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I have a ring $R$ which is known to be a Dedekind domain, but not necessarily a Euclidian domain, and a nonzero ideal generated by one or two elements in this ring. How can I check if this ideal is a prime ideal?

For ideals generated by one element, this is equivalent to being a prime element in the ring. I programmed a clone of the Erathosthenes sieve for my ring, but this only shows me which elements are irreducible, which does not necessarily imply primality.

To be more concrete, my ring $R$ is the ring of integers in an imaginary quadratic number field $\mathbb{Q}(d)$ (for $d$ a negative squarefree integer).

These rings are of the form $R = \mathbb{Z}[\sqrt{-d}]$ or $R = \mathbb{Z}[\frac{1+\sqrt{-d}}{2}]$, depending on $d$. These are Dedekind domains, and every ideal there is generated by either one or two elements.

For some $d$ (like $-1$, $-2$, $-3$), these are Euclidean domains, where I can use the division with remainder to find the greatest common divisor of two numbers, meaning that each ideal is generated by one element.

For others (like $d = -7$), there is no division with remainder, and we can't use the Euclidean algorithm.

But even here it is not yet clear for me: how do I distinguish irreducible from prime elements? The most famous example: for $d = -5$ we have $R = \mathbb{Z}[\sqrt{-5}]$, and the element $2$ is not prime here, since $2 \cdot 3 = 6 = (1+\sqrt{-5})\cdot(1-\sqrt{-5})$ and 2 is not a factor of either term on the right side.

By testing multiples, I can find numbers with different decompositions, and then know that those factors can't be prime. But when can I stop searching, if I don't find any?

In this particular ring, the first elements not yet shown as either units, irreducible-not-prime or composite (after checking all elements until about $\pm 50 \pm 50i$) are $\pm \sqrt{-5}$, $\pm 4 \pm 2\sqrt{-5}$, $\pm 6 \pm \sqrt{-5}$. Could I be sure here that there really are prime?

Another example, $d = -7$, $R = \mathbb{Z}[\frac{1+\sqrt{-7}}{2}] = \mathbb{Z}[X]/(X^2+X+2)$: Here I find lots of irreducible-non-prime elements, and the only small "irreducible and not yet shown as non-prime" are $\pm \frac{1-\sqrt{-7}}{2}$. (The principal ideal generated by these elements look quite same as the one created by $\frac{1+\sqrt{-7}}{2}$, though - I think there is simply a glitch in my program.)

This ring is non-Euclidean, and I suppose most (if not all but the zero ideal) prime ideals are generated by two elements here. If I have a candidate pair of elements, how do I find out if it is really a nonzero prime (= maximal) ideal?

Any ideas here would helpful, since I'm a bit stuck here. (Most literature about computational ideal theory I found only works in polynomial rings over fields (and makes heavy use of this fact), thus it does not really help here.)

(I'm writing a program which should then be able to work on any of these rings $R$ (and any ideals there), thus facts that are only valid in a small number of those rings are less helpful. But feel free to mention them nevertheless, maybe some of them can be generalized.)

KCd
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Paŭlo Ebermann
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2 Answers2

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To check if an ideal is prime, one way is to look at the quotient $R / I$; the ideal $I$ is prime if and only if $R / I$ is an integral domain.

To expand a bit on that: say you have two elements $(x, y)$, and you want to know if the ideal $I$ they generate is prime. This ideal $I$ certainly contains the elements $xx^*$ and $y y^*$ (where * is the Galois "conjugation" automorphism of R). These are in $\mathbb{Z}$, so they have a highest common factor; let's say that is $k$.

Then $kR$ is an ideal of $R$ contained in $I$. Moreover, the ring $R / k R$ is finite, and you can now describe $R / I$ by a finite computation, because it's a quotient of $R / kR$. In particular, you can check whether or not $R / I$ is a field.

David Loeffler
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  • Looks good. I'll have to see how this fits in my program. – Paŭlo Ebermann Jun 20 '11 at 17:14
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    This is false for the zero ideal. The correct general statement is that $I$ is a prime ideal if and only if $R/I$ is an integral domain, and $I$ is a maximal ideal if and only if $R/I$ is a field. It just so happens that for number rings, every nonzero ideal $I$ has the property that $R/I$ is finite, hence is an integral domain if and only if it is a field, so every nonzero prime ideal is maximal. – Qiaochu Yuan Jun 20 '11 at 22:35
  • Why would $ kR$ be an ideal of $R$ contained in $I$? It is the smallest ideal containing the ideals $(x)$ and $(y)$ right? – Conjecture Nov 03 '19 at 03:04
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$\newcommand{\Z}{\mathbb{Z}}$

This works for principal ideals pretty well, and it was how I was shown to compute the class group. Let me see if I remember this right...

One way is to translate the problem. You are looking to see if the ideal $I$ is prime in an imaginary quadratic ring. For $d \equiv 1,2$ modulo $4$, this is $\frac{\Z[x]}{x^2+d}$; for $d \equiv 3$ modulo $4$, it's $\frac{\Z[x]}{x^2 + x + \frac{d+1}{4}}$.

The ideal $I$ is prime if $R/I$ is an integral domain (no zero divisors). The trick then is to interchange the quotients of the rings. So for example, in $d=5$, say you're checking to see if $I=(2)$ is prime. Then $R=\frac{\Z[x]}{x^2+5}$. Now by interchanging the quotients this is equivalent to seeing whether $(x^2+5)=(x^2+1)$ is prime in $\frac{\Z[x]}{2}$. In this case it is not, since $(x^2+1) = (x+1)^2$. If you translate this back, since $x$ "represents" $\sqrt{-5}$, you get that the ideal $(2) = (1+\sqrt{-5})^2$.

Michael Chen
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