# Determine the generator of an ideal of ring of integers

I am trying to find the generators of the ideal $$(3)$$ in the ring of integers of $$\mathbb{Q}[\sqrt{-83}]$$ the ring of integers is $$\mathbb{Z}\left[\frac{1+\sqrt{-83}}{2}\right]$$ I evaluated the Minkowski constant and it is $$2/\pi \sqrt{83} \sim 5.8;$$ then I checked the minimal polynomial of $$x^2-x+84/2,$$ which is reducible module $$3,$$ hence the ideal $$(3)$$ is generated by two elements, right? Did I miss something? I want to say the ring of integers is not a UFD.

## 2 Answers

You seem to have touched upon several different ideas here.

Generators of the ideal $$(3)$$. Usually, when you talk about the generators of $$(3)$$, you mean its generators as an abelian group.

Define $$\theta := \tfrac 1 2 (1 + \sqrt{-83})$$. We know that the ring of integers $$\mathbb Z[\theta]$$ is generated by $$\{1, \theta \}$$ as an abelian group, so $$(3)$$ is generated by $$\{3, 3\theta \}$$ as an abelian group.

But having read the remainder of your question, it looks like this is not what you're after! What you're really interested in are the generators of the ideal class group for $$\mathbb Z[\theta]$$. (In fact, to prove that $$\mathbb Z [\theta]$$ is not a UFD, we don't even need to explicitly identify a set of generators for the ideal class group - it's enough to show that the ideal class group is non-trivial.)

Minkowski constant. The fact that $$\frac 4 \pi \times \frac {2!} {2^2} \sqrt{83} \approx. 5.8$$ implies that the ideal class group is generated by prime ideals that are factors of $$(2)$$, $$(3)$$ or $$(5)$$.

Dedekind's criterion. Dedekind's criterion is a way of factorising $$(3)$$ as a product of primes.

As you pointed out, we have the factorisation $$x^2 - x + \frac {84}{2} \equiv x(x-1) \mod 3,$$

so Dedekind's criterion says that $$(3) = (3, \theta)(3, \theta - 1)$$ is the prime factorisation of $$(3)$$ in $$\mathbb Z[\theta]$$.

Whether $$\mathbb Z[\theta]$$ is a UFD. This is equivalent to asking whether the ideal class group is trivial, i.e. whether all ideals are principal.

Why don't we check whether $$(3, \theta)$$ is principal? The answer must be "no". Note that $$(3, \theta)$$ has norm $$3$$. If it was principal, then there would exist $$x, y \in \mathbb Z$$ such that $$(3, \theta) = (x + y\theta)$$. But then $$3 = N(3, \theta) = N(x + y\theta) = (x + \tfrac 1 2 y)^2 + 83(\tfrac 1 2 y)^2,$$ which is impossible.

So $$\mathbb Z[\theta]$$ is not a principal ideal domain, and hence, is not a unique factorisation domain.

• Thanks that helpful, that what I was looking for – Ameryr Apr 22 at 19:57

The minimal polynomial of $$\alpha=\frac{1+\sqrt{-83}}{2}$$ is $$f = x^2 - x + 84/4 = x^2 - x + 21$$. Modulo $$3$$ this factors as $$f \equiv x^2 - x = x(x-1) \pmod 3,$$ so by the Kummer-Dedekind theorem the ideal $$(3)$$ factors into primes in $$\mathbb{Z}\bigg[\frac{1+\sqrt{-83}}{2}\bigg]$$ as $$(3) = (3, \alpha)(3, \alpha-1)$$.

The ideal $$(3)$$ is principal, generated by $$3$$. You can show that the prime factors are not principal, e.g. using the field norm:

If $$(3,\alpha) = (\beta)$$ then $$N(\beta)$$ divides both $$N(3) = 3^2$$ and $$N(\alpha) = f(0) = 21 = 3\cdot 7$$, so $$N(\beta) = 3$$.

If we write $$\beta = a+b\alpha$$ then $$N(\beta)=(a+b\alpha)(a+b(1-\alpha)) = \ldots = a^2+ab + 21b^2.$$

The equation $$a^2+ab + 21b^2 = 3$$ is an ellipse in the $$(a,b)$$-plane without integral points, so there is no such $$\beta$$.

• How could you determine that the ellipse has no integral points? – Ameryr Apr 22 at 20:05
• I plotted it (with a grid in the background). It's a very small ellipse, so there are very few candidates for integral points, and you can see that it misses them all. – Ricardo Buring Apr 22 at 20:11