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Prime numbers in $\mathbb{Z}$ are numbers where its only factors are 1 and itself.

Let R be a commutative ring with identity.

Definition of a prime element for commutative rings with identity: A nonzero nonunit p $\in$ R is called a prime element if whenever p $\vert$ ab in R, either p $\vert$ a or p $\vert$ b.

Does the same statement hold then, that an element r of R is prime if its only factors are 1 and itself? It seems to be true as if p = 7, 7 $\vert$ 1 or 7 $\vert$ 7.

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  • $\begingroup$ Note that $-1$ and $-7$ are factors of $p=7$ in $\mathbb{Z}$ to. You should consider factors up to multiplication by a unit (and the units in $\mathbb{Z}$ are $1$ and $-1$). In general, rings can have some pretty complicated unit structure. $\endgroup$ Nov 21 '19 at 8:12
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Recall the following definition:

Let $R$ be a (unital) ring. An non-zero and non-unit element $x\in R$ is called irreducible if it is not the product of two non-units.

In an integral domain (no zero divisors) $R$, every prime element is automatically irreducible. However, in general irreducible elements need not be prime.

In many rings (such as unique factorization domains) irreducible elements coincide with prime elements.

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  • $\begingroup$ I understand that Primeness and Irreducibility are linked, but my question is for a commutative ring with identity R, are the prime elements where its only factors are 1 and itself? $\endgroup$
    – yagayeet
    Nov 21 '19 at 8:12
  • $\begingroup$ The answer to that is automatically no, even in $\mathbb{Z}$. For example $7$ is divisible by $-1$ as well. You should relax that condition and you'll end up with the notion of an irreducible element. Then the answer is still no, there are examples of non-prime irreducible elements. For example, $2\in \mathbb{Z}[\sqrt{-5}]$. Note that this ring is unital and commutative, so this would then answer your question. $\endgroup$ Nov 21 '19 at 8:17

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