The Integral Domain Hierarchy, Part 1

Here is a list of some of the subsets of integral domains, along with the reasoning (a.k.a proofs) of why the bullseye below looks the way it does. Part 2 of this post will include back-pocket examples/non-examples of each.   

Integral Domain: a commutative ring with 1 where the product of any two nonzero elements is always nonzero

Unique Factorization Domain (UFD): an integral domain where every nonzero element (which is not a unit) has a unique factorization into irreducibles

Principal Ideal Domain (PID): an integral domain where every ideal is generated by exactly one element

Euclidean Domain: an integral domain RR with a norm NN and a division algorithm (i.e. there is a norm NN so that for every a,bRa,bR with b0b0, there are q,rRq,rR so that a=bq+ra=bq+r with r=0r=0 or N(r)<N(b)N(r)<N(b))

Field: a commutative ring where every nonzero element has an inverse

Because... We can just choose the zero norm: N(r)=0N(r)=0 for all rFrF.

Proof: Let FF be a field and define a norm NN so that N(r)=0N(r)=0 for all rFrF. Then for any a,bFa,bF with b0b0, we can write a=b(b1a)+0.a=b(b1a)+0.

Because... If IRIR is an arbitrary nonzero ideal in the Euclidean domain RR, then I=(d)I=(d), where dIdI such that dd has the smallest norm among all elements in II. Prove this using the division algorithm on dd and some aIaI.

 Proof: Let RR be a Euclidean domain with respect to the norm NN and let IRIR be an ideal. If I=(0)I=(0), then II is principle. Otherwise let dIdI be a nonzero element such that dd has the smallest norm among all elements in II. We claim I=(d)I=(d). That (d)I(d)I is clear so let aIaI. Then by the division algorithm, there exist q,rRq,rR so that a=dq+ra=dq+r with r=0r=0 or N(r)<N(d)N(r)<N(d). Then r=adqIr=adqI since a,dIa,dI. But my minimality of dd, this implies r=0r=0. Hence a=dq(d)a=dq(d) and so I(d)I(d).

Because...Every PID has the ascending chain condition (acc) on its ideals!* So to prove PID UFD, just recall that an integral domain RR is a UFD if and only if 1) it has the acc on principal ideals** and 2) every irreducible element is also prime.

 Proof: Let RR be a PID. Then 1) RR has the ascending chain condition on principal ideals and 2) every irreducible element is also a prime element. Hence RR is a UFD.

Because... By definition.

 Proof: By definition.

*Def: In general, an integral domain RR has the acc on its principal ideals if these two equivalent conditions are satisfied:

  1. Every sequence I1I2I1I2 of principal ideals is stationary (i.e. there is an integer n01n01 such that In=In0In=In0 for all nn0nn0).
  2. For every nonempty subset XRXR, there is an element mXmX such that whenever aXaX and (m)(a)(m)(a), then (m)=(a)(m)=(a).

 **To see this, use part 1 of the definition above. If I1I2I1I2 is an acsending chain, consider their union I=n=1InI=n=1In. That guy must be a principal ideal (check!), say I=(m)I=(m). This implies that mm must live in some In0In0  for some n01n01 and so I=(m)In0I=(m)In0. But since II is the union, we have for all nn0nn0 (m)=IInIn0=(m).(m)=IInIn0=(m). Voila!

Miscellany

Every field FF is a PID 

because the only ideals in a field are (0)(0) and F=(1)F=(1)! And every field is vacuously a UFD since all elements are units. (Recall, RR is a UFD if every non-zero, non-invertible element (an element which is not a unit) has a unique factorzation into irreducibles).

In an integral domain, every maximal ideal is also a prime ideal. 

(Proof: Let RR be an integral domain and MRMR a maximal ideal. Then R/MR/M is a field and hence an integral domain, which implies MRMR is a prime ideal.)

Butut the converse is not true (see counterexample below). However, the converse is true in a PID because of the added structure! 

(Proof: Let RR be a PID and (p)R(p)R a prime ideal for some pRpR. Then pp is a prime - and hence an irreducible - element (prime irreducible in PIDs). Since in an integral domain a principal ideal is maximal whenever it is generated by an irreducible element, we conclude (p)(p) is maximal.)         

This suggests that if you want to find a counterexample - an integral domain with a prime ideal which is not maximal - try to think of a ring which is not a PID:   In Z[x], consider the ideal (p) for a prime integer p. Then (p) is a prime ideal, yet it is not maximal since (p)(p,x)Z[x].            

If F is a field, then F[x] - the ring of polynomials in x with coefficients in F - is a Euclidean domain with the norm N(p(x))=degp(x) where p(x)F[x]

By the integral domain hierarchy above, this implies every ideal in F[x] is of the form (p(x)) (i.e. F[x] is a PID) and every polynomial can be factored uniquely into a product of prime polynomials (just like the integers)! The next bullet gives an "almost converse" statement.

If R[x] is a PID, the R must be a field.

To see this, simply observe that RR[x] and so R must be an integral domain (since a subset of a integral domain inherets commutativity and the "no zero divisors" property). Since R[x]/(x)R, it follows that R[x]/(x) is also an integral domain. This proves that (x) is a prime ideal. But prime implies maximal in a PID! So R[x]/(x) - and therefore R - is actually a field.

  • This is how we know, for example, that Z[x] is not a PID (in the counterexample a few bullets up) - Z is not a field!

Reference: D. Dummit and R. Foote, Abstract Algebra, 3rd ed., Wiley, 2004. (ch. 8-9)

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