# Prime elements in a commutative ring with identity

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.

• 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. Nov 21 '19 at 8:12

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.
• 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. Nov 21 '19 at 8:17