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

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 $(19)$.

• Yes, you've definitely made progress. Next up: Each ideal class contains a prime ideal, and you have 7 prime ideal classes in front of you. So your remaining challenge is to decide which ones are in the same class. – Cam McLeman Oct 31 '14 at 15:44
• Thanks for the support :) . But I must say I'm not quite at ease with operations on ideals. I always fear to do things I can't do. Now I know what to do but not how to do it, sorry. – Nerra Oct 31 '14 at 16:00
• Yup, should have $r_2=0$. – Cam McLeman Nov 3 '14 at 20:55

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 $$29^2=2^2\cdot223-51.$$

• I don't have the words to describe my thanks. It helps far beyond the scope of my question. – Nerra Nov 2 '14 at 19:19
• Glad to hear that this helped. It has been a while since I calculated anything of the sort, so thank you for giving me the opportunity to refresh a few things. I feel that there is a more systematic way of simplifying those product ideals. May be work with basis over $\Bbb{Z}$, where the algorithm for computing the Smith normal form should give something useful? My calculations were very ad hoc. – Jyrki Lahtonen Nov 2 '14 at 19:25
• Either Cam McLeman or Will Jagy (or ...) might be able to describe a more deterministic way. For a different point of view you may also take a look at Will Jagy's on-site blog post. – Jyrki Lahtonen Nov 2 '14 at 19:28
• Very nice! +1. And comment filler. – Cam McLeman Nov 3 '14 at 14:30
• this question, in disguise initially, at math.stackexchange.com/questions/1383345/… which I did answer my way... – Will Jagy Aug 3 '15 at 19:36

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?

• Thank you for your answer but it doesn't help me at all. I'm almost new to this domain of mathematics and I don't have the results/theorems/definitions in my head. – Nerra Nov 2 '14 at 15:20
• Well, you're welcome. And if you're learning the material for the first time, you're not expected to have all of it in your head -- that's why you have a textbook? In any case, I think you might have better luck if you can pinpoint what exactly you don't understand. For example, it sounds like you might be uncomfortable with multiplication of ideals. Is that true? Your original question made it seems like you were adept at ideal factorization. – Cam McLeman Nov 2 '14 at 16:22
• Yep I thought I was not too bad but I'm still unsure of what I can do and cannot do. And again, that's not easy to come by some nice examples. I'll take a look at the godlike answer of Jyrki and see if I can swipe my doubts away. – Nerra Nov 2 '14 at 19:15