# The relation between irreducible and primes

Am I right in thinking that the conventional definition of a prime integer (can only be written as itself times $$1$$ and has no other factors) is actually the definition for irreducible?

Is it true that this is given because prime and irreducible are equivalent in a ring and it the above definition is so much easier than explaining that a number $$p$$ is prime if and only if “$$p$$ divides $$ab$$” implies either “$$p$$ divides $$a$$” or “$$p$$ divides $$b$$”.

Why are these equivalent?

Are there any simple examples of numbers that are either prime or irreducible but not both?

• In a ring, we have irreducible elements and prime elements. Every prime element is irreducible, but an irreducible element need not be prime. In the ring $\mathbb Z$ however, an element is a prime element if and only if it is irreducible. – Peter Oct 7 '18 at 10:47

The set of integers is Unique Factorization Domain, that is, every element can be uniquely factorized. In UFD, a non-unit element is prime if and only if it is irreducible. So in $$\mathbb{Z}$$, there is no counterexample.

In $$\mathbb{Z}[\sqrt{5}]$$, which is a set of elements of the form $$a+b\sqrt{-5}$$ where $$a$$ and $$b$$ are integers, $$9$$ can be written in two forms, $$9=3^2 = (2-\sqrt{-5})(2+\sqrt{-5})$$ and $$3$$ divides $$(2-\sqrt{-5})(2+\sqrt{-5})$$ but does not divide neither $$2-\sqrt{-5}$$ nor $$2+\sqrt{-5}$$. So $$3$$ is not a prime here. You can also see that $$\mathbb{Z}[\sqrt{-5}]$$ is not UFD. However, it is an irreducible element. This can be shown by solving the following equation $$3=(a+b\sqrt{-5})(c+d\sqrt{-5})$$ for integers $$a,b,c$$ and $$d$$. Long and uninteresting calculation will show that $$b=d=0$$ and either one of $$a$$ and $$c$$ is $$1$$, and the other one is $$3$$.

Conversely, in integral domain $$R$$, being prime always implies that it is an irreducible element. Suppose a prime $$p$$ is reducible, so there are two non-unit elements $$a,b$$ such that $$p=ab$$. Since $$p$$ divides $$ab=p$$, by the definition of the primality, $$p$$ must divide $$a$$ or $$b$$. We may assume that $$p$$ divides $$a$$, and there is $$c\in R$$ such that $$pc=a$$. Thus $$p=ab=pcb \implies cb=1$$ so it contradicts that $$b$$ is non-unit. The cancellation works because we are working in the integral domain.

You're pretty much right: In a general commutative ring $$R$$, the property \begin{align*} p = ab \Rightarrow a \in R^* \vee b \in R^* \end{align*}

is called irreducibility. Primality on the other hand is the property

\begin{align*} p\mid ab \Rightarrow p\mid a \vee p \mid b. \end{align*}

If $$R$$ is a UFD (e.g. $$R = \mathbb{Z}$$), an element has one of these two properties iff it has the other one, so the two notions coincide. Therefore a prime element in $$\mathbb{Z}$$ can be defined as a number that is irreducible.

BUT: Primality and irreducibility are not, in general, the same: While primes are always irreducible in integral domains, the number $$3 \in \mathbb{Z} \hspace{-0.1cm} \left[ \sqrt{-5} \right]$$ is an example of an irreducible element which isn't prime.

The definitions are equivalent in a unique factorization domain like $$\mathbb Z$$. In a non-UFD, there is still a set of prime numbers and a set of irreducible numbers, but they don't overlap completely.

For example, $$-37$$ is not divisible by any number in $$\mathbb Z$$ closer to 0 than itself, other than $$-1$$ and 1. Hence it is irreducible in $$\mathbb Z$$. But it is also prime, because in any product $$ab$$ that is divisible by $$-37$$ we will find that either $$a$$ or $$b$$ is a multiple of $$-37$$, or maybe they both are.

Compare to $$74$$. That's obviously not prime, and it's not irreducible either, it's reducible. We see that $$74 \mid 148$$, but if we express 148 as $$-4 \times -37$$, we see that 74 divides neither the $$a$$ nor the $$b$$.

Now let's take a look at an integral domain like $$\mathbb Z[\sqrt{-70}]$$. In that domain, $$-37$$ is still irreducible, because it's not divisible by any number closer to 0 other than $$-1$$ and 1.

But now we see that $$-37 \mid (2 - \sqrt{-70})(2 + \sqrt{-70})$$, yet neither $$2 - \sqrt{-70}$$ nor $$2 + \sqrt{-70}$$ is divisible by $$-37$$. Numbers like 3 and 13 are still irreducible and prime in $$\mathbb Z[\sqrt{-70}]$$, though.

As for a number that is prime but not irreducible, consider $$\mathbb Z_6$$, which consists only of 0, 1, 2, 3, 4, 5 (addition and multiplication "wrap around" to keep things within the ring). Verify that $$3 = 3^2 = 3^3 = 3^4 = \ldots = 3 \times 5 = 3 \times 5^2 = \ldots$$ but also that the only way to get 3 as a product in this ring is to include 3 at least once as a multiplicand.

P.S. I chose $$\mathbb Z[\sqrt{-70}]$$ rather than $$\mathbb Z[\sqrt{78}]$$ because even though we have to deal with complex numbers, things are in some ways much simpler.

• Beware Your last example is meaningless without any definition of irreducible in rings with zero-divisors. There are various incompatible definitions in use, and what you wrote is false with some common definitions, e.g. see the paper linked here, where (Corollary 2.7) idempotents are irreducible $\iff$ prime (note $3 = 3^2,$ i.e. $3$ is idempotent in $\Bbb Z_6)$ I'd recommend adding a remark so as not to mislead readers. – Bill Dubuque Oct 8 '18 at 1:25
• Ah, good to have you back, @Bill, I missed you. – Robert Soupe Oct 8 '18 at 1:38

In a ring in general the definitions are not the same; being prime is rather the stronger property, this is literally true when there are no zero-divisors. (Note that the definition of irreducible needs to be modified a bit to account for invertible elements other than $$1$$.)

Consider for example the ring of polynomials without linear term, that is polynomials of the form:

$$a_0 + a_2X^2 + a_3X^3 + \dots + a_nX^n$$

In this structure $$X^2$$ is irreducible, yet not prime since $$X^2$$ divides $$X^3 \times X^3$$ yet not $$X^3$$.

However, the definitions are equivalent in a principal ideal domain or more generally in a unique factorization domain, or in even more general structures.

Since the integers are such a domain, the definitions are equivalent in that context.

You can also construct examples in the integers by restricting to sub-semigroups. For example if you consider only the positive integers whose decimal digit expansion end in $$1$$, then this is a multiplicative subsemigroup with identity.

In that structure $$21$$ is irreducible, yet not prime: $$21$$ divides $$81 \times 2401$$, the numbers on the right being $$3^4$$ and $$7^4$$ respectively.

In a domain, every prime is irreducible. A standard example that the converse does not always hold is $$3 \cdot 3 = 9 = (2+\sqrt{-5})(2-\sqrt{-5})$$ in the ring $$\mathbb Z[\sqrt{-5}]$$. Then $$3$$ and $$2\pm\sqrt{-5}$$ are irreducible but not prime.