# Abstract Algebra/The hierarchy of rings
## Commutative rings

Definition 11.1:
A ring $R$ with multiplication $\cdot$ is called commutative if and only if $a\cdot b=b\cdot a$ for all $a,b\in R$.
Examples 11.2:
* The whole numbers $\mathbb {Z}$ are commutative.
* The matrix ring $\mathbb {R} ^{n\times n}$ of $n$-by-$n$ real matrices with matrix multiplication and component-wise addition is not commutative for $n\geq 2$.
In commutative rings, a left ideal is a right ideal and thus a two-sided ideal, and a right ideal also.

## Integral domains

Definition 11.3:
An integral domain is defined to be a commutative ring (that is, we assume commutativity by definition) such that whenever $ab=0$ ($a,b\in R$), then $a=0$ or $b=0$.
We can characterize integral domains in another way, and this involves the so-called zero-divisors.
Definition 11.4:
Let $R$
Thus, a ring is an integral domain iff it has no zero divisors.

## Unique factorisation domains

Theorem 11.?:
Suppose that $R$ is a commutative ring

## Principal ideal domains

Due to its importance in algebra, we'll briefly give the definition of noetherian rings, which is a fairly exhaustive class of rings for which many useful properties hold. The theory of noetherian rings is well-studied, powerful and extensive, and we'll only study it in detail in the wikibook on Commutative Algebra. The reason that we give the definition here is that principal ideal domains are noetherian rings, which will imply that they are, in fact, unique factorisation domains.
Definition 11.?:
Let $R$ be a commutative ring. $R$ is called noetherian iff for every sequence of ideals $(I_{n})_{n\in \mathbb {N} }$ of $R$ such that
    $I_{1}\subseteq I_{2}\subseteq I_{3}\subseteq \cdots \subseteq I_{k}\subseteq \cdots$
there exists an $N\in \mathbb {N}$ such that $I_{N}=I_{N+1}=I_{N+2}=\cdots$.
This condition can be interpreted to state that every ascending chain of ideals stabilizes. Noetherian rings are named in honour of Emmy Noether.
Theorem 11.?:
Every PID is Noetherian.
Proof:
We observed earlier that the set of all ideals of a ring is inductive, with an explicit description of. If therefore we are given an ascending chain of ideals
Theorem 11.?:
Every PID is a UFD.
Proof:
Let $R$ be a PID, and let $a\in R$.

## Euclidean domains

Example 11.? (Gaussian integers):
We have already seen that $\mathbb {Z}$ is a Euclidean domain. Now consider the ring
    $\mathbb {Z} [i]:=\{a+ib|a,b\in \mathbb {Z} \}\subseteq \mathbb {C}$
with addition and multiplication induced by that of $\mathbb {C}$. We'll see in the exercises that this is indeed a commutative ring with identity. Furthermore, on it we define a Euclidean function as thus:
    $N(a+ib):=a^{2}+b^{2}$
This is indeed a Euclidean function, the units of $\mathbb {Z} [i]$ are $\{+1,-1,+i,-i\}$ and furthermore we may precisely describe the prime elements of $\mathbb {Z} [i]$ and set them in relation to the prime elements of $\mathbb {Z}$:
* If $p\in \mathbb {Z}$ is a prime in $\mathbb {Z}$, then either it is already a prime in $\mathbb {Z} [i]$, or there is a prime $\pi \in \mathbb {Z} [i]$ in the Gaussian primes such that $p=\pi {\overline {\pi }}$.
* If $\pi \in \mathbb {Z} [i]$ is a Gaussian prime, then set $n:=\pi {\overline {\pi }}$. Either we have that $n$ is a prime in $\mathbb {Z}$, or $n=p^{2}$, where $p$ is a prime in $\mathbb {Z}$.
* In 1., if $p\neq 2$, the former case happens if and only if $p\equiv 1\mod 4$ and the latter if and only if $p\equiv 3\mod 4$.
Proof:
First, the proof of multiplicativity of $N$ is relegated to the exercises, that is, you'll show in the exercises that
    $N((a+ib)(c+id))=N(a+ib)N(c+id)$.
Then we have to prove that division with remainder holds. Let thus $\sigma :=a+ib$ and $\tau :=c+id$ be elements of $\mathbb {Z} [i]$.
Due to $1=(-1)^{2}=i^{4}=(-i)^{4}$, $1,-1,i,-i$ are units. Any other unit would have to have the form $a+ib$, where $|a|+|b|\geq 2$. Let $c+id$ be its inverse. Then $1=N((a+ib)(c+id))=N(a+ib)N(c+id)=(a^{2}+b^{2})(c^{2}+d^{2})\geq 2$, a contradiction.
Finally, let's prove the statements about the relation of the Gaussian primes to the integer primes.
* Since $\mathbb {Z} [i]$ is a Euclidean domain, we have a decomposition of $p$ into prime elements of $\mathbb {Z} [i]$, say $p=u\pi _{1}\cdots \pi _{n}$, where $u\in \{+1,-1,+i,-i\}$ is a unit in $\mathbb {Z} [i]$. If $n=1$, we are done. If $n\geq 2$, observe that $p^{2}=N(p)=N(\pi _{1})\cdots N(\pi _{n})$, and since $p$ is prime, uniqueness of prime factorisation in the integers implies that at most two of $N(\pi _{1}),\ldots ,N(\pi _{n})$ are not one and those that are are either $p$ or $p^{2}$. If one is $p^{2}$, there is exactly one prime factor of $p^{2}$ in $\mathbb {Z} [i]$, which is absurd since $p^{2}$ is obviously not irreducible. If $$,

## Exercises

* Prove that the Gaussian integers as defined above do form a commutative ring with identity. Use your knowledge on complex numbers (cf. the corresponding chapter in the wikibook on complex analysis).