Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Show that the ideal $ I = (2, 1 + \sqrt{-7} ) $ in $ \mathbb{Z} [\sqrt{-7} ] $ is not principal.

My thoughts so far:

Work by contradiction. Assume that $ I $ is principal, i.e. that it is generated by some element $ z = a + b\sqrt{-7} \in \mathbb{Z}[\sqrt{-7}] $. I'm really not sure what to consider though - I can't really 'see' what $ I $ looks like.

Any help would be greatly appreciated. Thanks

share|improve this question
3  
Maybe this answer can help you since it is a similar problem. –  Adrián Barquero May 6 '11 at 18:44
1  
Let me point out that the class number of $K=\mathbb{Q}(\sqrt{-7})$ is 1, so the ring of integers $\mathcal{O}_K$ is a PID. However, $\mathbb{Z}[\sqrt{-7}]$ is only an order of conductor $2$ in $\mathcal{O}_K$. –  Jiangwei Xue May 12 '11 at 10:23

4 Answers 4

Consider the map $N\colon\mathbb{Z}[\sqrt{-7}]\to\mathbb{Z}$ given by $$N(a+b\sqrt{-7}) = a^2+7b^2.$$ This map is multiplicative, so if $z\in\mathbb{Z}[\sqrt{-7}]$ divides $2$, then $N(z)$ divides $N(2) = 4$. So $N(z)=1$, $N(z)=2$, or $N(z)=4$. Check the possibilities, and see if any of them divides $1+\sqrt{-7}$; those are your possible generators (note that the ideal $(2,1+\sqrt{-7})$ is principal if and only if $(2,1+\sqrt{-7})=(z)$ for some $z$, which implies that $z$ divides both $2$ and $1+\sqrt{-7}$). Not check to see if any of the possible generators are actually generators.

The map $N$ is called the "norm map". It is given by taking an element of $\mathbb{Z}[\sqrt{-7}]$, and multiplying all its images under the different embeddings of its field of fractions $\mathbb{Q}(\sqrt{-7})$ into $\mathbb{C}$; it is a standard tool for studying divisibility and ideals in orders, such as $\mathbb{Z}[\sqrt{-7}]$.

share|improve this answer
    
Thanks. If $ N(\alpha) = 1 $, we are forced to have $ \alpha = 1 $. The ideal generated by $ \alpha $ is therefore the whole ring. Why can't this be the case? –  user938272 May 6 '11 at 18:43
    
@user938272: Because $(2,1+\sqrt{-7})$ is not the entire ring. Verify that the square of any element in that ideal is necessarily a multiple of $2$, so the ideal cannot contain $1$. –  Arturo Magidin May 6 '11 at 18:45
    
@user938272: Alternatively, notice that $(2,1+\sqrt{-7})^2 = (4,2+2\sqrt{-7},-6+2\sqrt{-7})$, so every element of the square of the ideal is a multiple of $2$, hence we cannot have $(2,1+\sqrt{-7}) = (1)$. Also: $N(\alpha)=1$ implies $\alpha=\pm 1$, and not only $\alpha=1$. –  Arturo Magidin May 6 '11 at 19:05

Here is a picture of the elements of $\mathbb Z[\sqrt{-7}]$ (in green) embedded into the plane in such a way that their distance form the origin (blue) is equal to their norm. The ideal is drawn on top in purple.

lattices

Now here are some pictures of principal ideas.

$$(5)$$

lattice

$$(3 - \sqrt{-7})$$

lattice

$$(1 - 2\sqrt{-3})$$

lattice

$$(1 + \sqrt{-7})$$

lattice

$$(1+3\sqrt{-7},10+2\sqrt{-7})$$

lattice

You could probably see immediately that last one is not a principal ideal! The density of points is too strong for it to be principal. I am not sure how to turn this "density" concept into mathematical proof but I'm sure it can be done.

I the gnuplot command used

plot './lattice.txt' with points pointsize 0.4 pt 20 lt 2 notitle, './lattice2.txt' with points pointsize 0.5 pt 20 lt 4, './origin.txt' with points pointsize 0.5 pt 20

to draw these.

share|improve this answer
1  
the inverse of the density is the size of the quotient ring A/I, and it's the norm of the ideal I. So you are using a norm argument. –  mercio May 6 '11 at 20:58
1  
I don't know what that means. –  quanta May 12 '11 at 12:38

To prove this note that $1 \not \in I$ so $I \not = (1)$. Then suppose $I = (\alpha)$, that implies that $\alpha = 2$ or $\alpha = 1 + \sqrt{-7}$ since those are both irreducibles, but neither of them can hold since one irreducible is not a multiple of another.

share|improve this answer
2  
You can't simply claim $\: I\ne 1\:$ That requires proof. Indeed, it's the only nontrivial part of the proof. –  Bill Dubuque May 7 '11 at 1:19

Put $\rm\: w = 1+\sqrt{-7}\:.\:$ By norms $2\:$ is irreducible. So, if principal, $\rm\: (2,w) = (1)\ $ (not $(2)$ since $\rm\:2\nmid w$)

so $\rm\ 2\ |\ 2\:w,\:w\:w'\Rightarrow\ 2\ |\ (2\:w',\:ww')\ =\ (2,w)\:(w')\: =\ (w')\:,\ $ so $\rm\ 2\ |\ w'\:,\ $ a contradiction. $\ \: $ QED

This is a special case of the fact that the failure of an irreducible element to be prime (or a failure of Euclid's Lemma) immediately yields a nonexistent gcd and nonprincipal ideal - see my post here.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.