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).

Why do we re­quire \(R\) to be an in­te­gral do­main?

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Examples

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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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