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I have a ring:

$${\bf{O}}[\sqrt{5}] = \{c_1 + c_2\sqrt{5}: (c_1 \in \mathbb{Z} \wedge c_2 \in \mathbb{Z}) \lor (c_1 + \frac{1}{2} \in \mathbb{Z} \wedge c_2 + \frac{1}{2} \in \mathbb{Z}) \}.$$

I then need to

  • Show that is $n \in \mathbb{Z}$ and $c_1 + c_2\sqrt{5}$ divides $n$ in ${\bf{O}}[\sqrt{5}]$ then so does $c_1 - c_2\sqrt{5}$. You may assume that $\theta: {\bf{O}}[\sqrt{5}] \rightarrow {\bf{O}}[\sqrt{5}]$ defined by

$$\theta(c_1 + c_2\sqrt{5}) = c_1 - c_2 \sqrt{5}$$

is a well-defined ring isomorphism, in particular, that $\theta(cd) = \theta(c)\theta(d)$ for all $c$ and $d$ in $\mathbb{Z}[\sqrt5]$.

  • Show that $c_1 + c_2 \sqrt{5}$ is a unit in ${\bf{O}}[\sqrt{5}]$ - that is, a divisor of $1$ - if and only if $c_1 + c_2\sqrt{5}$ is a unit, and if and only if $c_1^2 - 5c_2^2 = \pm 1$.

For the first bit, would it suffice to say that the way the map $\theta$ is defined tells us that $c_1 - c_2\sqrt{5}$ divides $n$ as for any value of $(c_1 + c_2 \sqrt{5}) \mapsto (c_1 - c_2\sqrt{5})$ and if $c_1 + c_2\sqrt{5} \mid n \implies c_1 - c_2\sqrt{5} \mid n$?

For the second bit, I have already shown that $c_1^2 - 5c_2^2 \in \mathbb{Z}$ whenever $(c_1 + \frac{1}{2}), (c_2 + \frac{1}{2}) \in \mathbb{Z}$ but I don't see how this is helping me answer this question.

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4 Answers 4

If $d\in\mathbf O[\sqrt 5]$ such that $(c_1+c_2\sqrt 5)\cdot d = n$ (i.e. $d$ witnesses that $c_1+c_2\sqrt 5$ is a divisor of $n$), then $\theta(c_1+c_2\sqrt 5) \cdot \theta(d)=\theta(n)$. Since $\theta(n)=n$ and $\theta(c_1+c_2\sqrt5)=c_1-c_2\sqrt 5$, the claim follows.

For the second note that the product of units is a unit, hene $(c_1+c_2\sqrt 5)\cdot\theta(c_1+c_2\sqrt 5)=(c_1+c_2\sqrt 5)(c_1-c_2\sqrt 5)=c_1^2-5c_2^2$ is a unit. A unit in $\mathbf O[\sqrt 5]$ that happens to be $\in\mathbb Z$, must be a unit in $\mathbb Z$ (beause $\mathbf O[\sqrt 5]\cap \mathbb Q=\mathbb Z$), i.e. $=\pm 1$.

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Or directly:

$$(a+b\sqrt 5)\,\mid\,n\implies n=(a+b\sqrt 5)(c+d\sqrt 5)=(ac+5bd)+(ad+bc)\sqrt 5\implies$$

$$\begin{cases}da+cb=0\\ca+(5d)b=n\end{cases}\;\;\implies a=-\frac{cn}{5d-c}\;,\;b=\frac{dn}{5d-c}$$

The above is obtained simply applying Cramer and assuming $\,n\neq0\,$ (otherwise the claim's boring) , so...

For the second one:

$$a+b\sqrt 5\;\;\text{is a unit}\iff (a+b\sqrt 5)(c+d\sqrt 5)=1\iff (a^2-5b^2)(c^2-5d^2)=1$$

the last equality following by multiplying by the conjugates of both factors on the left (why is this possible and why the equality remains?)

