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conditions for integral domain to be a PID

oblixps

Member
May 20, 2012
38
Let R be an integral domain. Then R is a PID if the following 2 conditions hold:

1) any 2 elements [tex] a, b \in R [/tex] have a greatest common divisor which can be written as ra + sb for some [tex] r, s \in R [/tex].
2) If [tex] a_1, a_2, ... [/tex] are nonzero elements of R such that [tex] a_{i+1}|a_i [/tex] for all i, then there is a positive integer N such that [tex] a_n [/tex] is a unit times [tex] a_N [/tex] for all [tex] n \geq N [/tex].

Letting I be any ideal in R with the ordering [tex] x \leq y [/tex] iff y|x, condition #2 implies that each chain of elements in I has an upper bound and thus by Zorn's Lemma, there is a maximal element in I with respect to this ordering, namely, an element that divides every other element in I.

what i am having trouble understanding is why condition #1 is necessary. it seems that if we say a is the element in I that divides every other element, then can't we conclude that I = (a), since (a) is clearly in I and I is in (a) since every element x of I must be contained in (a)?

if there was another such element b in I not equal to a, then since b and a are both maximal elements with respect to this ordering, a|b and b|a which means b = ac for some c and a = bd for some d, which implies that cd = 1 since R is an integral domain. Thus c and d are units so a and b are associates so (a) = (b). Why is the condition that the gcd of a and b exist necessary for R to be a PID?
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Disclaimer: I'm very bad at algebra but I believe attempting to answer your question may enlighten us both. Let us try! (Handshake)

I assume that by PID you mean principal ideal domain. The definition of PID is that every ideal is a principal ideal. In other words, this means that every ideal is generated by a multiple of one element.

That being said, I do not understand your conditions. The first states that you have an euclidean division algorithm such that you can have a greatest common divisor, whereas the second appears to be, loosely speaking, mentioning that every ascending chain in a PID will eventually stabilize, i.e., be essentially the same.

I do not see these as necessary conditions for an integral domain to be a PID. Where did you find this definition?
 

oblixps

Member
May 20, 2012
38
this was actually an exercise in Dummit and Foote. in case you have a copy, it is #4 in section 8.2.
 

oblixps

Member
May 20, 2012
38
after thinking about it some more, it seems to me that condition #1 follows from condition #2 by taking the ideal I = (a, b) generated by 2 elements a and b. Then from the argument above using condition #2 and Zorn's Lemma, we have I = (a, b) = (c) for some element c that divides all elements of I. But then, it is easy to see from here that gcd(a, b) exists and c = gcd(a, b). We can thus write c = ra + sb for some r, s in R.

so unless I made a mistake somewhere, it seems to me that condition #1 is really not necessary and I'm not sure why it was part of the exercise.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
no. condition #2 is not enough.

consider the ring Z[x], the ring of polynomials in x with integer coefficients.

suppose we have a sequence of polynomials:

p1, p2,...., with pj+1|pj for each j, with p1 ≠ 0

well p1 is of finite degree (since its non-zero), so it only has a finite number of irreducible factors. so the degree of the pj's can only go down a finite number of times.

but, you might object: maybe there's an infinite string of pj's where the degree "stays the same". but such a string must occur "at the tail" of the sequence. in that case, since the pj all have the same degree after pN, then for such pj, if:

pj(x) = k(x)pj+1(x), then deg(k) = 0, that is k(x) is a constant, k(x) = aj.

this in turn gives a sequence of integers of the leading coefficients of the pN+i:

a1​, a2,.... where ai+1|ai.

now Z IS a PID, and it is clear that we do have some M for which ai = ±a for all i greater than or equal to M.

so, going back to our original sequence of polynomials, after N+M terms (at most), we have pj = ±pN+M, for all j greater than or equal to N+M. so condition #2 is satisfied for Z[x].

HOWEVER, i defy you to find two polynomials in Z[x] such that:

2p(x) + xq(x) = 1

(it is clear that gcd(2,x) = 1, the only divisors of 2 are 1 and 2 (up to sign, that is multiplication by a unit), and 2 does not divide x).

what went wrong? well it turns out the the only element c with c|2 and c|x is 1 (or -1). but (1) = Z[x] (the entire ring), which is not the same as (2,x). in short: it may be that if c = gcd(a,b), that the ideal generated by c is a LOT bigger than the ideal generated by (a,b), because c may not be a R-linear combination of a and b.

THAT is why you need condition #1.
 

oblixps

Member
May 20, 2012
38
thanks for the counterexample! that helped me see what could happen if that condition wasn't met.

i'm still trying to figure out what went wrong in my argument though. By using Zorn's Lemma, i was able to find an element a in I that divides every element in I so it follows that I is contained in (a). but since a itself is contained in I, we also have that (a) is contained in I and thus I = (a). so it seems to follow that every ideal I can be generated by this "minimal" divisible element.

you've shown that Z[x] satisfies condition #2 so from my above argument, every ideal in Z[x] should have such a minimal divisible element. looking at (2, x) in Z[x] however, I see that the only possible element that could divide every element in (2, x) is 1 (or -1). but (2, x) doesn't contain 1. so i'm suspecting that my argument with Zorn's lemma wasn't careful enough, but i can't seem to figure out where my logic went wrong. could you help me find where I went wrong in my argument?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
your argument uses condition #1 in an essential way, you assert:

c = ra + sb (GCD domains in which this is true are called bezout domains. not ALL GCD domains are bezout domains).

you showed that there does indeed exist an element c such that any ideal (a,b) is contained in (c).

but if R is not principal, then not every ideal is of the form (c).

condition #2 is also known as "the ascending chain condition on principal ideals" (ACCP). we can summarize this like so:

A) GCD domain + ACCP = UFD
B) bezout domain + ACCP = PID

as in my counter-example, any UFD which is NOT a PID will satisfy condition 2, but will not satisfy condition 1.

the condition d = gcd(a,b) = ra + sb is exactly what we need to conclude (a,b) = (d).