Show that $\mathbb{Z}[\sqrt{223}]$ has three ideal classes.

Question

Well the question is the title. I tried to grab at some straws and computed the Minkowski bound. I found 19,01... It gives me 8 primes to look at. I get

$2R = (2, 1 + \sqrt{223})^2 = P_{2}^{2}$

$3R = (3, 1 + \sqrt{223})(3, 1 - \sqrt{223}) = P_{3}Q_{3}$

$11R = (11, 5 + \sqrt{223})(11, 5 - \sqrt{223}) = P_{11}Q_{11}$

$17R = (17, 6 + \sqrt{223})(17, 6 - \sqrt{223}) = P_{17}Q_{17}$

where I denote $R$ the ring of integers $\mathbb{Z}[\sqrt{223}]$. Others are prime.

EDIT : Ok, so I tried to find some answers on various pdf. Unfortunately, it's always at best half cryptic. That is what I understood I can do.

I computed the norm for a general element of $\mathbb{Z}[\sqrt{223}]$ : $N(a + b\sqrt{223}) = a^{2} - 223b^{2}$. I checked which norms were available among $\pm 2, \pm 3, \pm 11, \pm 17$. Only 2 is (with $a = 15$ and $b = 1$). This tells me that $P_{2}$ is principal and the other six are not.

And I'm stuck here. And please, don't tell me to think about some hint. I already have, I can't think of anything. I just need one example to understand the method and so helping me understand the theory behind.

$\text{EDIT}^2$ : I think I made a mistake with the Minkowski's bound. For remainder, it is $M_{K} = \frac{n!}{n^{n}}\left(\frac{4}{\pi}\right)^{r_{2}}\sqrt{|\text{disc}(K)|}$. I took $r_{2} = 1$, which is wrong, it should be $r_{2} = 0$ (I think), thus giving $M_{K} = 14, \dots$. We don't need to consider $(17)$ and $(19)$.

Answer 1

Cam would know this better, but let me give a few pointers for pencil & paper calculations. IIRC calculations like this use heavily the following rules. In what follows $(a_1,a_2,\ldots,a_n)$ is the ideal generated by the listed elements.

We have the rules

• $(a_1,a_2,\ldots,a_n)=(a_1,a_2,\ldots,a_{n-1},a_n-ra_1)$ for all $r\in R$. In other words we can subtract from one generator a multiple of another. The validity of this rule is hopefully clear (the process is reversible). This allows to replace an uncomfortably large generator with a smaller one. Also, if $a_n$ happens to be a multiple of $a_1$, the remainder is zero, and we can drop that altogether, as zero won't generate much.
• The product of the ideals follows the rule: $$(a_1,a_2,\ldots,a_n)\cdot(b_1,b_2,\ldots, b_m)=(a_ib_j|1\le i\le n, 1\le j\le m).$$ Because the operation in the ideal class group is the product of representatives you will be doing this a lot.

You calculated that the ideal classes are among the listed six (and the class of principal ideals). The goal of proving that the class group is of order three is a powerful hint. Among other things it implies that the ideals $P_3^3$ and $Q_3^3$ should both be principal. These are both of index $3^3=27$ in $R$, so a search for principal ideals of norm $27$ is natural. We don't need to look further than $$ 223-27=196=14^2 \implies (27)=(14-\sqrt{223})(14+\sqrt{223}). $$ Because $(27)=P_3^3Q_3^3$ this actually already implies the claim that $P_3^3$ and $Q_3^3$ are the principal ideals $J_1=(14+\sqrt{223})$ and $J_2=(14-\sqrt{223})$. This is because we easily see that $J_1+J_2$ contains both $27$ and $28$, hence also $1$, so the ideals $J_1$ and $J_2$ are coprime. As $P_3$ and $Q_3$ are the only primes lying above $3$, we must have $P_3^3=J_1$ and $Q_3^3=J_2$ or the other way round.

