I have seen many proofs about a prime element is irreducible, but up to now I am thinking whether this result is true for any ring.

Recently, I got this proof:

Suppose that $a$ is prime, and that $a = bc$. Then certainly $a\mid bc$, so by definition of prime, $a\mid b$ or $a\mid c$, say $a \mid b$. If $b = ad$ then $b = bcd$, so $cd = 1$ and therefore $c$ is a unit. (Note that $b$ cannot be $0$,for if so, $a = bc = 0$, which is not possible since $a$ is prime.) Similarly, if $a\mid c$ with $c = ad$ then $c = bcd$, so $bd = 1$ and $b$ is a unit. Therefore $a$ is irreducible.

I think with the above proof we do not need the ring to be an integral domain. If this is the case then I will stop doubting, else, I am still in it.

Can somebody help me out?

  • 19
    $\begingroup$ The proof is incorrect if the ring is not a domain because $b=bcd$ does not imply $1=cd$, even if you assume $b\neq 0$. $\endgroup$ – Georges Elencwajg Oct 3 '11 at 9:49
  • $\begingroup$ If I get you Georges, primes are irreducible only in a domain. $\endgroup$ – Hassan Muhammad Oct 3 '11 at 9:52
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    $\begingroup$ It is worse than that, Hassan: I don't even think that the notion "irreducible" is a good one in rings with zero-divisors. On the other hand the notion of prime element, with the definition you used in your post, seems to me quite reasonable, even in a ring which isn't a domain. Of course this is somewhat subjective, but I have the feeling that I'm expressing a rather widely held consensus among algebraists. $\endgroup$ – Georges Elencwajg Oct 3 '11 at 10:48
  • $\begingroup$ Notes, In ring $\mathbb R \times \mathbb R$, we have $(b,0) = (b,0)(1,c)(1,d)$, obviously $(1,c)(1,d) \ne \mathbf 1$ $\endgroup$ – Zang MingJie Mar 7 '19 at 8:11
  • $\begingroup$ $b=bcd \Rightarrow b(cd-1) = 0 \Rightarrow b=0 \text{ or } cd=1$ Here requires $R$ is a domain $\endgroup$ – Zang MingJie Mar 7 '19 at 8:35

Notice that your proof assumes that $\rm\: b\ne 0\ \Rightarrow\ b\:$ is cancellable, so it fails if $\rm\:b\:$ is a zero-divisor. Factorization theory is more complicated in non-domains. Basic notions such as associate and irreducible bifurcate into a few inequivalent notions. See for example

When are Associates Unit Multiples?
D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles.
Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828.

Factorization in Commutative Rings with Zero-divisors.
D.D. Anderson, Silvia Valdes-Leon.
Rocky Mountain J. Math. Volume 28, Number 2 (1996), 439-480

  • $\begingroup$ Interesting papers mentioned. Thanks $\endgroup$ – Bhaskar Vashishth Aug 4 '15 at 10:05
  • $\begingroup$ If i could i would give + 100 :) $\endgroup$ – mick Aug 20 '16 at 20:20
  • $\begingroup$ When I try to download the PDFs from those links, I get 'Invalid or corrupted PDF file' , in both cases. $\endgroup$ – evaristegd May 20 '19 at 0:54
  • $\begingroup$ @evaristegd Both work fine for me. $\endgroup$ – Bill Dubuque May 20 '19 at 2:26
  • $\begingroup$ You're right. It was probably a problem related to the fact that I was using Brave Browser. $\endgroup$ – evaristegd May 20 '19 at 3:51

If you choose the definition of $a$ is irreducible if $a=bc$ implies that $(a)=(b)$ or $(a)=(c)$ then it is true actually. For instance the proof is as follows:

Let $p\in R$ be a non-zero, non-unit. Suppose $p=bc$. We clearly have $b\mid p$ and $c \mid p$ since $b$ and $c$ are factors of $p$. On the other hand, $1\cdot p=bc$ implies that $p \mid bc$ and $p$ being prime implies $p \mid b$ or $p \mid c$. Thus $(p)=(b)$ or $(p)=(c)$ showing $p$ is irreducible by this definition.

Unfortunately, prime is quite different than any of the other possible definitions of irreducible when zero-divisors are present that I am familiar with. The definition above is the weakest choice for irreducible that I am aware of.


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