Notes  Rings and Modules HT24, Overview of results and relationships in rings

Integral domains:
 No zero divisors (definition)
 $R$ is finite implies $R$ is a field, [[Notes  Rings and Modules HT24, Fields]]^{U}
 $\langle a \rangle = \langle b \rangle \iff$ $a$ and $b$ are associates, [[Notes  Rings and Modules HT24, Ideals]]^{U}
 $\text{char}(R) = 0$ or a prime, [[Notes  Linear Algebra MT23, Integral domains]]^{U}
 Have the cancellation property, if $ac = bc$ and $c \ne 0$, then $a = b$, [[Notes  Linear Algebra MT23, Integral domains]]^{U}
 If $R$ is an integral domain, then $R[x]$ is also an integral domain (proof: a zero divisor in $R[x]$ would imply $a _ n b _ m = 0$ where $a _ n$, $b _ m$ are leading coefficients of polynomials)

Every UFD is an integral domain (by definition), and for UFDs:
 Every nonzero nonunit can be factored into a product of irreducibles unique up to reordering and multiplication by units (definition)
 $R$ is a UFD $\iff$ all irreducibles are prime and every nonzero nonunit is a product of irreducibles $\iff$ every nonzero nonunit is a product of primes, [[Notes  Rings and Modules HT24, Factorisation]]^{U} (this lemma basically says that if we can factor all elements into primes, then we don’t have to worry about uniqueness).
 If $d\ne 0$ then $\langle d \rangle$ is prime ideal $\iff$ $d$ is irreducible $\begin{array}{c} \centernot\implies \ \impliedby \end{array}$ $\langle d \rangle$ is a maximal ideal, [[Notes  Rings and Modules HT24, Prime and maximal ideals]]^{U}
 If $R$ is a UFD, then $R[x]$ is a UFD, [[Notes  Rings and Modules HT24, Factorisation in polynomial rings]]^{U} (it’s only proved that $\mathbb Z[x]$ is a UFD, but the proof actually works in general by considering the field of fractions for $R$)
 Not every UFD is a PID (e.g. $\mathbb Z[x]$ is a UFD, but not a PID), but some UFDs are PIDs

Every PID is an integral domain (by definition), and for PIDs:
 Every ideal is principal (definition)
 $R$ is Noetherian, i.e. every ascending chain of ideals stabilises, [[Notes  Rings and Modules HT24, Principal ideal domains]]^{U}
 If $d \ne 0$ then $\langle d \rangle$ is prime ideal $\iff$ $\langle d \rangle$ is maximal $\iff$ $d$ is irreducible, [[Notes  Rings and Modules HT24, Prime and maximal ideals]]^{U}
 $R$ is a UFD, [[Notes  Rings and Modules HT24, Factorisation]]^{U}
 $R[x]$ is not necessarily a PID (e.g. $\mathbb Z$ is a PID, but $\mathbb Z[x]$ is not)
 The $\text{hcf}$, $\text{lcm}$ always exists, [[Notes  Rings and Modules HT24, Divisibility]]^{U}
 Bezout lemma applies: if $a$ and $b$ have $\text{hcf}$ $c$, then $\exists s, t$ such that $as + bt = c$, [[Notes  Rings and Modules HT24, Divisibility]]^{U}
 Matrices over $R$ can be put in Smith Normal Form, [[Notes  Rings and Modules HT24, Smith normal form]]^{U} (the proof for this assumes that $R$ is actually a Euclidean domain, but this is not necessary)
 Every finitely generated module over $R$ is a direct sum of cyclic modules ( [[Notes  Rings and Modules HT24, Structure theorems]]^{U})

Every Euclidean domain is a PID:
 $R$ can be equipped with a Euclidean function (definition)
 There is a division algorithm, [[Notes  Rings and Modules HT24, Euclidean domains]]^{U}
 If $R$ is a Euclidean domain, it doesn’t follow $R[x]$ is also a Euclidean domain (e.g. $\mathbb Z$ is a Euclidean domain, but $\mathbb Z[x]$ is not even a PID)

Every field is a PID, and for fields: (quick proof: there are two ideals in any field, and these are both generated by a single element, namely $0$ and $1$).
 Every nonzero element has a multiplicative inverse (definition)
 Every ideal is either $\{0\}$ or $R$ (a ring has two ideals iff it is a field)
 $R[x]$ is a Euclidean domain (use the degree as the norm function)