# 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.

A slick way to do this goes via the norm $$N(2)$$ of the com­plex num­ber $$2$$; namely $$4$$.

If $$2 = ab$$ then $$N(2) = N(a)N(b)$$ be­cause 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 $$\pm 1$$). <div><div>

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 $$R$$ is a UFD if and only if it has “all ir­re­ducibles are prime (in the sense of ring the­ory)” and “ev­ery $$r \in R$$ 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 $$\mathbb{Z}$$ 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 $$\mathbb{Z}$$.

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