Euclidean Algorithm 
\(\mathbb{Z}\left[i\sqrt{5}\right]\)
After having seen a generalization of the division algorithm and of the Euclidean algorithm to \(\mathbb{Z}\left[i\right]\)
and \(\mathbb{Z}\left[\sqrt{2}\right]\),
one can think of generalizing to other cases. For example, can we do the same for \(\mathbb{Z}\left[\sqrt{3}\right]\)?
\(\mathbb{Z}\left[i\sqrt{2}\right]\)?
The answer is yes: in \(\mathbb{Z}\left[\sqrt{3}\right]\)
one can define \(v\left(a+b\sqrt{3}\right)=\left|a^{2}-3b^{2}\right|\),
in \(\mathbb{Z}\left[i\sqrt{2}\right]\)
one can define \(v\left(a+bi\sqrt{2}\right)=a^{2}+2b^{2}\)
and things are very similar. Can we then do the same in \(\mathbb{Z}\left[\sqrt{n}\right]\)
and \(\mathbb{Z}\left[i\sqrt{n}\right]\),
for every positive integer \(n\)?
Here the answer is no; it is quite easy to see why doing things the same way doesn't work, but it is less easy to be sure that there is no way that works.
Concretely, it is easy to see that a certain function \(v\)
doesn't give a division algorithm, but it is more difficult to justify why there is no
\(v\) function
which gives a division algorithm.
Here we will see only an idea of such a justification for the case of \(\mathbb{Z}\left[i\sqrt{5}\right]\).
If there were a division algorithm in \(\mathbb{Z}\left[i\sqrt{5}\right]\),
then there would be an Euclidean algorithm for such division algorithm. But then every two elements of \(\mathbb{Z}\left[i\sqrt{5}\right]\)
would have a greatest common divisor (in the sense of a common divisor which is a multiple of all the others).
But \(6\) and \(2+2i\sqrt{5}\) don't have a \(gcd.\)
The divisors
of \(6\)
in \(\mathbb{Z}\left[i\sqrt{5}\right]\) are: \[\begin{array}{c}
1,2,3,6,1+i\sqrt{5},1-i\sqrt{5},\\
-1,-2,-3,-6,-1-i\sqrt{5},-1+i\sqrt{5}
\end{array}.\]
The divisors of
\(2+2i\sqrt{5}\) in \(\mathbb{Z}\left[i\sqrt{5}\right]\) are:
\[\begin{array}{c} 1,2,1+i\sqrt{5},2+2i\sqrt{5},\\ -1,-2,-1-i\sqrt{5},-2-2i\sqrt{5}
\end{array}.\]
Hence the common divisors of \(6\) and \(2+2i\sqrt{5}\) are: \[1,2,1+i\sqrt{5},-1,-2,-1-i\sqrt{5}\]and none of them is a multiple of (all) the others.
