# Class group of a field

Given a field $k = \mathbb{Q}(\sqrt{-14})$, compute the class group of $k$. Use the Minkowski bound $M_{k}=\dfrac{n!}{n^{n}}(\dfrac{4}{\pi})^{s}\sqrt{|d_{k}|}=4.76$

I know there is a theorem says in every ideal class, we can always find a ideal $I$ such that its norm is bounded by $M_{k}$ and in each ideal ideal we can find an element $a$ such that its norm is bounded by $M_{k}Norm(I)$. Now $Norm(a)$ is bounded by $M_{k}^2=22.66$. So I searched the numbers in $\mathcal{O}_{k}$ with norm less than or equal to $22$. I know the # of classes must be bounded by # of these numbers. But then how do I continue? How can I decide which and which goes into one class?

Another possible approach is to find the prime ideals in $\mathcal{O}_{k}$, and use prime factorization. But I also don't know how to classify these ideals with norm less than $M_{k}$.

Any help is appreciated!

I know there is a theorem says in every ideal class, we can always find a ideal $I$ such that its norm is bounded by M_k.

Well this is all you should need. The only primes that could have norm less than $5$ are any primes dividing $(2)$ or $(3)$. These primes will then generate all elements of the class group by unique factorization of ideals.

$x^2+14 \equiv x^2 \pmod 2$, so $(2)=(2,\sqrt {-14})^2$, and $x^2+14 \equiv (x-1)(x+1) \pmod 3$, so $(3) = (3,\sqrt {-14}+1)(3,\sqrt{-14}-1)$.

Thus the primes $\mathfrak p_1 = (2,\sqrt {-14}),\mathfrak p_2 = (3,\sqrt {-14}+1),\mathfrak p_3 = (3,\sqrt{-14}-1)$ generate the class group. I will let you calculate the relations between them (in the end $\mathfrak p_2$ will generate the class group and will be order $4$).

This isnt the method you have in mind but there is a way to compute class groups all day every day.

A quadratic form $ax^2+bxy+cy^2$ (henceforth abreviated as $(a,b,c)$) has discriminant $D=b^2-4ac$. The discriminant of $\mathbb{Q}(\sqrt{-14})$ is $-56$. Thus look at forms of discriminant $D=-56$. Forms correspond to ideals: $(a,b,c)$ corresponds to the ideal $$\left(a, \frac{-b+\sqrt{D}}{2}\right)$$ And equivalent forms (under linear transforms of $x$ and $y$) correspond to equivalent ideal. Now reduced forms for $D<0$ are those for which $$|b|\leq a\leq c$$ and have the property that $$|b|\leq \sqrt{\frac{D}{3}}$$

We can now enumerate all the ideal classes. If $D=-56$ then $|b|\leq 4$ so

$$b=0, \pm 2$$

If $b=0$ we get $$(1,0,14)$$ $$(2,0,7)$$ if $b=\pm 2$ we get

$$(3,2,5)$$ $$(3,-2,5)$$

So the class number is $4$ and the ideal classes are represented by $$(1)$$ $$(2,\sqrt{-14})$$ $$(3,1-\sqrt{-14})$$ $$(3,1+\sqrt{-14})$$

One can indeed go further and show that the ideal class group is $\mathbb{Z}_4$ generated by $(3,1+\sqrt{-14})$.