To gain a bit more experience and to decide which is which let us calculate using the above rules. I will abbreviate $u=\sqrt{223}$ to spare some keystrokes :-) $$ \begin{aligned} P_3^2&=(3,1+u)(3,1+u)=(9,3+3u,3+3u,(1+u)^2)=(9,3+3u,224+2u)\\ &=(9,3+3u,224+2u-9\cdot25)=(9,3+3u,-1+2u)\\ &=(9,(3+3u)+3(-1+2u),-1+2u)=(9,9u,-1+2u)\\ &=(9,-1+2u). \end{aligned} $$ Therefore $$ \begin{aligned} P_3^3&=(9,-1+2u)(3,1+u)=(27,-3+6u,9+9u,(-1+2u)(1+u)=445+u)\\ &=(27,-3+6u,9+9u,445+u-16\cdot27=13+u)\\ &=(27,-3+6u,9+9u+3(-3+6u),13+u)=(27,-3+6u,27u,13+u)\\ &=(27,-3+6u,13+u)\\ &=(27-2(13+u),-3+6u,13+u)=(1-2u,-3(1-2u),13+u)\\ &=(1-2u,13+u)=(1-2u,14-u). \end{aligned} $$ Here $1-2u$ has norm $-891=-33\cdot27$. The expectation is now that $P_3^3=(14-u)=J_2$, so we know what to try! $$ \begin{aligned} \frac{1-2u}{14-u}&=\frac{(1-2u)(14+u)}{14^2-u^2}=-\frac{14-2u^2+27u}{27}\\ &=-\frac{14-2\cdot223+27u}{27}=-\frac{-16\cdot27+27u}{27}=16-u. \end{aligned} $$ Therefore $P_3^3=(1-2u,14-u)=(14-u)$.

As $P_3$ itself is not principal we have now shown that it is of order three in the class group. Also, $Q_3$ has to be a representative of the inverse class, as $P_3Q_3$ is principal.

What about the remaining four ideals? The goal is surely to prove that the ideals $P_{11},Q_{11},P_{17},Q_{17}$ are all in the same class as either $P_3$ or $Q_3$ for otherwise the class group would be larger. Can we find principal ideals of norm $3\cdot11=33$? Sure! $$ 33=256-223=16^2-223=(16-u)(16+u). $$ Thus we have high hopes that either $P_3P_{11}$ or $Q_3P_{11}$ would be one of $(16-u)$ or $(16+u)$ (the other being the product of $Q_{11}$ and one of the norm three prime ideals). Let's try! $$ \begin{aligned} P_3P_{11}&=(3,1+u)(11,5+u)=(33,11+11u,15+3u,228+6u)\\ &=(33,11+11u,15+3u,-3+6u)\\ &=(33,11+11u,15+3u+5(-3+6u),-3+6u)=(33,11+11u,33u,-3+6u)\\ &=(33,11+11u,-3+6u)=(33,17-u,-3+6u)\\ &=(-1+2u,17-u,3(-1+2u))=(-1+2u,17-u)\\ &=(-1+2u,16+u). \end{aligned} $$ A calculation similar to the preceding one shows that $-1+2u$ is a multiple of $16+u$, so $P_3P_{11}=(16+u)$. I got lucky! It might have just as well happened that $P_3Q_{11}$ is the principal one!

Anyway, this shows that the classes of $P_3$ and $P_{11}$ are inverses to each other in the class group.

Leaving the norm 17 ideals for you with the hint (sorry!) that $$ 29^2=2^2\cdot223-51. $$

Answer 2

Hints/relevant information:

1) If the ideal class group has order 3, then it can't have any elements of order 2. That tells you something about your factorization of $2R$.

2) One of your ideal classes is always the principal ideal class.

3) That leaves two ideal classes, $A$ and $B$. Two ideals are in the same ideal class if one is a principal ideal times another. For each of your factorization of 3, 11, and 17, one of the factors is in $A$ and one in $B$. Can you figure out which ones go together?