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Consider the ideal $I = \langle 3, 1 + \sqrt{223} \rangle \subset \textbf Z[\sqrt{223}]$.

Q: How do I show $I$ is non-principal? This is related to an exercise in Number Fields by Daniel Marcus. I do not think there should be any advanced machinery showing up.

I tried the following. Consider $I = \langle a \rangle$ for some $a \in \textbf Z[\sqrt{223}]$. The ideal norm gives $3$ for both. Then $x^2 - 223y^2 = -3$ has a solution by $a$. Quadratic residue by modding out $223$ or $3$ does not help. I am aware there is the following links. However $-3$ is never showing up in a trivial way.

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

Or by continued partial fractions. (This is not used in Marcus's Number Fields for sure. I guess there should be a much naïve way of solving this.)

How to prove there are no solutions to $a^2 - 223 b^2 = -3$.

Or A diophantine equation

Robert Soupe
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user45765
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  • Which exercise is this in Marcus ? – Rene Schipperus Apr 04 '18 at 00:08
  • @ReneSchipperus Chpt 5. Ex 8. Show $Z[\sqrt{223}]$ has 3 ideal classes. – user45765 Apr 04 '18 at 00:18
  • How do you know that this is the solution ? There are several primes less than $\frac{1}{2}\sqrt{disc}$, maybe one of the others is easier to verify. – Rene Schipperus Apr 04 '18 at 00:25
  • @ReneSchipperus I assumed nothing on the solution. Yes, there are some primes with negative signs. Some you can use quadratic residue to rule out. I was caught in this part of computation and figured that I could not see why this is not principal. – user45765 Apr 04 '18 at 00:29
  • One of the answers on the questions linked, point out that $-3$ is represented in the principal genus, if you dont understand this it translates to meaning that the question is hard to solve. So my guess is that another solution is intended. How many primes are there ? The classes must form a group, so pick one of the non-principal classes and compute its powers. – Rene Schipperus Apr 04 '18 at 00:31
  • @ReneSchipperus Yes. I am aware of that and that method is not shown in marcus. I assumed there should be much simpler/naiver method. – user45765 Apr 04 '18 at 00:33
  • What do you know about the factorization of $2$ ? Is it principal? – Rene Schipperus Apr 04 '18 at 01:14
  • Can you prove that any of the factorisations are non principal ? – Rene Schipperus Apr 04 '18 at 02:59
  • @ReneSchipperus 2 clearly ramifies here. It is going to be $(2,1+\sqrt{223})^2$. If the order is 3, I would expect this becomes principal. – user45765 Apr 04 '18 at 03:17
  • Yeah, I think its principal too. But, you know, I cant find a single ideal I can prove is non principal. – Rene Schipperus Apr 04 '18 at 03:19
  • @ReneSchipperus Non principalness should always comes from diophantine equation solution in general and it falls into this category this time. This requires existence of solution to diophantine equations which I do not know. There should be much simpler method to solve this. – user45765 Apr 04 '18 at 03:23
  • Marcus says look at norms, but how to show $a^2-223b^2=\pm n$ has no solutions I dont see how. – Rene Schipperus Apr 04 '18 at 03:26
  • @ReneSchipperus I would try quadratic residue or naive modulo operation first. Some of them you could rule out at the first. You only have to look at prime factors. If prime factors has such solution, their product will have solutions as well. I do not know by norm method naively. – user45765 Apr 04 '18 at 03:30
  • Do mean modulo $n$ ? If $n$ is prime, then this is the same as the requirement for $n$ to split, and you assume this holds, so you dont get anywhere. And modulo $223$ here, you have the $\pm$ sign and one will give a residue. – Rene Schipperus Apr 04 '18 at 03:36
  • I always got the feeling with Marcus that there were things going on in the background he wasnt telling us about. – Rene Schipperus Apr 04 '18 at 03:37
  • @ReneSchipperus Mod 223 gives rise to quadratic residue. However, you might have quadratic residue without integer solutions. – user45765 Apr 04 '18 at 03:37

2 Answers2

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Here is an algorithm for solving this type of problem. It is not in Marcus, but then I think Marcus makes some assumptions as to background, I dont know what he has in mind as a solution, but at least this method is simple and general. It is related to continued fractions and ultimately quadratic forms.

