Abstract Algebra/The hierarchy of rings

Commutative rings

Examples 11.2:

• The whole numbers $\mathbb{Z}$ are commutative.
• The matrix ring ${ℝ}^{n\times n}$ of $n$-by-$n$ real matrices with matrix multiplication and component-wise addition is not commutative for $n\ge 2$.

In commutative rings, a left ideal is a right ideal and thus a two-sided ideal, and a right ideal also.

Integral domains

We can characterize integral domains in another way, and this involves the so-called zero-divisors.

Thus, a ring is an integral domain iff it has no zero divisors.

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.

This condition can be interpreted to state that every ascending chain of ideals stabilizes. Noetherian rings are named in honour of Emmy Noether.

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

Proof:

Let $R$ be a PID, and let $a\in R$.

Euclidean domains

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 $|a|+|b|\ge 2$. Let $c+id$ be its inverse. Then $1=N((a+ib)(c+id))=N(a+ib)N(c+id)=(a^2+b^2)(c^2+d^2)\ge 2$, a contradiction.

Finally, let's prove the statements about the relation of the Gaussian primes to the integer primes.

1. 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 $n\ge 2$, 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 $N({\pi }_1),\dots ,N({\pi }_n)$ 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 ${\displaystyle }$,

Exercises

1. 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).