How do the definitions of "irreducible" and "prime" elements differ?

In my commutative algebra lecture notes it says:

A non-zero element $p$ of a ring $R$ which is not a unit of $R$ is called a prime element if $p=ab$ implies $a$ is a unit or $b$ is a unit.

Is this not the definition of an irreducible element?? Everywhere else I've read that a non-zero, non-unit element $p$ is a prime element if $p|ab $ implies $p|a$ or $p|b$

Thanks :)

Solution 1:

An element in an integral domain is irreducible if it is neither a unit nor a product of non-units.

An element $p$ of a commutative ring $A$ (not necessarily an integral domain) is prime if $A / p A$ is an integral domain, that is, is non-zero (that is $p$ is not a unit) and has no zero-divisor, that is, whenever $ab \in pA$ then $a$ or $b$ is in $p A$ which is the same as what you wrote.)

Now, in an integral domain, if an element is prime, it is irreducible. Indeed : consider $p$ a prime that is reducible : $p=ab$. Then $p \mid ab$ implies that $p \mid a$ or $p \mid b$. Say $p \mid a$, then $a = pc$, then we have: $p=ab=pcb$ which implies that $p(1-cb)=0$. Because $A$ is an integral domain we have: $cb=1$. So $b$ is a unit and $p$ is irreducible.

The implication "irreducible implies prime" is false in general, for instance in the ring $A = \mathbf{Z}[\sqrt{-5}]$ the irreducible (exercise !) element $3$ is not prime as it divides the element $9 = (2 - \sqrt{-5}) (2 + \sqrt{-5})$ but does not divide either of the two elements $2 \pm \sqrt{-5}$.

The implication "irreducible implies prime" is true in integral domains in which any two non-zero elements have a greatest common divisor. This is for instance the case of unique factorization domains. In the previous expression the decomposition $3 \times 3 = (2 - \sqrt{-5}) (2 + \sqrt{-5})$ is indeed an example of an element having two factorizations, so that $A = \mathbf{Z}[\sqrt{-5}]$ is not a unique factorization domain.