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,b∈Ra,b∈R with b≠0b≠0, there are q,r∈Rq,r∈R 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 r∈Fr∈F.
Proof: Let FF be a field and define a norm NN so that N(r)=0N(r)=0 for all r∈Fr∈F. Then for any a,b∈Fa,b∈F with b≠0b≠0, we can writea=b(b−1a)+0.a=b(b−1a)+0.
Because… If I◃RI◃R is an arbitrary nonzero ideal in the Euclidean domain RR, then I=(d)I=(d), where d∈Id∈I such that dd has the smallest norm among all elements in II. Prove this using the division algorithm on dd and some a∈Ia∈I.
Proof: Let RR be a Euclidean domain with respect to the norm NN and let I◃RI◃R be an ideal. If I=(0)I=(0), then II is principle. Otherwise let d∈Id∈I 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 a∈Ia∈I. Then by the division algorithm, there exist q,r∈Rq,r∈R so that a=dq+ra=dq+r with r=0r=0 or N(r)<N(d)N(r)<N(d). Then r=a−dq∈Ir=a−dq∈I since a,d∈Ia,d∈I. 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:
- Every sequence I1⊂I2⊂⋯⊂⋯I1⊂I2⊂⋯⊂⋯ of principal ideals is stationary (i.e. there is an integer n0≥1n0≥1 such that In=In0In=In0 for all n≥n0n≥n0).
- For every nonempty subset X⊂RX⊂R, there is an element m∈Xm∈X such that whenever a∈Xa∈X and (m)⊂(a)(m)⊂(a), then (m)=(a)(m)=(a).
**To see this, use part 1 of the definition above. If I1⊂I2⊂⋯I1⊂I2⊂⋯ is an acsending chain, consider their union I=⋃∞n=1InI=⋃n=1∞In. That guy must be a principal ideal (check!), say I=(m)I=(m). This implies that mm must live in some In0In0 for some n0≥1n0≥1 and so I=(m)⊂In0I=(m)⊂In0. But since II is the union, we have for all n≥n0n≥n0(m)=I⊃In⊃In0=(m).(m)=I⊃In⊃In0=(m).Voila!
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 M◃RM◃R a maximal ideal. Then R/MR/M is a field and hence an integral domain, which implies M◃RM◃R 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 p∈Rp∈R. 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]Z[x], consider the ideal (p)(p) for a prime integer pp. Then (p)(p) is a prime ideal, yet it is not maximal since(p)⊂(p,x)⊂Z[x].(p)⊂(p,x)⊂Z[x].
If FF is a field, then F[x]F[x] – the ring of polynomials in xx with coefficients in FF – is a Euclidean domain with the norm N(p(x))=degp(x)N(p(x))=degp(x) where p(x)∈F[x]p(x)∈F[x].
By the integral domain hierarchy above, this implies every ideal in F[x]F[x] is of the form (p(x))(p(x)) (i.e. F[x]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]R[x] is a PID, the RR must be a field.
To see this, simply observe that R⊂R[x]R⊂R[x] and so RR must be an integral domain (since a subset of a integral domain inherets commutativity and the “no zero divisors” property). Since R[x]/(x)≅RR[x]/(x)≅R, it follows that R[x]/(x)R[x]/(x) is also an integral domain. This proves that (x)(x) is a prime ideal. But prime implies maximal in a PID! So R[x]/(x)R[x]/(x) – and therefore RR – is actually a field.
- This is how we know, for example, that Z[x]Z[x] is not a PID (in the counterexample a few bullets up) – ZZ is not a field!
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*Credit for article given to Tai-Danae Bradley*