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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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    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 $\mathcal{O}_K$.2011-05-12

4 Answers 4

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Put $\rm\, w = 1\!+\!\sqrt{-7}.\,$ By norms $2\,$ is irreducible so, if principal $\rm\,\color{#c00}{ (2,w) = (1)}\ $ [not $(2)$ by $\rm\,2\nmid w$]

so $\rm\ 2\mid 2w,\,ww'\Rightarrow\ 2\mid (2w',ww') = \color{#c00}{(2,w)}(w') = (w'),\ $ so $\rm\ 2\mid w',\ $ 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 this answer.

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    @user1952009 $\ $ Suppose it's principal $\,(2,w) = (a).\,$ Then $\,a\mid 2.\,$ Since $2$ is irreducible, $\,a\,$ is associate to $\,2\,$ or $\,1,\,$ i.e. $\,(a) = (1)\,$ or $\,(a) = (2).\,$ But $\,(2,w)=(a)=(2)\,\Rightarrow\, 2\mid w,\,$ contra $\,w/2\not\in\Bbb Z[\sqrt{-7}].\,$ Thus $\,(2,w)=(a)= (1).\,$ The rest of the proof is essentially [Euclid's Lemma in ideal form](http://math.stackexchange.com/a/690282/242), as explained in [the linked answer.](http://math.stackexchange.com/questions/37085/integral-domain-with-two-elements-that-do-not-have-a-gcd/37089#37089)2016-12-16
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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.

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    I don't know what that means.2011-05-12
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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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    @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 $\alpha=1$.2011-05-06
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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.2011-05-07