Irreducibles elements in $\mathbf Z[\sqrt{-3}]$

Question

The ring $A:=\mathbb Z[\sqrt{-3}]$ is the prototype of the rings usually used in a first algebraic number theory course to show the difference between prime and irreducible elements.

I was wondering if I could list all prime and irreducible elements of this ring, instead of just giving examples showing they are different sets.

Since every element $x$ in $A$ divides a non zero integer (its norm $N(x):=x \cdot \overline{x}$), one can restrict the search to divisors of integers.

For prime elements the situation is easy: almost by definition, we find all primes elements in $A$ as divisors of the primes in $\mathbb Z$. On sees that a primes $p\in \mathbb Z$ remains prime in $A$ iff $-3$ is not a square mod $p$ (this can be made more explicit using quadratic reciprocity...). If $-3$ is a square, we get two prime factors.

For irreducible elements, can we make the list (as explicit as possible) of all such elements? Of course there are all elements whose norm is not a (non trivial) product of elements of the form $a^2+3b^2$. It's not clear to me that these are the only ones. Can we make it explicit?

Answer 1

You're referring to all numbers of the form $a + b \sqrt{-3}$, where $a, b \in \mathbb{Z}$ (or $\mathbf{Z}$ if you prefer that notation)? I suppose because you haven't explicitly said so no one has pointed out that this is not an integer closure (not exactly the right term, but I think you know what I mean).

Notice for example that $$\omega^3 = \left(-\frac{1}{2} + \frac{\sqrt{-3}}{2}\right)^3 = 1,$$ which means that this $\omega$ number is an algebraic number of at most cubic degree. But since $\omega^2 + \omega + 1 = 0$, it is in fact an algebraic integer of quadratic degree.

Then I think that what you're looking for is for a number that is prime in $\mathbb{Z}$, can't be expressed as a product of numbers of the form $a + b \sqrt{-3}$ but can be expressed as a product of numbers of the form $a + b \omega$.

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EDIT: I think I got completely confused on that last paragraph yesterday. It seems to me now that there exist no numbers of the sort I described, and the reason should be perfectly obvious but it eludes me. I would withdraw my answer if it weren't for the reference in Mr. Soupe's answer.

Answer 2

Mr. Brooks was on the right track for the first two paragraphs of his answer. What he needed to do next, after clarifying the part about this subdomain not being integrally closed, was to clarify the difference between irreducibles that are also prime and irreducibles that are not prime.

If whenever $p = ab$ either $a$ or $b$ is a unit, then $p$ is irreducible. But if also whenever $p \mid ab$ either $p \mid a$ or $p \mid b$ holds true (maybe both), then $p$ is also prime.

So we see that in $\mathbb{Z}[\sqrt{-3}]$ (a subdomain of $\mathbb{Z}[\omega]$) the number 2 is irreducible. It is also irreducible in $\mathbb{Z}[\omega]$, as are all primes $p$ of $\mathbb{Z}^+$ satisfying $p \equiv 2 \pmod 3$.

And yet $4 = 2^2 = (1 - \sqrt{-3})(1 + \sqrt{-3})$, but $$\frac{1 - \sqrt{-3}}{2} \not\in \mathbb{Z}[\sqrt{-3}]$$ and likewise $$\frac{1 + \sqrt{-3}}{2} \not\in \mathbb{Z}[\sqrt{-3}].$$

But both of these numbers are in $\mathbb{Z}[\omega]$, one is $-\omega$, the other is $-\omega^2$. And since $N(\omega) = 1$ (as Mr. Brooks already showed), then $(1 - \sqrt{-3})(1 + \sqrt{-3})$ is an incomplete factorization of 4 in $\mathbb{Z}[\omega]$.

So what I think you're actually looking for is numbers of the form $a^2 + 3b^2$ that are composite in $\mathbb{Z}$ and divisible by some positive purely real integer $p \equiv 2 \pmod 3$.

My hunch is that only 2 fits this bill, since, for example, $5^2 + 3 \times 1^2 = 28 = 2^2 \times 7$, but $7 \equiv 1 \pmod 3$ and $(2 - \sqrt{-3})(2 + \sqrt{-3}) = 7$. I will continue to ponder this as I wait for further clarification from you.