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Suppose $p$ is a prime element in a commutative ring. Does this imply its only divisors are $1,p$? By definition $a\mid p\implies \exists b:ab=p\implies p\mid a\text{ or }p\mid b$.

If $p\mid a$ then we have $a\mid p\text{ and }p\mid a$, but this only gives $ab=p,pc=a$ and I don't see how to show $p=a$ (maybe it's not even true).

If $p\mid b$ then also $a\mid b$ but I don't see where to go from here.

  • Did you try in $\mathbb{Z}$? – Zhen Lin Nov 19 '15 at 09:08
  • @ZhenLin in $\mathbb Z$ I know prime decompositions are unique, but I'm trying to prove they are unique in general. –  Nov 19 '15 at 09:09
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    To follow @Zhen Lin's comment, does $-1$ divide $p$? Also (to make sure that we are on the same page) what is the definition of "prime"you are using? – Clément Guérin Nov 19 '15 at 09:10
  • @Exterior Do you understand how uniqueness follows from existence of prime decompositions for $\mathbb Z,\mathbb C[x]$? The idea is if you have distinct decompositions, you respectively look at them $\mod p$ and evaluate at each root. "Looking at them $\mod p$" and "evaluating at each root" is nothing more than going to the quotient by the prime ideal generated respectively by $p$ and your root. The essential fact is that for primes, this quotient is always a domain, hence an equation $0=\prod_i a_i$ for nonzero $a_i$ would be a contradiction. – Arrow Nov 19 '15 at 15:21
  • @Exterior the contents of my comment are worked our in more details in this blog post by Qiaochu Yuan. – Arrow Nov 19 '15 at 15:21

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No, it doesn't imply that. As a simple example, consider that $5$ is prime in $Z$, and among its divisors is $-5$ which is neither $1$ nor $5$. It's easy to "forget" about the negative numbers when you're thinking of divisibility, but they're still there.

In more general terms, to understand divisibility in a commutative ring you need to be aware of what units are. A unit is an element $u$, which has an inverse, so there exists $v$ such that $uv=1$. Units divide anything: if $u$ is a unit and $a$ is any element, than $u|a$, because $u*u^{-1}*a = a$.

In $Z$, the only units are $1$ and $-1$, but in other commutative rings, there are often infinitely many units. So if $p$ is a prime element, you cannot hope to prove that its only divisors are $1$ and $p$; at the very least all units are.

In a sense, units "don't count", because if you multiply anything by a unit, you can always get back by multiplying again by its inverse. This corresponds to the intuition that in $Z$, $p$ and $-p$ are "the same" as far as divisibility is concerned. If $a,b$ are such that $b=a*u$ where $u$ is a unit, we say that they are associates or, informally, that they're the same "up to associates".

Suppose $p$ is prime and I want to claim that $p$ only has divisors $1,p$ "up to associates": what this means is that if $a$ is any divisor, then $a$ is either a unit or "the same as p". So in $Z$ for example, I would be claiming that $a$ is either a unit ($1$ or $-1$), or an associate of $p$ ($p$ itself or $-p$). This claim is true in $Z$ when $p$ is prime. But it's still not true in commutative rings in general. It is only true in integral domains.

AnatolyVorobey
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The answer is yes for integral domains (of course, up to associates), but can be negative in general. In this example $x$ is prime, $xy\mid x$, and $xy$ is not associate to $1$ or $x$.

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