Irreducible element (ring theory)

Ring the­ory can be viewed as the art of tak­ing the in­te­gers $$\mathbb{Z}$$ 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 $$R$$ which is an in­te­gral do­main, we say that an el­e­ment $$x \in R$$ is ir­re­ducible if, when­ever we write $$r = a \times b$$, it is the case that (at least) one of $$a$$ or $$b$$ is a unit (that is, has a mul­ti­plica­tive in­verse).

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Re­la­tion­ship with pri­mal­ity in the ring-the­o­retic sense

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

If $$p$$ is prime, then $$p \mid ab$$ im­plies $$p \mid a$$ or $$p \mid b$$ by defi­ni­tion. We wish to show that if $$p=ab$$ then one of $$a$$ or $$b$$ is in­vert­ible.

Sup­pose $$p = ab$$. Then in par­tic­u­lar $$p \mid ab$$ so $$p \mid a$$ or $$p \mid b$$. As­sume with­out loss of gen­er­al­ity that $$p \mid a$$; so there is some $$c$$ such that $$a = cp$$.

There­fore $$p = ab = cpb$$; we are work­ing in a com­mu­ta­tive ring, so $$p(1-bc) = 0$$. Since the ring is an in­te­gral do­main and $$p$$ is prime (so is nonzero), we must have $$1-bc = 0$$ and hence $$bc = 1$$. That is, $$b$$ is in­vert­ible. <div><div>

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

• In $$\mathbb{Z}$$, it is a fact that ir­re­ducibles are prime. In­deed, it is a con­se­quence of Bé­zout’s the­o­rem that if $$p$$ is “prime” in the usual $$\mathbb{Z}$$-sense (that is, ir­re­ducible in the rings sense) noteI’m sorry about the no­ta­tion. It’s just what we’re stuck with. It is very con­fus­ing., then $$p$$ is “prime” in the rings sense. (Proof.)

• In the ring $$\mathbb{Z}[\sqrt{-3}]$$ of com­plex num­bers of the form $$a+b \sqrt{-3}$$ where $$a, b$$ are in­te­gers, the num­ber $$2$$ is ir­re­ducible but we may ex­press $$4 = 2 \times 2 = (1+\sqrt{-3})(1-\sqrt{-3})$$. That is, we have $$2 \mid (1+\sqrt{-3})(1-\sqrt{-3})$$ but $$2$$ doesn’t di­vide ei­ther of those fac­tors. Hence $$2$$ is not prime.

%%hid­den(Proof that $$2$$ is ir­re­ducible in $$\mathbb{Z}[\sqrt{-3}]): A slick way to do this goes via the [norm_complex_number norm]$$N(2)$$of the complex number$$2$$; namely$$4$$. If$$2 = ab$$then$$N(2) = N(a)N(b)$$because [norm_of_complex_number_is_multiplicative the norm is a multiplicative function], and so$$N(a) N(b) = 4$$. But$$N(x + y \sqrt{-3}) = x^2 + 3 y^2$$is an integer for any element of the ring, and so we have just two distinct options:$$N(a) = 1, N(b) = 4$$or$$N(a) = 2 = N(b)$$. (The other cases follow by interchanging$$a$$and$$b$$.) The first case:$$N(x+y \sqrt{3}) = 1$$is only possible if$$x= \pm 1, y = \pm 0$$. Hence the first case arises only from$$a=\pm1, b=\pm2$$; this has not led to any new factorisation of$$2$$. The second case:$$N(x+y \sqrt{3}) = 2$$is never possible at all, since if$$y \not = 0$$then the norm is too big, while if$$y = 0$$then we are reduced to finding$$x \in \mathbb{Z}$$such that$$x^2 = 2$$. Hence if we write$$2$$as a product, then one of the factors must be invertible (indeed, must be$$\pm 1$$). %% In fact, in a [-principal_ideal_domain], "prime" and "irreducible" are equivalent. ([5mf Proof.]) # Relationship with unique factorisation domains It is a fact that an integral domain$$R$$is a [unique_factorisation_domain UFD] if and only if it has "all irreducibles are [5m2 prime] (in the sense of ring theory)" and "every$$r \in R$$may be written as a product of irreducibles". ([alternative_condition_for_ufd Proof.]) This is a slightly easier condition to check than our original definition of a UFD, which instead of "all irreducibles are prime" had "products are unique up to reordering and multiplying by invertible elements". Therefore the relationship between irreducibles and primes is at the heart of the nature of a unique factorisation domain. Since$$\mathbb{Z}$$is a UFD, in some sense the [fundamental_theorem_of_arithmetic Fundamental Theorem of Arithmetic] holds precisely because of the fact that "prime" is the same as "irreducible" in$$\mathbb{Z}\$.

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