# Integral Domains and GCDs

#### Peter

##### Well-known member
MHB Site Helper
I am reading Dummit and Foote Sections 9.3 Polynomial Rings that are UFDs.

I have a problem understanding what D&F say regarding GCDs on page 306 at the end of Section 9.3 (see attached)

D&F write:

======================================================================================

"we saw earlier that if R is a Unique Factorization Domain with field of fractions F and [TEX] p(x) \in R[x] [/TEX], then we can factor out the greatest common divisor d of the coefficients of p(x) to obtain p(x) = dp'(x) where p'(x) is irreducible in both R[x] and F[x]. Suppose now that R is an arbitrary integral domain with field of fractions F. In R the notion of greatest common divisor may not make sense, however, one might still ask if, say, a monic polynomial which is irreducible in R[x] is still irreducible in F[x] (i.e. whether the last statement in Corollary 6 is true). ... ...

=======================================================================================

My question is as follows: Why do D&F say "Suppose now that R is an arbitrary integral domain with field of fractions F. In R the notion of greatest common divisor may not make sense"? D&F's definition of GCD on page 274 (see attached) gives the definition for a GCD of two ring elements a and b for any commutative ring - and there are no conditions on the existence of the GCD - so why for an integral domain would we have a situation where the GCD does not make sense???

Can anyone clarify this for me?

Peter

#### Poirot

##### Banned
I think a gcd between every two elements is only guarenteed in a euclidean domain.

#### Opalg

##### MHB Oldtimer
Staff member
To expand a bit on Poirot's answer, you can define a gcd for two elements in any commutative ring. But that does not mean that such a gcd actually exists. D&F give the example of the ring $\mathbb{Z}[2i]$, which is an integral domain but not a UFD. In that ring, the elements $8 = 2*4 = (2+2i)(2-2i)$ and $12+4i = 2(6+2i) = (2+2i)(4-2i)$ have the common divisors $2$ and $2+2i$. But neither of them is a multiple of $2(2+2i) = 4+4i$, so they do not have a greatest common divisor.