The chain of inclusions
\[ \text{fields} \subset \text{Euclidean domains} \subset \text{PIDs} \subset \text{UFDs} \subset \text{integral domains} \subset \text{commutative rings} \]
Inclusions among fields, Euclidean domains, PIDs, UFDs, and integral domains (Tag PR-LLBRB)
Every field is a Euclidean domain, every Euclidean domain is a principal ideal domain, every principal ideal domain is a unique factorization domain, every unique factorization domain is an integral domain, and every integral domain is a commutative ring: \[ \text{Fields} \subset \text{Euclidean domains} \subset \text{PIDs} \subset \text{UFDs} \subset \text{Integral domains} \subset \text{Commutative rings}. \]
Each inclusion is proper: \(\ZZ\) is a Euclidean domain that is not a field; \(\ZZ\big[\tfrac{1+\sqrt{-19}}{2}\big]\) is a PID that is not Euclidean; \(\ZZ[x]\) is a UFD that is not a PID; \(\ZZ[\sqrt{-5}]\) is an integral domain that is not a UFD; and \(\ZZ/4\ZZ\) is a commutative ring that is not an integral domain.
(Proofs of the inclusions)
Field \(\implies\) Euclidean domain. With the norm \(d(x)=0\) for \(x\neq0\), write \(x = qy+r\) with \(q = \inverseof{y} x\) and \(r=0\).
Euclidean domain \(\implies\) PID. Let \(I\neq 0\) be an ideal and \(a\in I\) a nonzero element with \(d(a)\) minimal. If \(b\in I\), write \(b = aq+r\) with \(r=0\) or \(d(r) < d(a)\); then \(r = b-aq \in I\), so minimality forces \(r=0\), and \(I = \gens a\).
PID \(\implies\) UFD. A PID is Noetherian, so a nonzero nonunit that is not a product of irreducibles would give a strictly ascending chain \(\gens a \subsetneq \gens{a_1}\subsetneq\cdots\) of principal ideals; hence factorizations exist. In a PID every irreducible element \(p\) generates a maximal ideal, so \(p\) is prime, and two factorizations into primes agree up to order and units by cancelling one prime at a time.
Strictness of the inclusions
(Each inclusion is strict)
Euclidean domain, not a field. \(k[x]\) for a field \(k\): the division algorithm makes it Euclidean, and \(x\) is not invertible.
PID, not a Euclidean domain. \(\ZZ\left[\frac{1 + \sqrt{-19}}{2}\right]\).
UFD, not a PID. \(\ZZ[x]\). Since \(\ZZ\) is a UFD, so is \(\ZZ[x]\). The ideal \(\gens{2,x} = \ts{\sum r_ix^i \st r_0 \in 2\ZZ}\) is proper and not principal: a constant generator \(\pm2\) generates only polynomials with even coefficients and misses \(x\), and a generator of degree at least one misses \(2\).
Integral domain, not a UFD. \(\ZZ[\sqrt{-5}]\), where \((2+\sqrt{-5})(2-\sqrt{-5}) = 9 = 3\cdot 3\) and all four factors are irreducible and pairwise nonassociate, by the multiplicativity of the norm \(N(a+b\sqrt{-5})=a^2+5b^2\).
Commutative ring, not an integral domain. \(\ZZ/4\), where \([2]^2 = [0]\).
$R[x]$ is a PID if and only if $R$ is a field (Tag E-XH2QU)
Show that \(R[x]\) a PID \(\iff R\) is a field.
Hint
Take \(r\in R\), then \(\gens{r, x} = \gens{f}\) for some \(f\).
Solution
Let \(R\) be a commutative ring with identity.
<1>1. If \(R\) is a field, then \(R[x]\) is a PID.
Division with remainder in \(R[x]\) shows that a nonzero ideal is generated by any of its nonzero elements of least degree.
<1>2. If \(R[x]\) is a PID, then every \(0\neq r\in R\) is a unit.
\(R[x]\) is a domain, hence so is \(R\subseteq R[x]\). Write \(\gens{r,x}=\gens{f}\), so \(r=fp\) and \(x=fq\) for some \(p,q\in R[x]\). In the domain \(R[x]\), degrees add, so \(\deg f=0\) from \(r=fp\); write \(f=c\in R\). Then \(x=cq\) forces \(q=ax+b\) with \(ca=1\), so \(c\) is a unit and \(\gens{r,x}=R[x]\). Hence \(1=rg(x)+xh(x)\) for some \(g,h\in R[x]\), and evaluating at \(x=0\) gives \(1=rg(0)\).
<1>3. Q.E.D.
Steps <1>1 and <1>2, together with \(1\neq0\) in the domain \(R\).
Quotients
Maximal, prime, and radical ideals via their quotient rings (Tag PR-76BDN)
Let \(R\) be a commutative ring, and let \(\mfm\), \(\mfp\), and \(J\) be ideals of \(R\).
\(R/\mfm\) is a field if and only if \(\mfm\) is a maximal ideal.
\(R/\mfp\) is an integral domain if and only if \(\mfp\) is a prime ideal.
(Maximal and prime ideals through quotients)
For an ideal \(I\) of a commutative ring \(R\), \[ I \text{ maximal} \iff R/I \text{ is a field}, \qquad I \text{ prime} \iff R/I \text{ is an integral domain}. \] Every field is an integral domain, so every maximal ideal is prime, and \(\mspec R \subseteq \spec R\).
For a field \(k\) and a nonconstant \(f\in k[x]\), the following are equivalent: \(\gens f\) is maximal in \(k[x]\); \(k[x]/\gens f\) is a field; \(f\) is irreducible.
Transporting properties
Implications among fields, PIDs, UFDs, and Noetherian rings (Tag PR-GEHJF)
Let \(R\) be a commutative ring with \(1\).
If \(R\) is a commutative division ring, then \(R\) is a field.
If \(R\) is a finite integral domain, then \(R\) is a field.
\(R\) is a field if and only if \(R[x]\) is a principal ideal domain.
\(R\) is a field if and only if \(R\) is a simple ring with \(1 \neq 0\).
\(R\) is a unique factorization domain if and only if \(R[x]\) is a unique factorization domain.
If \(R\) is a principal ideal domain, then \(R[x]\) is a unique factorization domain.
If \(R\) is a principal ideal domain, then \(R\) is Noetherian.
If \(\mfm\subseteq R\) is a maximal ideal and \(x\in R\sm\mfm\), then \(\mfm + Rx = R\), so \(1 = m + rx\) for some \(m\in\mfm\) and \(r\in R\).
Gorenstein rings
Gorenstein ring (Tag D-DU4UQ)
A Noetherian local ring \(R\) is Gorenstein if \(R\), as a module over itself, has finite injective dimension. A commutative Noetherian ring \(R\) is Gorenstein if its localization \(R_{\mathfrak p}\) is a Gorenstein local ring for every prime ideal \(\mathfrak p\subseteq R\) [@Eis95].
A commutative Noetherian ring of finite injective dimension over itself is Gorenstein; conversely, a Gorenstein ring of finite Krull dimension has finite injective dimension over itself [@Eis95].
(Finite-dimensional graded Gorenstein algebras)
Let \(k\) be a field and \(R = R_0 \oplus R_1 \oplus \cdots \oplus R_n\) a graded commutative \(k\)-algebra with \(R_0 = k\), \(\dim_k R < \infty\), and \(R_n\neq0\). Then \(R\) is Gorenstein if and only if \(R\) satisfies Poincaré duality: \(\dim_k R_n = 1\), and for each \(i\) the multiplication pairing \(R_i \times R_{n-i} \to R_n\) is perfect.