${\bf Theorem.}$The solutions of $x^2-dy^2=N$ where $|N|<\sqrt{d}$ are given by $$p_n^2-dq_n^2=(-1)^{n-1}a_{n+1}$$ where $\frac{p_n}{q_n}$ are the continued fraction approximations.

To find the continued fraction expansion Build a sequence of triples $(a_n,b_n,c_n)$ such that $b_n^2-4a_nc_n=D$. In the case of $d=223$, we have $D=4\cdot 223=892$. The $a_n$ will give the possible values for $N$. Further

1)$a_{n+1}=-c_n$

2)$2c_n|b_n+b_{n+1}$

3) $\sqrt{D}-2|c_n|<b_{n+1}<\sqrt{D}$

So in the case of $d=223$ here is the calculation. $\sqrt{892}=29.8\cdots$ Start with

$$(a_0,b_0,c_0)=(223,0,-1).$$ To get the next triple, set, $$(223,0,-1)(1,b_1,c_1)$$

we must have $2|0+b_1$ and $27.8<b_1<29.8$ so $b_1=28$ and since $D=b_1^2-4a_1c_1$, $c_1=27$.

To get the next one, $$(-1,28,27)(-27,b_2,c_2)$$ and $2\cdot 27| 28+b_2$ with $0\leq b_2<29.8$ , thus $b_2=26$. The complete sequence is $$(223,0,-1)$$ $$(-1,28,27)$$ $$(-27,26,2)$$ $$(-2,26,27)$$ $$(-27,28,1)$$ $$(-1,28,27)$$

thus the last is the same as the second, so we have repeat. Thus we see that the only numbers $\leq 14$ that are represented are $223, 1, 27, 2$ so only $1,2$ amongst the candidate numbers are represented and split into principal ideals. Thus $3,11,13$ split into non principal ideals.

If further you calculate the differences $\delta_n=\frac{b_n+b_{n+1}}{2|c_n|}$ you get the continued fraction expansion, here we have

$$\delta_0=14$$ $$\delta_1=1$$ $$\delta_2=13$$ $$\delta_3=1$$ $$\delta_4=28$$ giving $$\sqrt{223}=[14,\overline{1,13,1,28}]$$

${\bf Additional \ Explanation:}$ If $\sqrt{d}=[k_0, k_1, k_2, \cdots ]$ is a continued fraction expansion, and let

$$\tau_n=[k_n, k_{n+1}, k_{n+2}, \cdots ]$$

Then $$\tau_n=\frac{b_n+\sqrt{D}}{2c_n}$$ where $c_n,b_n$ are the numbers above.

Rene Schipperus
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Assume there exists $\alpha$ such that $\alpha \cdot \bar{\alpha} =\pm 3$. Consider a unit $\omega = 224 + 15 \sqrt{223}$, ( a fundamental unit ). There exists an integer $k$ such that $|\omega^k|< |\alpha| < |\omega|^{k+1}$. Substituting $\alpha$ with $\frac{\alpha}{\omega^k}$ we may assume $1< |\alpha| < |\omega|$. We conclude $\frac{3}{\omega}<|\bar{\alpha}|<3$. If $\alpha = a + b \sqrt{223}$ we get $$|a + b \sqrt{223}| < \omega\\ |a- b\sqrt{223}| < 3$$

From the above we get $|a|,\sqrt{223}|b| < \frac{\omega+3}{2}$. This reduces the problem to a finite check, that can done by computer.

Note: This method uses the different embeddings of the field $\mathbb{Q}(\sqrt{223})$ into $\mathbb{R}$. In general, for a number field $K$ and an order of $K$ one can find in this way elements of a prescribed norm, if we have a fundamental system of units ( or at least a system of maximal rank). This is treated nicely in the book Number Theory by Borevich and Shafarevich.

orangeskid
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