# Ideal class group of $\mathbb{Q}(\sqrt{-65})$

I am trying to show that the ideal class group, $Cl(A)$ of $K:=\mathbb{Q}(\sqrt{-65})$, where $A$ is the ring of integers of $K$, is isomorphic to the product two cyclic groups of order 2 and 4, respectively. I am going to share the process I have followed and enumerate the questions I have come up with.

By now, since $-65\equiv 3\mod(4)$ the discriminant is $\Delta=-4\cdot 65$, $A=\mathbb{Z}[\sqrt{-65}]$, the group of units is $U(K)=\{\pm 1\}$, and since we are working with a complex quadratic extension $r_1=0$, $r_2=1$ , and the Minkowski bound is $\mathcal{M}<11$. Therefore, every ideal class must have a prime ideal of norm less or equal than $10$.

Since every ideal class must have a prime ideal of norm less or equal than 10, the only primes we have to study are those generated by $2,3,4,5,7,$ and $9$ ($\textbf{1.}$ Why is it that all prime ideals must have norm $p$ or $p^2$?).

Now, since $2$ and $5$ are the only primes that ramify (because $2,5|\Delta$), then there doesn't exist any prime ideal of norm $4$ ($\textbf{2.}$ Why?). Also, $\left(\frac{-65}{3}\right)=1$, hence $3A=\mathfrak{p}_3\mathfrak{p}_3'$ thus there doesn't exist any prime ideal of norm $9$ ($\textbf{3.}$ Why?). Finally, $7A$ is inert, and therefore principal so $[7A]=1$

All things considered, $Cl(A)=\langle[\mathfrak{p}_2],[\mathfrak{p}_3], [\mathfrak{p}_5]\rangle$.

In addition to this, since $$N(4+\sqrt{-65})=3^4\Rightarrow(4+\sqrt{-65})A=\mathfrak{p}_3^4$$ $$N(5+\sqrt{-65})=2\cdot 3^2\cdot 5\Rightarrow(4+\sqrt{-65})A=\mathfrak{p}_2\mathfrak{p}_3^2\mathfrak{p}_5$$ and there doesn't exist any element of order $10$, $\mathfrak{p}_2\mathfrak{p}_5$ is not a principal ideal, $\mathfrak{p}_3^2$ is not principal and $\mathfrak{p}_3$ has order 4. ($\textbf{5.}$ I am not sure at all of what is going on in this paragraph, I do not follow the logic behind this).

And this is were I get stuck, when I try to find the order of each of the elements of $Cl(A)$.

• See Example $6$ here for all further details. Jan 3, 2017 at 19:16
• Answer to question $1$: see here, etc. For the question you have you need some basics of algebraic number theory, and the above text of Keith Conrad. Jan 3, 2017 at 19:18
• My source is Samuel, who does not say that every class contains a prime ideal of norm less than $\mathscr M$, but just an ideal. Also, $13$ is ramified, though this may not affect your argument. Jan 4, 2017 at 19:15

Let me show you my computations: I agree that the M-bound is between $10$ and $11$, so we need only look at ideals of norm $\le10$. To make things easier for myself typographically, I’ll call $\sqrt{-65}=\rho$, our integer ring $\Bbb Z[\rho]=R$, and $C$ the class group. Since $\{1,\rho\}$ is an integral basis, the norm form is $m^2+65n^2$. Thus there are no elements of norm $2$, $3$, $5$, $6$, $7$, $8$, nor $10$.

The prime $7$ is easiest to deal with, since $\bigl(\frac{-65}7\bigr)=\bigl(\frac57\bigr)=-1$, quadratic residue symbol. which means that $(7)$ doesn’t split, i.e. is still a prime in our field, with norm $49$. (To answer your question about prime ideals having norm $p$ or $p^2$, every prime ideal $\mathfrak p$ of $R$ contracts to a prime ideal of $\Bbb Z$, so contains an integral prime $p$. If $(p)$ is prime in $R$, then it’s equal to $\mathfrak p$, and has norm $p^2$. Otherwise $p$ ramifies or splits in $R$, product of two proper ideals, each necessarily of norm $p$, and therefore prime.)

For the other primes $\le10$, $2$ ramifies and thus we have $(2)=\mathfrak p_2^2$, where we may take $\mathfrak p_2=(2,1+\rho)$.

Since $\bigl(\frac{-65}3\bigr)=\bigr(\frac13\bigl)=1$, $3$ splits, and we may write $(3)=\mathfrak p_3\overline{\mathfrak p_3}$, with $\mathfrak p_3=(3,1+\rho)$, $\overline{\mathfrak p_3}=(3,1-\rho)$, each of norm $3$ of course.

And $5$ ramifies too, with $(5)=\mathfrak p_5^2$, and we may take $\mathfrak p_5=(5,\rho)$.

To find the structure of $C$, we must determine the order (period) of $\mathfrak p_3$, and see how the two elements $\mathfrak p_2$ and $\mathfrak p_5$, both of order two, play with each other. And then see what other relations among all these there may be.

First, I claim that $\mathfrak p_3$ is of order four: $\mathfrak p_3^2=(3,1+\rho)^2=(9,3+3\rho, -64+2\rho)=(9,3+3\rho,-1+2\rho)=(9,3+3\rho,-4-\rho)$. In this last expression for $\mathfrak p_3^2$, call the second and third generators $a$ and $b$ respectively. Then since $a+3b=-9$, we have $\mathfrak p_3^2=(9,4+\rho)$. And it’s easy to see that $(9,4+\rho)^2$ is principal. Indeed, it’s equal to $(81,36+9\rho,-49+8\rho)=(81,85+\rho,-49+8\rho)=(81,4+\rho,-49+8\rho)=(4+\rho)$. So there you have two elements of $C$ of order four, since certainly $\mathfrak p_3$ and $\overline{\mathfrak p_3}$ are inverses and not equal to each other.

All right, we now are staring at four elements of order two, namely $\mathfrak p_2$, $\mathfrak p_5$, $\mathfrak p_2\mathfrak p_5$, and $\mathfrak p_3^2$. But if $C$ really is isomorphic to $\Bbb Z/(2)\oplus\Bbb Z/(4)$, there can be only three. In fact, I claim that $\mathfrak p_2\mathfrak p_5\mathfrak p_3^2$ is principal, equal to $(5-\rho)$. You can check that $\mathfrak p_2\mathfrak p_5=(2\rho,5+\rho)$, so we get: \begin{align} (2\rho,5+\rho)(9,4+\rho)&=(18\rho,45+9\rho,-130+8\rho,-45+9\rho)\\ &=(18\rho,-130+8\rho,-45+9\rho)\\ &=(18\rho,130+10\rho,-45+9\rho)\\ &=(18\rho,85+19\rho,-45+9\rho)\\ &=(18\rho,85+\rho,-45+9\rho)\\ &=(18\rho,-5+19\rho,-45+9\rho)\\ &=(18\rho,5-\rho,45-9\rho)=(5-\rho)\,, \end{align} since $18\rho$ is a multiple of $5-\rho$. It follows that $\mathfrak p_2\mathfrak p_5\sim\mathfrak p_3^2$, and I think you can work out all other relations among $\mathfrak p_2$, $\mathfrak p_3$, and $\mathfrak p_5$ that are necessary.