Some Results in $\mathbb{Z} [\sqrt{10}]$ This is a question from an old Oxford undergrad paper on calculations in $\mathbb{Z} [\sqrt{10}]$. We equip this ring with the Eucliden function $d(a+b\sqrt{10})=|a^2-10b^2|$. I want to prove the following results:


*

*If $d(x)=1$, then $\frac{1}{x} \in \mathbb{Z} [\sqrt{10}]$

*Any non-zero element of $\mathbb{Z} [\sqrt{10}]$ which is not a unit can be expressed as a product of finitely many irreducibles in $\mathbb{Z} [\sqrt{10}]$

*The ideal generated by $2$ and $\sqrt{10}$ is not principal in $\mathbb{Z} [\sqrt{10}]$


Thought so far


*

*Suppose $x=a+b\sqrt{10}$. Clearly if $x$ is a unit then $d(x)=1$, though I'm not sure if this helps. Are we OK simply to note that $\frac{1}{x}=\frac{a-b\sqrt{10}}{a^2-10b^2}$ and since $d(x)=1$ then the deonminator is either $1$ or $-1$.

*I know this is true in general in a principal ideal domain and every Euclidean ring is a principal ideal domain, but this proof is lengthy. Is there any calculation one can perform in $\mathbb{Z} [\sqrt{10}]$ to demonstrate this property more quickly.

*Any help would be appreciated; I'm not actually too sure what this ideal looks set. Could someone put it in a set notation for me?
Many thanks.
 A: It seems to me that (1) and (3) are dealt with adequately in the comments (but I would be happy to incorporate that here if you need). For (2), the key point is that the ring is Noetherian, which is effectively a consequence of the Hilbert basis theorem, implying that a polynomial ring $\mathbb{Z}[x]$ is Noetherian, and the fact that a quotient ring of a Noetherian ring is Noetherian. Given this, every element can be written as a product of irreducibles, since otherwise you obtain an infinite ascending chain of ideals.
Here are some more details for (3): First, observe that $d(xy)=d(x)d(y)$ for all $x,y$ and that $d(2)=4$ and $d(\sqrt{10})=10$. Thus any common divisor $x$ has $d(x)|2$. If $x$ is a non-unit this implies $d(x)=2$. Thus assuming a non-unit common divisor $x$ exists, there are integers $a,b$ with
$$\pm 2=a^2-10b^2.$$ Consider this equation modulo $10$. It implies that either $2$ or $8$ is a square mod $10$, but the squares modulo $10$ are just $0,1,4,9,6,5$. Hence there is no non-unit common divisor $x$. On the other hand, the ideal is proper since its elements are all of the form $a+b\sqrt{10}$ with $a$ even. Therefore it is not principal.
A: The set Z[10] makes use of half-primes, which must occur in even numbers.  This is pretty normal for many of these sorts of systems.  These half-primes live in some other part that have this same ratio, ie $a\sqrt{5}+b\sqrt{2}$.  In some systems, like z[3], there are even uits in there, eg $(\sqrt{6}+\sqrt{2})/2$. 
In the case of Z[10] (and many other schemes), there are subspaces that hold primes, but there is no unit to move to that space.  
An example is that of Z[210], which has three subspaces $a\sqrt{3}+b\sqrt{70}$ and $a\sqrt{2}+b\sqrt{105}$ and $a\sqrt{6}+b\sqrt{35}$, and a unit $\sqrt{15}+\sqrt{14}$, which transitions to another set of subspaces.  This group has no fewer than eight sub-spaces, linked in pairs, and primes can be individually found in any of the main group, or the three subgroups (3), (2), (6).  A number appears in the main group, if the product of the special primes of the subgroup, multiply up to a square.  For example the prime factors for 29 are $\sqrt{35}+\sqrt{6}$. and for 7, one can find it in $\sqrt{105}+7\sqrt{2}= \sqrt{7}*(\sqrt{15}+\sqrt{14})$, but since 6*2 gives 12, one needs a further '3' type prime to make it appear in Z[210].
That is pretty much the order of the day for composite numbers in this type.
