Primes and Irreducibles in Rings

In school, you were likely taught that a prime number is a number with precisely two positive factors (itself and $1$). You were likely also taught a useful property of primes: if $p \mid ab$, that is if a prime $p$ is a factor of some product of two numbers, then it is a factor of one of those numbers. If $a \times b = 8645$, then either $a$ or $b$ is a multiple of $5$. This is generally not true for composite numbers: $4 \mid 60 = 6 \times 10$, even though $4$ is a factor of neither $6$ nor $10$.

Unfortunately, this definition is backwards! The latter property is in fact the true essence of a prime number—the former actually describes an irreducible number. These coincide among the integers we are familiar with, but they are subtly different. Let’s explore these two definitions and see where they come apart.

An element $r$ in an integral domain (a ring with no nontrivial zero divisors: $a \cdot b = 0_R$ implies $a = 0_R$ or $b = 0_R$) $R$ is called irreducible if $r \neq 0_R$, $r$ is not a unit (that is, $\forall a \in R, a \cdot r \neq 1_R$), and if $r = xy$ then either $x$ or $y$ is a unit.

Similarly, an element $r$ is called prime if $r \neq 0_R$, $r$ is not a unit, and if $r \mid xy$ then either $r \mid x$ or $r \mid y$. In a ring, an element $d$ divides an element $a$, written $d \mid a$, if there exists some element $r \in R$ such that $d \cdot r = a$.

In the ring of integers $\mathbb{Z}$, the primes are in fact exactly the irreducibles. This is not generally true, as we will see in the main example. However, we do have the following result.

Primes Are Always Irreducible

1. Suppose $r \in R$ is a prime, and $r = a \cdot b$ for two non-unit elements $a$ and $b$.
2. Since $r = 1_R \cdot r$, we have $r \mid r = a \cdot b$. Then, as $r$ is prime, $r \mid a$ or $r \mid b$.
3. Without loss of generality (we’re only considering commutative rings here), suppose $r \mid a$, ie. $r \cdot d = a$ for some $d \in R$.
4. Then, $r = a \cdot b = (r \cdot d) \cdot b = r \cdot (d \cdot b)$.
5. Then, as we are in an integral domain, $d \cdot b = 1_R$, which means $b$ is a unit.
6. Thus $r$ is irreducible.

But Not Vice Versa

Now, we look at the ring $R = \mathbb{Z}[\sqrt{-5}] = \{a + b \sqrt{-5} : a, b \in \mathbb{Z}\} \leq \mathbb{C}$. As this is a subring of a field (a ring in which every nonzero element is a unit, like $\mathbb{C}$), it is an integral domain.

Consider the norm function given by

$$N: R \to \mathbb{N}_0, a + b \sqrt{-5} \mapsto a^2 + 5b^2.$$

Note that this maps $z \in \mathbb{C}$ to $|z|^2$, and satisfies $N(z \cdot w) = N(z) \cdot N(w)$.

This is a really useful property! If $r \cdot s = 1$, then

$$N(r \cdot s) = N(r) \cdot N(s) = N(1) = 1 \implies N(r) = N(s) = 1.$$

So any unit $a + b \sqrt{-5}$ has to satisfy $N(a + b \sqrt{-5}) = a^2 + 5b^2 = 1 \implies a = \pm 1, b = 0$.

Thus the only possible units are $1$ and $-1$, which are obviously both units (as $1 \cdot 1 = -1 \cdot -1 = 1$), so they are precisely the two units of $R$.

Note that there are clearly no elements of norm $2$ or $3$: any such element $r$ would be nonzero and require

$$5b^2 \leq N(r) < 5 \implies b < 1 \implies b = 0, \quad \text{and} \quad N(r) < 4 \implies |a| < 2 \implies a = \pm 1$$

but then these are just the units again, so there are no such elements.

With this in mind, we claim that $2 \in R$ is irreducible. Suppose $2 = a \cdot b$. Then

$$N(a) \cdot N(b) = N(2) = 4 \implies N(a) = 1, \; N(b) = 4 \quad \text{ (in some order).}$$

Thus $a$ or $b$ has norm $1$ and so is a unit, hence $2$ is irreducible as required.

In the same way, we can demonstrate that the elements $3, 1 + \sqrt{-5},$ and $1 - \sqrt{-5}$ are irreducible too.

Importantly, $2$ does not divide either of $1 \pm \sqrt{-5}$. Suppose $2 \cdot r = 1 \pm \sqrt{-5}$. Then we would have $4 \cdot N(r) = 6 \implies N(r) = 1.5$, which is not possible, as the norm function $N$ maps to $\mathbb{N}_0$.

Now, note that

$$(1 + \sqrt{-5}) \cdot (1 - \sqrt{-5}) = 6 = 2 \cdot 3$$

So $2 \mid 6 = (1 + \sqrt{-5}) \cdot (1 - \sqrt{-5})$, but neither $2 \mid (1 + \sqrt{-5})$ nor $2 \mid (1 - \sqrt{-5})$. Thus by definition, $2$ cannot be prime in $R$.

So in general, irreducible elements in integral domains are not necessarily prime.

(Also, factorisation of ring elements into irreducibles is not necessarily unique. The rings in which this property holds are called unique factorisation domains.)