatest' means 'largest in value'.