Irreducible element (ring theory)

Written by Patrick Stevens, et al. last updated

Ring theory can be viewed as the art of taking the integers and extracting or identifying its essential properties, seeing where they lead. In that light, we might ask what the abstracted notion of "prime" should be. Confusingly, we call this property irreducibility rather than "primality"; "prime" in ring theory corresponds to something closely related but not the same.

In a ring which is an integral domain, we say that an element is irreducible if, whenever we write , it is the case that (at least) one of or is a unit (that is, has a multiplicative inverse).

Why do we require to be an integral domain?

Examples

Relationship with primality in the ring-theoretic sense

It is always the case that primes are irreducible in any integral domain.

Show proof

If is prime, then implies or by definition. We wish to show that if then one of or is invertible.

Suppose . Then in particular so or . Assume without loss of generality that ; so there is some such that .

Therefore ; we are working in a commutative ring, so . Since the ring is an integral domain and is prime (so is nonzero), we must have and hence . That is, is invertible.

However, the converse does not hold (though it may in certain rings).

  • In , it is a fact that irreducibles are prime. Indeed, it is a consequence of Bézout's theorem that if is "prime" in the usual -sense (that is, irreducible in the rings sense) [1], then is "prime" in the rings sense. (Proof.)
  • In the ring of complex numbers of the form where are integers, the number is irreducible but we may express . That is, we have but doesn't divide either of those factors. Hence is not prime.

Proof that is irreducible in $\mathbb{Z}[\sqrt{-3}]

A slick way to do this goes via the norm of the complex number ; namely .

If then because the norm is a multiplicative function, and so . But is an integer for any element of the ring, and so we have just two distinct options: or . (The other cases follow by interchanging and .)

The first case: is only possible if . Hence the first case arises only from ; this has not led to any new factorisation of .

The second case: is never possible at all, since if then the norm is too big, while if then we are reduced to finding such that .

Hence if we write as a product, then one of the factors must be invertible (indeed, must be ).

In fact, in a principal_ideal_domain, "prime" and "irreducible" are equivalent. (Proof.)

Relationship with unique factorisation domains

It is a fact that an integral domain is a UFD if and only if it has "all irreducibles are prime (in the sense of ring theory)" and "every may be written as a product of irreducibles". (Proof.) This is a slightly easier condition to check than our original definition of a UFD, which instead of "all irreducibles are prime" had "products are unique up to reordering and multiplying by invertible elements".

Therefore the relationship between irreducibles and primes is at the heart of the nature of a unique factorisation domain. Since is a UFD, in some sense the Fundamental Theorem of Arithmetic holds precisely because of the fact that "prime" is the same as "irreducible" in .

  1. ^︎

    I'm sorry about the notation. It's just what we're stuck with. It is very confusing.

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