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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?

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The proof is incorrect if the ring is not a domain because $b=bcd$ does not imply $1=cd$, even if you assume $b\neq 0$. –  Georges Elencwajg Oct 3 '11 at 9:49
If I get you Georges, primes are irreducible only in a domain. –  Hassan Muhammad Oct 3 '11 at 9:52
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. –  Georges Elencwajg Oct 3 '11 at 10:48

2 Answers 2

up vote 11 down vote accepted

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

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Interesting papers mentioned. Thanks –  Bhaskar Vashishth Aug 4 at 10:05

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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This deserves more upvotes. –  goblin Jan 6 at 11:18

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