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 $R$ with a norm $N$ and a division algorithm (i.e. there is a norm $N$ so that for every $a,b\in R$ with $b\neq 0$, there are $q,r\in R$ so that $a=bq+r$ with $r=0$ or $N(r)< N(b)$)
Field: a commutative ring where every nonzero element has an inverse
Because... We can just choose the zero norm: $N(r)=0$ for all $r\in F$.
Proof: Let $F$ be a field and define a norm $N$ so that $N(r)=0$ for all $r\in F$. Then for any $a,b\in F$ with $b\neq 0$, we can write $$a=b(b^{-1}a)+0.$$
Because... If $I\triangleleft R$ is an arbitrary nonzero ideal in the Euclidean domain $R$, then $I=(d)$, where $d\in I$ such that $d$ has the smallest norm among all elements in $I$. Prove this using the division algorithm on $d$ and some $a\in I$.
Proof: Let $R$ be a Euclidean domain with respect to the norm $N$ and let $I\triangleleft R$ be an ideal. If $I=(0)$, then $I$ is principle. Otherwise let $d\in I$ be a nonzero element such that $d$ has the smallest norm among all elements in $I$. We claim $I=(d)$. That $(d)\subset I$ is clear so let $a\in I$. Then by the division algorithm, there exist $q,r\in R$ so that $a=dq+r$ with $r=0$ or $N(r)< N(d)$. Then $r=a-dq\in I$ since $a,d\in I$. But my minimality of $d$, this implies $r=0$. Hence $a=dq\in (d)$ and so $I\subset (d)$.
Because...Every PID has the ascending chain condition (acc) on its ideals!* So to prove PID $\Rightarrow$ UFD, just recall that an integral domain $R$ 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 $R$ be a PID. Then 1) $R$ has the ascending chain condition on principal ideals and 2) every irreducible element is also a prime element. Hence $R$ is a UFD.
Because... By definition.
Proof: By definition.
*Def: In general, an integral domain $R$ has the acc on its principal ideals if these two equivalent conditions are satisfied:
- Every sequence $I_1\subset I_2\subset \cdots \subset \cdots$ of principal ideals is stationary (i.e. there is an integer $n_0\geq 1$ such that $I_n=I_{n_0}$ for all $n\geq n_0$).
- For every nonempty subset $X\subset R$, there is an element $m\in X$ such that whenever $a\in X$ and $(m)\subset (a)$, then $(m)=(a)$.
**To see this, use part 1 of the definition above. If $I_1\subset I_2\subset \cdots $ is an acsending chain, consider their union $I=\bigcup_{n=1}^{\infty}I_n$. That guy must be a principal ideal (check!), say $I=(m)$. This implies that $m$ must live in some $I_{n_0}$ for some $n_0\geq 1$ and so $I=(m)\subset I_{n_0}$. But since $I$ is the union, we have for all $n\geq n_0$ $$(m)=I\supset I_n\supset I_{n_0}=(m).$$ Voila!
Miscellany
Every field $F$ is a PID
because the only ideals in a field are $(0)$ and $F=(1)$! And every field is vacuously a UFD since all elements are units. (Recall, $R$ 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 $R$ be an integral domain and $M\triangleleft R$ a maximal ideal. Then $R/M$ is a field and hence an integral domain, which implies $M\triangleleft R$ 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 $R$ be a PID and $(p)\triangleleft R$ a prime ideal for some $p\in R$. Then $p$ is a prime - and hence an irreducible - element (prime $\Leftrightarrow$ irreducible in PIDs). Since in an integral domain a principal ideal is maximal whenever it is generated by an irreducible element, we conclude $(p)$ 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 $\mathbb{Z}[x]$, consider the ideal $(p)$ for a prime integer $p$. Then $(p)$ is a prime ideal, yet it is not maximal since $$(p)\subset (p,x)\subset \mathbb{Z}[x].$$
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 $N(p(x)) = \deg p(x)$ where $p(x)\in F[x]$.
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 $R\subset R[x]$ and so $R$ must be an integral domain (since a subset of a integral domain inherets commutativity and the "no zero divisors" property). Since $R[x]/(x)\cong R$, 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 $\mathbb{Z}[x]$ is not a PID (in the counterexample a few bullets up) - $\mathbb{Z}$ is not a field!
Reference: D. Dummit and R. Foote, Abstract Algebra, 3rd ed., Wiley, 2004. (ch. 8-9)