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Sorry, why is $n = (a + b \sqrt{5}) \times (c + d\sqrt{5})$? Does that $c + d\sqrt5$ come into play because of the definition of the ring, as opposed to $n = (a + b \sqrt{5} \times k$ for some integer $k$? And so, after that, as $(a + b\sqrt{5}) \mid n$, so $\frac{n}{a + b\sqrt{5}} = $ an integer, which is why ($ad + bc) = 0?$ If you don't mind, I was hoping to understand the first one properly before looking at the second one properly, instead of skimming over them both at the same time. –  Kaish Apr 15 '13 at 10:56
    
(1) That's the definition of "division" in a ring, @Kaish: $$r\,\mid\,s\iff \exists \,t\;\;s.t.\;\; s=tr$$ (2) you get $\,ad+bc=0\,$ because the ring you're working in is a free abelian group of rank 2 (ring of integral elements in a number field...) –  DonAntonio Apr 15 '13 at 11:00
    
So why does that mean that $ab + cd = 0?$ Also, what tells you its a free abelian group? Sorry, if these are silly questions, I've not actually done any Ring theory, this is all in my number theory module and we've just been told like the basics of rings, groups and fields. So groups is fine for me from last semester but I've never done rings before. –  Kaish Apr 15 '13 at 11:06
    
Well, then let's simplify it (which translated to my language means "make it look as if it were simpler but much less fun"): do you know that $\,x+y\sqrt 5=z+w\sqrt 5\iff x=z\,,\,y=w\,\,\text{in the ring}\;\;\mathcal O(\sqrt 5)\,$ ? Please do note that it was you who wrote about ring homomorphisms so I'm not very impressed by your claim that you haven't studied rings... –  DonAntonio Apr 15 '13 at 11:13
    
Sorry, I assumed they'd be the same as group ones. Yes, I did know that. –  Kaish Apr 15 '13 at 11:20

We can also adapt the view point of matrices: we represent $\sqrt 5$ by $A=\begin{pmatrix}2 &1\\1 &-2\end{pmatrix}$. Then an element in the ring is of the form $$cI_{2\times2}+dA=\begin{pmatrix}c+2d&d\\d&c-2d\end{pmatrix}$$. Its adjoint matrix is $A^*=\begin{pmatrix}c-2d&-d\\-d&c+2d\end{pmatrix}$, which satisfies: $$AA^*=A^*A=(c^2-5d^2)I_{2\times2}$$. In this case, an ordinary integer $n$ is represented by $nI_{2\times2}$, and the homomorphism in question $\theta$ is just to take the adjoint matrix. Hence, if there is another matrix $B$ of a similar form, such that $AB=nI_{2\times2}$, then $$B^*A^*=(nI_{2\times2})^*=nI_{2\times2}$$, i.e. $A^*$ also divides $n$, hence the first point.

The second point follows immediately from the first one, because, if $AB=I_{2\times2}$, then $A^*$ also divides the unity, on account of the first point, and hence $\theta(c+d\sqrt5)$ is also a unit. Moreover, from $AB=I$ and $A^*C=I$ we find that $AA^*BC=I$. But $$AA^*=\begin{pmatrix}c^2-5d^2&0\\0&c^2-5d^2\end{pmatrix}$$ is a diagonal matrix, thus $BC$ is also a diagonal matrix and $c^2-5d^2$ divides $1$. Since an integer divides $1$ if and only if it is $\pm1$, this finishes the proof.

Edit $\text{I}$
To decide the matrix that represents $\sqrt5$, notice that it must satisfy $A^2=5I_{2\times2}$. If $$A=\begin{pmatrix}\alpha&\beta\\ \gamma&\delta\end{pmatrix},$$ then
$\alpha^2+\beta\gamma=5$,
$\alpha+\delta=0$,
$\delta^2+\beta\gamma=5$. So $\alpha=-\delta$, and $\alpha^2+\beta\gamma=5$. I just chose the simplest solution, i.e. $\alpha=2$, $\beta=\gamma=1$, and $$A=\begin{pmatrix}2&1\\1&-2\end{pmatrix}$$.

