*(This is part of my series on algebraic number theory.)*

Since number theory is ultimately concerned with problems in the ring of integers $\mathbf{Z}$, we try to find rings which share as many properties with $\mathbf{Z}$ as possible. In the previous post we have seen commutative rings with unity, which share two key properties of $\mathbf{Z}$. In this post we present two additional properties of $\mathbf{Z}$ which are shared by the rings we will be ultimately interested in. Of course, the title of this post gives a hint of what we are after. ;) In this post and all the subsequent ones, when we speak of a ring, we mean a commutative ring with unity.

**3.1 Integral domains**

A familiar property of $\mathbf{Z}$ that we have used very often is that the product of two integers can only be $0$ if at least one of them is $0$. A ring wich shares this property is called an *integral domain* (or just a *domain* if there is no risk of ambiguity), thus if there are two elements $a$ and $b$ of an integral domain $R$ such that $ab = 0$, it must be that $a=0$ or $b=0$.

*Examples:*

- $\mathbf{Z}$ is an integral domain.
- Any field $F$ is an integral domain. We have often used that fact, perhaps without realizing it. Its proof is a good exercise for the interested reader.
- $R[x]$ inherits from $R$ the property of being an integral domain. In particular, $F[x]$ is an integral domain.
- The ring $\mathbf{Z}_n$ is not an integral domain if $n$ is not prime, because if $a$ and $b = n/a$ are two non-trivial divisors of $n$, then working in $\mathbf{Z}_n$ we have $a \ne 0$ and $b \ne 0$, but $ab = 0$.

Two properties which are satisfied by a field but are actually a consequence of its integral domain structure, and so are satisfied by any integral domain $R$, are the properties that the units of $R[x]$ are precisely the units of $R$, and the cancellation law for multiplication. Note that the cancellation law holds for multiplication in an integral domain even if the cancelled term has no multiplicative inverse (but it still must be non-zero). Indeed, if $ab = ac$ with $a \ne 0$ then we get $ab-ac = a(b-c) = 0$. Then since we are in an integral domain it must be that either $a$ or $b-c$ is zero. Since $a \ne 0$ we see that $b-c = 0$, and so $b=c$.

**3.2 Irreducible elements and unique factorisation**

Another familiar property of $\mathbf{Z}$ is that it contains some elements called *primes*, which have the following three properties (among others):

- Zero and the units ($\pm 1$) are not primes.
- If $p$ is a prime and $a$, $b$ are two integers such that $p=ab$, then $a$ or $b$ is a unit (
*i.e.*, is equal to $\pm 1$). For example $5$ is prime and can be written only in two ways as a product of two integers: $1 \times 5$ or $(-1) \times (-5)$. In both cases, one of the factors is a unit (they are $1$ and $-1$, respectively). - Every element which is not zero nor a unit can be expressed as a finite product of prime factors, and this factorisation is unique up to order and unit factors. For example $15$ can be written as $3\times 5$ or $(-3) \times (-5) = (-1) \times 3 \times (-1) \times 5$ among other ways. In the second factorisation we can remove the unit factors $-1$, and we obtain the first one.

In an integral domain, we define *irreducible elements* as in the first two properties above: an element $a$ of an integral domain which is neither $0$ nor a unit is said to be irreducible if whenever $a=bc$, $b$ or $c$ is a unit. Thus irreducible elements in an integral domain are a generalisation of prime numbers in $\mathbf{Z}$.

*Examples:*

- The irreducible elements of $\mathbf{Z}$ are precisely the prime numbers.
- From the definition of an irreducible polynomial in $F[x]$ and the definition of a unit in $F[x]$, we see that the irreducible elements of $F[x]$ are precisely the irreducible polynomials.
- All the elements of a field are either zero or units, and so a field has no irreducible elements.

An integral domain is said to be a *unique factorisation domain* (UFD for short) if it satisfies the third property above. That is, if every element which is neither zero nor a unit can be expressed as a finite product of irreducible elements, and if all such factorisations are identical up to order and unit factors.

- $\mathbf{Z}$ is a UFD. This is the so-called fundamental theorem of arithmetic.
- A field is a UFD by default, since it has no elements which are neither zero nor a unit.
- $R[x]$ inherits from $R$ the property of being a UFD. In particular, $F[x]$ is a UFD.

**Next:** Cyclotomic fields