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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.) enter image description here

user26857
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    That $I$ is not principal means that there is no $,a\in R$ with $,I = aR.,$ It doesn't matter if you prove it using norms or not. Once you've proved it you know it is true. – Bill Dubuque Mar 19 '22 at 20:09
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    A ring being "principal" does not depend on a norm - either all ideals can be generated by a single element or they can't. – Mark S. Mar 19 '22 at 20:09
  • @BillDubuque, Thanks, but I don't understand since bing principal not depending on norm, how can we be sure that this wouldn't happen- with one norm it's principal, and with another it's not? As in the following answer said it is linguistic and not mathematical, but I cannot see why it's just linguistic. –  Mar 19 '22 at 20:32

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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.

Elchanan Solomon
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  • Thanks, but I don't understand since bing principal not depending on norm, how can we be sure that this wouldn't happen- with one norm it's principal, and with another it's not? –  Mar 19 '22 at 20:14
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    The definition of being a PID does not involve any kind of norm. – Elchanan Solomon Mar 19 '22 at 20:15
  • Yes, but why defined without any norm prevent the case to happen?- with one norm it's principal, and with another it's not? –  Mar 19 '22 at 20:16
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    I think the confusion here is linguistic and not mathematical, so there's not much more I can say. – Elchanan Solomon Mar 19 '22 at 20:18
  • So "To show that it is not a PID, it suffices to find an ideal that is not principal." Don't need to prove it suffices to show that? –  Mar 19 '22 at 20:20
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    That is the definition of being a PID. You don't need a proof, you just need to look at the definition. – Elchanan Solomon Mar 19 '22 at 20:21
  • Ok, thanks, I understand the $f \neq cg$ example, but I just cannot see why the case in norm is similar. I must miss something. But still thanks! –  Mar 19 '22 at 20:22
  • Sorry may I say something more? Just like in the function example, at least in my textbook, it defined that if two function are the same, then they are equal in every point in the domain. But I don't have a definition clearly saying that if a ideal is a principal, then it is principal with any norm. –  Mar 19 '22 at 20:29
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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).

Bill Dubuque
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  • I am so sorry if I repeats the question, but don't we need to prove something about the additional structure before using it? Like to prove the original structure really does not change when introducing further structure? –  Mar 19 '22 at 20:43
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    @diiiiiklllllll Defining a norm (or any "definitional extension") doesn't change the ring operations (addition and multipication tables), so it has no impact at all on whether or not said equations all have roots in $R.\ \ $ – Bill Dubuque Mar 19 '22 at 20:45
  • OH, I see, thank you so much!!!!!! –  Mar 19 '22 at 20:46
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    @diiiiiklllllllq Another closely related example that might aid intuition is the persistance of GCDs in PID extension rings. – Bill Dubuque Mar 19 '22 at 20:48