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In some cases, it is quite straightforward to prove that a specific ideal cannot be principal. For example, in the ring of integers of $\mathbb{Q}(\sqrt{-5})$, the ideal $(2,1+\sqrt{-5})$ is not principal, by taking norms (since this is one of the ideals in the factorization of 2).

However, in that case, we used that the norm of a generating element would have to equal $\pm 2$.

Now, let $K=\mathbb{Q}(\sqrt{-39})$ and let $I=(2,\alpha-1)$ (where $\alpha$ the root of $x^2-x+10$, the minimal polynomial of $\sqrt{-39}$). I want to show that this ideal is not principal. (specifically, it is the square of one on the primes in the factorization of $2\mathcal{O}_K$).

Any suggestions?

Thanks.

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What is $\alpha$? –  Adrián Barquero Apr 12 '11 at 17:18
    
Ops, I forgot. I've added what $\alpha$ is now. Thanks. –  Fredrik Meyer Apr 12 '11 at 17:22
    
How does your original argument not work? I suppose you need to be a little more careful since the ring of integers is $\mathbb{Z}\left[\frac{1+\alpha}{2}\right]$, but you can still take norms and argue that the norm of a prospective generator would have to divide $N(2) = 4$ and $N(\alpha-1) = 40$. Then search for elements $(a + b\alpha)/2$, where $a \equiv b$ modulo $2$, such that $N((a + b\alpha)/2) = (a^2+40b^2)/4$ divides $4$. I think you need to check norms up to 16. –  Michael Chen Apr 12 '11 at 17:44
    
I don't think you're correct when you say $(2,\alpha-1)=Q^2$ for some prime $Q$ dividing 2 - if $2{\cal{O}}_K=Q^2P_1^{e_1}\cdots P_r^{e_r}$, then $$e(Q/2)f(Q/2)+\sum_{i=1}^re(P_i/2)f(P_i/2)=2f(Q/2)+\sum_{i=1}^r e_if(P_i/2)=2$$ forces all $e_i=0$, i.e. $Q^2=2{\cal{O}}_K$. –  Zev Chonoles Apr 12 '11 at 17:53
    
@Michael Chen: Note that we've developed some conflicting notation in the question and comments. Fredrik Meyer says that $\alpha$ is a root of $x^2-x+10$ (which is not, as he also writes, a minimal polynomial of $\sqrt{-39}$). So, we may take $\alpha = \frac{1+\sqrt{-39}}{2}$. Whichever root we take, the ring of integers is $\mathbb{Z}[\alpha]$ and the norm of $\alpha-1$ is $10$. (Is my interpretation correct, Fredrik Meyer?) –  Jonas Kibelbek Apr 12 '11 at 17:55

3 Answers 3

up vote 5 down vote accepted

$\rm (\beta)\: =\: (2,\:\alpha-1)\ \Rightarrow\ (\beta\:\beta')\: =\: (2)\ \Rightarrow\Leftarrow\ $ via $ \rm\ (2,\:\alpha-1)\ (2,\:\alpha'-1)\: =\: (4,\:10,\:2\alpha-2,\:2\alpha'-2)\: =\: (2)$

Simpler, avoiding (conjugate) ideals: $\rm\ \beta\ |\ 2,\:\alpha-1\ \Rightarrow\ N(\beta)\ |\ N(2),\:N(\alpha-1)\:,\:$ i.e. $\rm\:\beta\beta'\ |\ (4,10)= 2$

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I'm not sure if I understand your answer. What are $\beta^\prime$ and $\alpha^\prime$? –  Fredrik Meyer Apr 12 '11 at 23:10
    
@Fred: Conjugates. Multiply the first equation by its conjugate to get the second equation. I appended a simpler form avoiding conjugate ideals. –  Bill Dubuque Apr 13 '11 at 0:14
    
So the argument is this?: if $(2,\alpha-1)$ is principal, it would divide $2$ which is impossible since it is a power of a factor of two. –  Fredrik Meyer Apr 13 '11 at 0:17
    
@Fredrik: Could you clarify that last comment? I was thinking that, Bill's answer implies $N(\beta) = 2$, and you can show that there is no element with norm 2: $N\left(\frac{a+b\sqrt{-39}}{2}\right) = \frac{a^2+39b^2}{4} = 2$. –  Michael Chen May 7 '11 at 15:23

I think that again an approach by contradiction using norms does work. Since $\alpha$ is a root of $x^2 - x + 10$ then

$$\alpha = \frac{1 \pm \sqrt{-39}}{2}$$

So for example if you pick the negative sign then you want to show that the ideal $$I = \left \langle 2, \frac{1 - \sqrt{-39}}{2} - 1 \right \rangle = \left \langle 2, \frac{1 + \sqrt{-39}}{2} \right \rangle$$

is not principal. So if you suppose it is principal then it may be of the form $I = \langle a + b\sqrt{-39} \rangle$ for $a, b \in \mathbb{Z}$ or $I = \left \langle \frac{a + b\sqrt{-39}}{2} \right \rangle$ with $a, b \in \mathbb{Z}$.

Then in the first case by taking norms you get $(a^2 + 39b^2) | 2$ because it divides $\mathrm{\textbf{N}}(2) = 4 $ and $\mathrm{\textbf{N}} \left ( \frac{1 + \sqrt{-39}}{2} \right ) = 10$. This case is impossible because the corresponding diophantine equation has no solutions in integers.

And well, in the other case the only difference is that you get

$$\frac{a^2}{4} + \frac{39b^2}{4} | 2 \implies a^2 + 39b^2 | 8$$

And again a case by case analysis shows that this is not possible. So the ideal is not principal.

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Let me just say that in principle you should be able to decide whether any ideal $I$ in an imaginary quadratic number ring $R = \mathbb{Z}[\sqrt{-d}]$ is principal or not by looking at norms. Indeed, because the unit group of $R$ is finite (usually just $\pm 1$ in fact), there are going to be at most finitely many elements $\alpha \in R$ with $N(\alpha) = N(I)$ and you could write a short piece of code to enumerate them all. Assuming that $I = \langle \beta_1,\ldots,\beta_n \rangle$ (we can arrange for $n = 2$, but never mind that) we just check whether $\beta_i/\alpha \in R$ for all $i$. If so, then $\langle \alpha \rangle \supset I$, and since they have the same norm they must be equal.

Added: My answer above is unnecessarily cautious. Of course for any $\alpha \in R$ and any unit $u$ of $R$, the principal ideals $\langle \alpha \rangle$ and $\langle u \alpha \rangle$ are equal. Thus the above argument works whenever you have a ring $R$ for which you can algorithmically determine all elements $\alpha$ with $\# R/\langle \alpha \rangle = N$ for any $N \in \mathbb{Z}^+$. All rings of integers of number fields have this property. Of course there are probably much more efficient algorithms than this: unfortunately my knowledge of the algorithmic aspects of algebraic number theory is very poor.

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