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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

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Maybe this answer can help you since it is a similar problem. – Adrián Barquero May 6 '11 at 18:44
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

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.


Now here are some pictures of principal ideas.



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


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


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




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.

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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
I don't know what that means. – quanta May 12 '11 at 12:38

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.

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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 $\mathbb{Z}[\sqrt{-7}]$.

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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

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.

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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

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