Show that the ideal $ (2, 1 + \sqrt{-7} ) $ in $ \mathbb{Z} [\sqrt{-7} ] $ is not principal 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
 A: 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. 
A: 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.
$$(5)$$

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

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

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

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

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