Rings and Modules HT24, IDs, PIDs and EDs hierarchy

Course - Rings and Modules HT24^U

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, Unique 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, Unique 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)