My understanding of this is as follows:

In the general case, one has a quadratic number field $F$, which is always of the form $\mathbb{Q}(\sqrt{d})$ for some square-free integer $d$.

Minkowski Bound Theorem states that every equivalence class in the ideal class group $C_F$ of an algebraic number field $F$ of degree $n$ over $\mathbb{Q}$, with $r_2$ complex embeddings, contains a non-zero ideal $I$ with norm $$N(I)\leq (\frac{4}{\pi})^{r_2}\frac{n!}{n^n}\sqrt{|d_F|}$$ where $d_F$ is the discriminant of $F$.

So once the Minkowski bound (RHS of inequality) is established, bearing in mind that $N(I)$ must be positive integer also, I have read that it is only necessary to factorise the principal ideals $(p)=pO_F$ into a product of prime/maximal ideals for $p\leq$ the Minkowski bound. Why is this? And is it the case that since $O_F$ is a Dedekind ring, that these ideal divisors of $(p)$ are prime iff maximal?

I am a bit confused about how to proceed from here in the case where the Minkowski bound is $\geq 2$ and primes have to be checked.

So one can determine whether $p$ splits in $F$, ramifies in $F$ or remains prime, for each prime $p$. How is this information used to compute the ideal class group $C_F$ and corresponding class number?

An illustrative example would be greatly appreciated.


1 Answer 1


Hopefully this will still be useful to you, or to someone in the future with a similar problem.

You have a number field $F$ and you want to find its ideal class group $C_F$. By Minkowski Bound Theorem every ideal class is represented by an ideal $I$ of norm $N(I) \leq c$ where $c$ is the Minkowski constant. So in order to find the elements of the class group, we need to find ideals of small norm in $O_F$.

There is a very important fact about ideals in rings of integers: $N(I) \in I$, so $I \mid (N(I))$. Now $N(I)$ is a natural number and can be factorised in the product of rational primes. So if we can factorise into primes all ideals $(p)$ with $p \leq c$, we will be able to find all ideals of small norm as their factors.

(Indeed prime and maximal ideals in Dedekind domains coincide.)

This is perhaps best illustrated by an example. Let $F=\mathbb{Q}(\sqrt{26})$. Then $O_F= \mathbb{Z}[\sqrt{26}]$, $n=2$, $r_2=0$ and $d_F=4\cdot 26 = 104$. The Minkowski bound is $c=\sqrt{26}<6$, so we need to find all prime ideals of norms $\leq 5$.

By Dedekind's Theorem for primes $2,3$ and $5$, we see that they factorise as

$(2) = (2, \sqrt{26})^2 =: P_2^2$ is a product of two prime ideals of norm $2$.

$(3)$ remains prime, so it has norm $9$, which is too large for our interest (ie the Minkowski Bound tells us that the same class is also represented by some ideal of smaller norm).

$(5)= (5, 1+\sqrt{26})(5, -1+\sqrt{26}) =: P_5 \cdot P_5'$ is a product of two distinct prime ideals of norm $5$.

Therefore, all ideals of norm $\leq 5$ in $O_F$ are $P_2, P_5$ and $P_5'$ and the ideal class group is generated by their classes $[P_2]$, $[P_5]$ and $[P_5']$. We have some relations between these already: $[P_5]$ and $[P_5']$ are inverses (because their product is a principal ideal) and $[P_2]$ has order $2$. Then we also observe that $(6-\sqrt{26})=P_2 \cdot P_5$ and so $[P_2] \cdot [P_5] = 1$ too and so $[P_2]=[P_5]=[P_5']$. After checking that the ideal $P_2$ is not principal, this means that the ideal class group has order $2$.

I hope this example was helpful and I'm happy to answer any further questions you may have.

  • $\begingroup$ Thanks for this explanation - I found it very useful. Is it possible to expand explicitly on using Dedekind's theorem to find the norm of the prime ideals? $\endgroup$
    – Sigma
    Aug 14, 2022 at 23:02

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