2 Attachment(s)

Integral Domains and GCDs

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:

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"we saw earlier that if R is a Unique Factorization Domain with field of fractions F and $\displaystyle p(x) \in R[x] $, 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). ... ...

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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

Re: Integral Domains and GCDs

Note that D&F did not specify the existence of such an element, only that such an element, if it exists, satisfies certain properties.

As for your question, it is because factorisation may not be unique. In the integral domain $\displaystyle \mathbb{Z}[\sqrt{-3}]$, note that

$\displaystyle 4 =2 \cdot 2 = (1+\sqrt{-3})(1-\sqrt{-3})$.

Now $\displaystyle 2$ and $\displaystyle 1-\sqrt{-3}$ divides both $\displaystyle x=4$ and $\displaystyle y=2\cdot (1-\sqrt{-3}) $ but neither can satisfy the second property of gcd$\displaystyle (x,y)$, since $\displaystyle 2$ does not divide $\displaystyle (1-\sqrt{-3})$ or vice versa. Note that $\displaystyle 1$ cannot be a gcd either because it divides both $\displaystyle 2$ and $\displaystyle (1-\sqrt{-3})$, which are both factors of $\displaystyle 4$ and $\displaystyle 2\cdot (1-\sqrt{-3}) $ and not the other way round.