As we’ve seen in All those primes, prime numbers can be very useful in both in theory and in applications. There is a fascination with the numbers, as they are generating set of the natural numbers, that is, we can determine the structure of the natural numbers using the primes as a building block. While the idea of prime comes up frequently, even in early mathematics, the idea of irreducible is much commonly mentioned. However, the fact that prime numbers are irreducible is the reason why we can use them as building blocks for the natural numbers. The reason irreducible are rarely mentioned, is that these two words coincide for the natural numbers. Why stick with prime then? I’m not sure, probably just because it sounds more fun. However, there is a reason to look at the difference between the two definitions. Seeing both provides more information about these numbers and increases the utility of knowing them.

**Definitions**

In All those primes I defined an irreducible number as a number that satisfies the property, if \(ab=p\), then \(a\) or \(b\) must be \(1\). However, we will need to work with a slightly different definition if we are not working over the natural numbers.

For example, let’s work with the integers. In the integers, we run into the situation \(2=(-1)(-2)\). According to the above definition, we would then get that \(2\) is reducible. However, so would every integer since everything can be written as \(-1\) multiplied by its negative. Therefore, we provide the following generalizations:

- \(u\) is a unit if \(u\) divides every element of the set.
- \(q\) is irreducible if whenever \(q=ab\), then \(a\) or \(b\) is a unit.
- \(p\) is prime if whenever \(p\) divides \(ab\), then \(p\) divides \(a\) or \(p\) divides \(b\).

In the natural numbers, 1, is the only unit, so the definition becomes equivalent to what we have above. However, in the integers, -1 is also a unit. Because of this we can write \(2=-2*-1\), so while \(2\) does not satisfy the first definition, both \(2\) and \(-2\) are considered irreducible over the integers because if they are written as a product, one of the terms must be a unit.

Here, we see that what is irreducible may change, and that the definitions of prime and irreducible look different. However, are there any cases were primes and irreducibles don’t coincide? If not, then separating the definition would be pointless. Therefore, before working specifically with natural numbers, let look at an example of an irreducible that is not prime.

**An irreducible that is not prime**

Suppose that we have the set \(A=\left\{ a+b\sqrt{5}: a,b \in \mathbb{Z}\right\}\). That is, the set of all numbers that can be written an integer plus an integer multiplied by the square root of 5. This is closed under the usual multiplication, so we can look at irreducibles and primes according to our definitions. Note that in order to check if something is irreducible, we have to show that if \((a+b\sqrt{5})(c+d\sqrt{5})=q\), then one of these terms must be a unit. In this way, we have to check if \(ac+5ef+(ad+bc)\sqrt{5}=q\) has a solution. While the following steps aren’t obvious, we can indeed show that

- \(2\) is irreducible,
- \(2\) does not divide \(1 \pm \sqrt{5}\),
- \((1-\sqrt{5})(1+\sqrt{5})=1-5=-4=(-2)(2)\),
- Hence \(2\) is not prime.

Therefore, it is indeed possible to have an irreducible that is not prime. Because we can come up with such an example, there is a reason to look at these elements as potentially different and use different definitions.

**Primes are Irreducible**

While we just saw an example of an irreducible that is not prime, it is not possible to have a prime that is not irreducible. To see this, we have

- Let \(p\) be prime,
- suppose that \(p=ab\),
- this implies \(p\) divides \(ab\),
- then \(p\) divides \(a\) or \(p\) divides \(b\),
- if \(p\) divides \(a\), then \(b\) is a unit,
- otherwise \(a\) is a unit,
- hence \(p\) is irreducible.

What we’ve just shown is that a number being prime is a more restrictive definition than a number being irreducible. That is, the primes are more special than just irreducibles. This is why it is so nice to know that in the natural numbers, these two coincide.

**The natural numbers**

Suppose that we are working in the natural numbers. Then let \(p\) be irreducible, and suppose that \(p\) divides \(a*b\). Then, if \(p\) divides \(a\), we have that \(p\) satisfies the definition of prime. If not, then the greatest common divisor of \(p\) and \(a\) must be one because the only divisors of \(p\) are itself and 1. Therefore, from the work we did in The Division Algorithm, there exist integers k and l such that

\[\begin{align*} kp+la=1 \text{ so } \\ bkp+bla=b \end{align*}\]

Since \(p\) divides \(p\) it divides \(bkp\) and since \(p\) divides \(ab\), \(p\) divides \(bla\). Hence \(p\) divides both terms on the left, so it divides the sum, which is b. Therefore, \(p\) is prime.

**Conclusion**

While it may seem that I am looking overly precise with definitions, the fact that if \(p\) divides \(a*b\) then \(p\) divides \(a\) or \(p\) divides \(b\) is extremely important. In fact, this is precisely the reason that the fundamental theorem of arithmetic holds. While I will save talking about this theorem for another time, I want to stress the importance of the theorem we have just proven. Furthermore, I would also like to point out that the most difficult part of the theorem was defining what I meant. Once we had the definitions, that every irreducible natural numbers is prime fell out directly from the work we did in the previous post The Division Algorithm. If you need to, just reread the section, **The natural numbers**, and you’ll see that our proof fell out in two sentences. This is indeed what I find so wonderful about this proof. There is a lot of build up while providing definitions and making distinctions, but, once you get done with that, the proof is… simple.

Thank you for reading, and I hope you are continuing to enjoy the proofs we are looking at. If you are, let me know by liking the page or sharing the post on Social Media. If you haven’t been enjoying it, or you’d prefer a different topic, you can also let me know by commenting below. Feedback is always appreciated.

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