Confused about norm and ideal not being principal On p.273 Dummit and Foote proved that $\mathbb Z[\sqrt{-5}]$ is not a quadratic ring of integers using field norm as in the following graph.
But in this proof, they only proved that with this norm, $\mathbb Z[\sqrt{-5}]$ is not a quadratic integer ring, and norm is somehow seemingly arbitrary defined and not unique. So why this proof can conclude that  $\mathbb Z[\sqrt{-5}]$ is not a quadratic integer ring with respect to any norm?
Or can be proved that if an ideal is not principal with respect to one norm, then it is not principal with all norm? (This feels odd, since being euclidean domain only requires one norm works, not all norms.)

 A: The structure of the proof is to show that $\mathbb{Z}[\sqrt{-5}]$ is not a PID, and therefore not a Euclidean Domain. To show that it is not a PID, it suffices to find an ideal that is not principal. The ideal $I = (3,2 +\sqrt{-5})$ is shown not to be principal by showing that $3$ and $2+\sqrt{-5}$ cannot be viewed as multiples (in the ring) of fixed element $a + b\sqrt{-5}$, and this is ruled out by using a norm argument. A different norm might have worked, what's important is that the norm has a nice symmetry with respect to multiplication, and this reduces the original problem to a problem with ordinary integers which is more easily shown to be unsolvable.
To take a simpler example, with vector spaces: suppose you have three vectors $u,v,w$ and you want to show $u \neq v+w$. It suffices to find at least one linear transformation for which $T(u) \neq T(v) + T(w)$, and this might be an easier than inspecting the original vectors if the vector space is very complicated.
An even simpler example: suppose you have two functions $f$ and $g$, and you want to show $f$ is not not a linear multiple of $g$, i.e. $f \neq cg$. What you can do is take two points $x,y$ and consider the system of equations:
$$ f(x) = cg(x)$$
$$ f(y) = cg(y)$$
If this system has no solution, you have shown $f \neq cg$. You might complain "hey! what if you tried other points? maybe $f = cg$ with respect to different points!" but (1) $f = cg$ is a general statement, it is not "with respect to particular points" and (2) finding one counterexample to $f=cg$ suffices.
A: That $I$ is principal ideal of $R$ means $I = aR$ for some $\,a\in R.\,$ Equivalently  it means that there is some $\,a\in I\,$ that divides every $\,i\in I;\,$ said equationally $\, ax_i = i\,$ has a root $\,x_i\in R\,$ for all $\,i\in I.\,$ Whether or not those equations all have roots $\,x_i\in R\,$ is an intrinsic property of the ring $R$ (depending only on its multiplication table). While you may find it helpful to introduce some additional structure on $R$ (e.g. norms) to (dis)prove such roots exist, that doesn't mean that the existence of such roots in $R$ depends on this additional structure. Rather, to decide whether or not those equations are solvable we need only browse the multiplication table of $R$ (which, of course, does not change when we introduce further defined maps such as a norm).
