Irreducible and Prime Integers

In the set of integers, prime numbers and irreducible numbers are one and the same.

However, this is not a general rule.

In algebraic theory, there are algebraic structures (such as rings) where prime elements do not necessarily coincide with irreducible ones.

Note. The notions of "prime" and "irreducible" are not generally equivalent. While every prime element is necessarily irreducible, the reverse implication does not always hold. Irreducibility is a more general property, and in some algebraic structures, an irreducible element may fail to be prime. However, in unique factorization domains, the two concepts coincide.

Irreducible Integers
Prime Integers
Irreducible and Prime Integers

Irreducible Integers

Definition

An integer $ a \in \mathbb{Z} $ is called irreducible if, whenever it can be expressed as a product $ bc $, where $ b $ and $ c $ are integers, one of the two factors must be a unit. $$ \forall a=bc \rightarrow b=u \lor c=u $$

By definition, neither zero nor units are considered irreducible.

Example 1

Consider the number 7. We can express it as a product of two integer factors:

$$ 7 = 7 \cdot 1 $$

$$ 7 = (-7) \cdot (-1) $$

There is no other way to write 7 as a product of two integers.

In both cases, one of the factors is a unit.

Therefore, 7 is an irreducible integer.

Example 2

The number 6 is not irreducible.

$$ 6 = 3 \cdot 2 $$

$$ 6 = (-3) \cdot (-2) $$

$$ 6 = 6 \cdot 1 $$

$$ 6 = (-6) \cdot (-1) $$

Since 6 can be factored into two non-unit integers, it is not irreducible.

Prime Integers

Definition

An integer $ a $ is prime if, whenever it divides a product $ bc $, where $ b $ and $ c $ are integers, it must divide at least one of the two factors. $$ \forall a \mid bc \rightarrow a \mid b \lor a \mid c $$

Example

Consider the integer $ a = 7 $ in $ \mathbb{Z} $.

The number 7 divides any product $ bc $ in which it appears as a factor.

$$ 7 | 7 \cdot 1 \\ 7 | 7 \cdot 2 \\ 7 | 14 \cdot 2 \\ 7 | (k \cdot 7) \cdot j \\ \vdots $$

Every time 7 is a divisor of a product $ bc $, it must also divide at least one of the two factors.

There are no exceptions.

Irreducible and Prime Integers

In the set of integers, every prime number is also irreducible.

Proof

Let $ a, b, c $ be three integers. The number $ a $ is irreducible if:

$$ a=bc \quad \text{where either } b=1 \text{ or } c=1 $$

By assumption, $ a $ is also prime:

$$ a \mid bc \quad \Rightarrow \quad a \mid b \lor a \mid c $$

So there exists an integer $ k $ (or $ j $) such that:

$$ a \mid b \Rightarrow b = ak \\ a \mid c \Rightarrow c = aj $$

If $ a \mid b $, then $ k $ must be equal to $ c $, which is defined as 1:

$$ \begin{cases} a=bc \quad \text{where } b=1 \text{ or } c=1 \\ b=ak \end{cases} $$

Similarly, if $ a \mid c $, then $ j $ must be equal to $ b $, which is also 1:

$$ \begin{cases} a=bc \quad \text{where } b=1 \text{ or } c=1 \\ c=aj \end{cases} $$

Thus, in the set of integers, every prime number is also irreducible.

Note. As mentioned earlier, this equivalence does not hold in general. It applies specifically to integers. For this reason, it is important to study the definitions of irreducible and prime elements separately. More broadly, the term "elements" is used rather than "numbers."

Every irreducible element in $ \mathbb{Z} $ is also a prime number.

Proof

Let $ p $ be an irreducible element in $ \mathbb{Z} $.

If $ p $ divides $ ab $ and also divides either $ a $ or $ b $, then it is prime.

$$ p \mid ab \rightarrow p \mid a \lor p \mid b $$

If $ p $ divides $ ab $, there exists an integer $ h $ such that:

$$ p \mid ab \rightarrow ab = ph $$

Now, assume that $ p $ does not divide $ a $.

Then, the greatest common divisor (GCD) of $ p $ and $ a $ is 1:

$$ \gcd(a, p) = 1 $$

By Bézout's identity, the GCD can be written as a linear combination of two integers $ j $ and $ k $:

$$ \gcd(a, p) = ja + kp $$

Since $ \gcd(a, p) = 1 $, there exist integers $ j $ and $ k $ such that:

$$ ja + kp = 1 $$

Multiplying both sides by $ b $, we get:

$$ b(ja + kp) = 1 \cdot b $$

$$ jab + kpb = b $$

Since $ p $ is a divisor of both $ ab $ and $ pb $, it must also divide $ b $.

Thus, we have proven that if $ p \mid ab $, then $ p \mid b $, completing the proof.