TITLE: What are the equivalents of primes in $\Bbb C$?
QUESTION [5 upvotes]: Note: I asked a similar question earlier about integers, and I was advised to create a new one for the question of the complex plane
The conventional concept of primes really has value only for the natural numbers, as given $\Bbb Z$, any prime $p$ can be broken down into $(1,p)(1,p)$ and $(−1,−p)(−1,−p)$. And when one includes irrationals, the concept breaks down even further.
My question is, are there numbers in $\Bbb C$ that can only be divided by 1 and themselves? Or is the concept not extendable beyond ℕ?

REPLY [4 votes]: The concept can be extended in a variety of ways, but not to $\mathbb{C}$ in an interesting way. (*)
You are trying to describe irreducibles, those numbers which cannot be factored (except by using units like $1,-1$).  There are many rings (like $\mathbb{Z}$) and semigroups (like $\mathbb{N}$) where this is an interesting concept.
Here is my favorite, easy, example: $2\mathbb{N}=\{2,4,6,8,\ldots\}$, the set of even natural numbers.  $6$ is irreducible, because the only way to factor it is $2\cdot 3$, which uses an "illegal" number, i.e. one not in the semigroup.  Hence $6$ is irreducible.  By similar logic, $10$ is irreducible.  It turns out that $30$ is also irreducible, because $30=2\cdot 3\cdot 5$, and there's only one $2$ available, so no matter how you factor $30$, only one of the two factors will get a $2$.
Hence, in this semigroup, you get two different factorizations of $60$ into irreducibles: $$6\cdot 10=2\cdot 30$$
You may have noticed that I didn't call these "primes", rather irreducibles.  The reason is that the two terms are used to mean slightly different things: a prime $p$ is a number that satisfies $$ \text{ if } p|xy\text{ then } p|x\text{ or } p|y$$
It turns out that every prime is always irreducible, but not every irreducible is prime.  In the example $2\mathbb{N}$ above, $2$ divides $2\cdot 30$, but $2$ does not divide either $6$ or $10$ (because if $2|6$, then $6=2\cdot 3$ would be a factorization in $2\mathbb{N}$, which it isn't).  Hence $2$ is irreducible but not prime in this semigroup.
(*)   The reason is that in $\mathbb{C}$, every nonzero number is a unit, so there are no irreducibles.