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Just started teaching myself some basic abstract algebra and I noticed that the euclidean function on $\mathbb{Z}[\sqrt k]$ for square-free $k$ giving euclidean domain is defined as $$N(a + b\sqrt{k}) = a^2 - kb^2$$

In particular, this euclidean function has a nice multiplicative property that is leveraged in many proofs.

My question might be an ill formed one, but I was wondering if there was any bigger reason a multiplicative euclidean function is desirable besides the fact that it is convenient. Must every euclidean function on $\mathbb{Z}[\sqrt k]$ be multiplicative or are there examples of other admissible euclidean function that satisfy the euclidean property on these quadratic rings?

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    Related for more general cases: https://math.stackexchange.com/questions/854771 – 2017-01-15
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    @Watson Thank you. That proof is a bit out of my depth currently but I will spend some time going over it. – 2017-01-15

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If instead of looking at $\Bbb Z[\sqrt{k}]$ you look at $\Bbb Q[\sqrt{k}]$ it is very easy to check that the norm of an element $x$ is nothing but the determinant of the $\Bbb Q$-linear map $$ m_x:\Bbb Q[\sqrt{k}]\longrightarrow\Bbb Q[\sqrt{k}] \qquad m_x(z)=xz. $$ Note that this tells you immediately that the norm is multiplicative, since the determinant of a composition of linear maps is the product of the determinants of the maps you are composing.

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    You don't have to pass to $\Bbb Q[\sqrt k]$. This argument works just as well in $\Bbb Z[\sqrt k]$ as a linear map over $\Bbb Z$. – 2017-01-15
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    @Arthur: of course it does. But since the OP is a beginner in algebra by his own admission, I thought wiser to stick to vector spaces than using linear algebra over free abelian groups. – 2017-01-15
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    There is nothing here that isn't vector spaces and linear algebra. It's just that vector spaces over $\Bbb Z$ can be a bit unusual if you've only worked with fields before. Especially that you can't scale vectors down. But we're not doing any of that here. – 2017-01-15
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    @AndreaMori I may be mistaken but I think this answer is slightly tangential to my question. You are telling me the norm is multiplicative because it is the determinant of the mapping. However, my question was about why the euclidean function on euclidean domains that are quadratic rings is as far as I have seen, always multiplicative. Such a euclidean function could be anything that satisfies the euclidean division property and thus doesn't need to be the determinant of the mapping (although that is one that works). – 2017-01-16
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    @AndreaMori It may have been my bad because I used the word norm in my question. To clarify, I am asking why is it that when we look for a euclidean function in quadratic rings that are euclidean domains, we seek a multiplicative euclidean function. Could we have a non multiplicative euclidean function that was also admissible? – 2017-01-16
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    @Arthur: Excuse me, are you explaining to me the difference between a vector space over a field and a possibly free module over a ring? Really? Btw, there are no such things as "vector spaces over $\Bbb Z$". – 2017-01-16
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    @gowrath: showing that the norm $N(x)$ is the determinant of the multiplication by $x$ map does not serve just to the purpose of proving its multiplicativity, but also to show that it carries some measure theoretic significance, thus enhancing its "naturalness". Said that, there are quadratic fields which are euclidean with respect to maps different from the norm, this is actually mentioned even in a wiki page, but never having really digged into these matters I cannot give a precise answer. – 2017-01-16
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    @AndreaMori Well, technically, vector spaces have to be over fields, you're right. But all of the machinery you need still works over $\Bbb Z$; matrices, determinants and linear maps are still well-defined, and inverse matrices only exists iff the determinant is invertible, so I don't see any reason not to use $\Bbb Z$. – 2017-01-16
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Maybe you've heard about the norm of an element in Galois theory. In a typical course of Field theory and Galois theory one studies field extensions like $\Bbb{Q}(\sqrt{2})/\Bbb{Q}$. Between the different kinds of field extensions we have the finite galoisian extensions. Well this kind of extensions are very useful because they have nice properties. Associated with a finite galoisian extension $F/K$ we have the Galois group $Gal(F/K)$ wich is the group of all the $K-$automorphisms of $F$, i.e., $Gal(F/K)=\{\sigma\in Aut(F):\sigma(\theta)=\theta,\;\forall\, \theta\in K\}$.

Now for a finite galoisian extension $F/K$ with Galois group $G=\{\sigma_1,\ldots, \sigma_n\}$ we can define the norm of an element $\alpha\in F$ as $$N(\alpha)=\prod_{\sigma\in G}\sigma(\alpha)=\sigma_1(\alpha)\cdots \sigma_n (\alpha).$$

From the above formula it's easy to prove that $N$ is multiplicative. The "nice" expresion that we've got it's a particular definition of the norm because we can define it for finite but not necessarily galosian extensions, see for example here.

On the other hand we have the quadratic fields $\Bbb{Q}[\sqrt{k}]$. We call them quadratic because the degree of the extension of $\Bbb{Q}[\sqrt{k}]/\Bbb{Q}$ is $2$, so this extension is finite, and it turn out to be a galoisian extension too. Moreover, we can determine the Galois group of $\Bbb{Q}[\sqrt{k}]/\Bbb{Q}$ and it's $G=\{\text{id}, \text{conj}\}$, where $\text{id}$ is the identity morphism and $\text{conj}$ is the conjugation, namely $\text{conj}(a+b\sqrt{k})=a-b\sqrt{k}$.

Then appying the above formula we deduce that for $\alpha=a+b\sqrt{k}\in \Bbb{Q}[\sqrt{k}]$: $$N(a+b\sqrt{k})=\text{id}(a +b\sqrt{k})\text{conj}(a+b\sqrt{k})=(a+b\sqrt{k})(a-b\sqrt{k})=a^2-kb^2.$$

This formula works in particular for the subring $\Bbb{Z}[\sqrt{k}]$ of $\Bbb{Q}[\sqrt{k}]$.

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An elementary justification is Brahmagupta's identity: $$(a^2-nb^2)(a'^2-nb'^2)=(aa'+nbb')^2-n(ab'+ba')^2.$$

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    That's just a restatement of the multiplicativity, surely not a "bigger reason" explaining why it is multiplicative. – 2017-01-15
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    @Bill Dubuque: I never said it was a bigger reason – only an elementary one, easy to check. – 2017-01-15
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    But the question explictly ask for a "bigger reason". Perhaps you missed that? – 2017-01-15
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    Yes, I've just checked this point. I probably overlooked ‘bigger’, read only ‘reason’ and thought the determinant explanation was a little tough for such a basic fact. – 2017-01-15