Is there another proof of $\mathbb{Z}[\sqrt{-6}]$ being not a UFD?

Question

I found out that $\mathbb{Z}[\sqrt{-6}]$ is not euclidean and all proofs I saw used a pretty simple idea: to show that this ring is not UFD, however they all used norm $\mathcal{N}(a+b\sqrt{-6}) = a^2+6b^2$. But why one can assume the norm to be such? Is there another way to prove?

Answer 1

From what I read into your question you are asking "why do we show that $\mathcal N(a+b\sqrt{-6})=a^2+6b^2$ is not an euclidean valuation" while we should prove that "there is no euclidean valuation at all". Or at least that is part of the question how to show that $\mathbb Z[\sqrt{-6}]$ is not an UFD.

To understand this we should take a step back and consider more generally quadratic extensions of $\mathbb Q$ of the form $\mathbb Q(\sqrt d)$ for some (squarefree) integer $d$. And then we note that every such extension field has a very natural associated norm function; namely $\mathcal N(a+b\sqrt d)=a^2-b^2d$. This function has some every desirable properties (for example it is valued entirely in $\mathbb Q$, the base field, and it is multiplicative).
In fact, if you know some Galois theory, this is the norm defined as product over all conjugates under the Galois group action. The conjugates of $a+b\sqrt d$ are in our case simply given by $a+b\sqrt d$ and $a-b\sqrt d$ whose product is precisely $a^2-b^2d$. So to define $\mathcal N(a+b\sqrt d)=a^2-b^2d$ is, indeed, a very, very natural thing.

Thus, every quadratic extension field comes with a norm function. We can even restrict this function to $\mathbb Z[\sqrt d]$ to examine this associated ring. Interestingly enough, this function turns out to actually be euclidean for specific values (for exampe $d=-2,-3$). But this does not hold in general ($d=-19$ is a notable counterexample altough we consider a slightly different ring then). In general this norm function is "just" a useful tool for examing the rings $\mathbb Z[\sqrt d]$. As and such tool it is typically used when showing that $\mathbb Z[\sqrt d]$ is not an UFD.

For example, you can show that an element $\alpha\in\mathbb Z[\sqrt{-6}]$ is a unit if and only if $\mathcal N(\alpha)\in\{\pm1\}$. Moreover, using this, you can show that an element whose norm is a prime integer has to be irreducible. This can then be applied to show that some elements in $\mathbb Z[\sqrt{-6}]$ admit different factorizations into irreducibles, which is impossible in an UFD.

Answer 2

For a person whose only background is abstract algebra, say a first course involving rings, there is very little technique available so you can't do much other than give a direct example of non-unique irreducible factorization. For someone farther along in their studies, there are other methods. Here is one such method.

Using concepts from algebraic number theory (Dedekind domains), it can be shown that if a quadratic ring $\mathbf Z[\sqrt{d}]$ is a UFD then it must be a PID, so finding a non-principal ideal in $\mathbf Z[\sqrt{-6}]$ proves this ring is not a UFD. (Be careful: many UFDs are not PIDs, like $\mathbf Z[x]]$, so it's very important that it's possible to prove quadratic rings can be UFDs if and only if they are PIDs first!)

Let $$ I = (2,\sqrt{-6}) = \mathbf Z 2 + \mathbf Z\sqrt{-6}. $$ That last equation is not a definition, but instead has to be proved: $I$ is defined to be the ideal generated by $2$ and $\sqrt{-6}$, so in principle you should be allowing coefficients in $\mathbf Z[\sqrt{-6}]$ and therefore it requires a proof that the $\mathbf Z$-linear combinations of $2$ and $\sqrt{-6}]$ already give you everything in the ideal. In any case, since $\mathbf Z[\sqrt{-6}] = \mathbf Z + \mathbf Z\sqrt{-6}$, the ideal $I$ has index 2 inside $\mathbf Z[\sqrt{-6}]$ (as abelian groups). To prove $I$ is not principal, it turns out that the index of a nonzero principal ideal $(a+b\sqrt{-6})$ inside $\mathbf Z[\sqrt{-6}]$ is $a^2 + 6b^2$ (oh look, there's the norm!) and since $a^2 + 6b^2$ is never 2 for integers $a$ and $b$, no principal ideal has index $2$ and thus $I$ is not a principal ideal.

Answer 3

It is possible for the integers in a quadratic number field to be Euclidean but not "norm-Euclidean"; see, e.g., Quadratic number field which is Euclidean but not norm Euclidean. BUT if you can show that the integers are not a UFD then you have shown that the ring is not Euclidean (since, if it's Euclidean, it is a fortiori a UFD), and there's nothing wrong with using the norm to prove it's not a UFD. You're not making any assumption about the ring being norm-Euclidean, you're just using the norm as a tool to prove it's not a UFD.