Edit $\text{II}$
I shall try to explain why one could have come up with the idea of using matrices, but I am not very good in explaining such things, so forgive me if my explanations sound weak and insufficient.
First, a classical topic in elementary number theory is Bramagupta's identity, and it has a nice interpretation via matrices.. This could be considered as the first reason why the idea of matrices ever came to me.
Second, there is a proof that finite-extensions of $\mathbb Q$ are algebraic: it uses the characteristic polynomials of algebraic numbers, or, more precisely, of the matrix associated to an algebraic number. This further confirms my primitive thought that there should be a matrix-approach.
Third, in the proof of quadratic reciprocity law$^{[1]}$, by Duke and Hopkins, one spots again the use of matrices in proving something extremely important in number theory.
The last but, hopefully, not the least, one could consider field extensions as algebra-extensions. By Artin-Wedderburn theorem, one might thus guess that algebraic numbers are in some matrix algebras. This last reason indeed sounds the weakest. But it is quite natural I suppose.

Edit $\text{III}$
As explained above, the matrix used to represent $\sqrt5$ is not unique, and the choice made here is not at all standard. Indeed, if we consider the regular representation of the algebraic number field $Q(\sqrt5)$, and taking as basis $\{1,\sqrt5\}$, then the matrix turns out to be $\begin{pmatrix}0&5\\1&0\end{pmatrix}$. And this matrix is just the companion matrix of the polynomial $x^2-5$. Everything is similar except for the difference in the chioce of the particular matrix. So your conclusion should not change of course.

Hope you like the edit. :D

Tell me if any error or inappropriate point occurs. Thanks in advance.
[1] I cannot find a direct link to the pdf, but any search of quadratic reciprocity in finite groups+Duke Hopkins should do.

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I think this is elementary enough, especially when the concept of rings is here replaced by that of matrices, which is much more elementary. Moreover, it bears the benefit of visulizing the operations that were abstractly defined. Regards. –  awllower Apr 15 '13 at 15:12
    
I'd agree this is elementary if you can explain (1) how did you come up with the idea to use matrices, and specially (2) how did you decide that $\,\sqrt 5\,$ is to be "represented" by that very particular matrix, otherwise one could see this as mathematical hokus-pokus. –  DonAntonio Apr 15 '13 at 16:06
    
@DonAntonio I came up with using matrices because $Q(\sqrt5)$ is a quadratic extension of $Q$. And field extensions of course are algebra extensions, that is to say, they are simple-algebra extensions. As these are matrices extensions, I thought that there ought to be a way of employing matrices. After this, everything fits well. As for how I decided the matrix form, I just used the condition that its square is $5I$. I shall put the second point into an edit. Thanks for your attention. –  awllower Apr 15 '13 at 16:15
    
@DonAntonio Hope you enjoy the elementary edit involving only with matrix operations. Thanks for pointing this out again. –  awllower Apr 15 '13 at 16:40
    
You say "as these are matrix extensions" but I don't think this has some definite meaning. What is true is that $\,\Bbb Q(\sqrt 5)/\Bbb Q\,$ is a field extension of dimension $\,2\,$ and then one can think of matrices as representing endomorphisms of this extension. Yet it isn't still clear to me how, why and from where you can see elements in this extension as matrices. Your explanation as why that matrix respresents $\,\sqrt 5\,$ is nice, thanks. –  DonAntonio Apr 15 '13 at 17:26

Using standard notation $\rm\:\alpha' = \theta(\alpha),\:$ let's check the multiplicative property

$\qquad\begin{eqnarray}\rm (a-b\sqrt{5})(c-d\sqrt{5}) &=&\rm\ \ ac+5bd - (ad+bc)\sqrt{5}\\ &=&\rm (ac+5bd + (ad+bc)\sqrt{5})'\\ &=&\rm ((a+b\sqrt{5})(c+d\sqrt{5}))'\end{eqnarray}$

i.e. $\rm\ \alpha'\beta' = (\alpha\beta)',\:$ so $\rm\:\alpha\mid n \:\Rightarrow\: \alpha\beta = n\:\Rightarrow\: \color{}{\alpha'\beta' = (\alpha\beta)'}= n'\Rightarrow\: \alpha'\mid n' = n,\:$ by $\rm\:n\in\Bbb Z$

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