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We know $R=\{a+bi\sqrt{5}: a,b \in \mathbb{Z}\}$ is not a UFD because, for example, you can factor

$$6=(1+i\sqrt{5})(1-i\sqrt{5})=(2)(3)$$

and these are two distinct factorizations into irreducibles. This is can be shown using the norm $N:R \to \mathbb{N}$ defined $N(a+bi\sqrt{5})=a^2+5b^2$ and the fact that it's multiplicative, i.e. $N(rs)=N(r)N(s)$.

Since $R$ is not a UFD, $R$ is not a Euclidean domain either. How do we find an example to prove that our norm is not a Euclidean function? That is, how do we find a pair $a,b \in R$ for which there does not exist a pair $q,r \in R$ such that $$b=aq+r$$

with $N(r)<N(a)$?

I would guess that the above factorization I gave is a place to start, but having looked at that for a bit, I still don't see where to begin.

This is a problem I'm working on in reviewing for a test, not a current homework problem.

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up vote 5 down vote accepted

Yes, as you surmised one can find an example using numbers mentioned above. Let $b=1+i\sqrt{5}$, and let $a=2$. The norm of $2$ is $4$, so we want to show there are no $q$ and $r$ such that $1+i\sqrt{5}=2q+r$, and the norm of $r$ is less than $4$. There are very few numbers $r$ of norm $<4$. They are $0$ and $\pm 1$. We could deal with them individually, but take a faster approach.

Let $q=x+iy\sqrt{5}$. Then if $1+i\sqrt{5}=2q+r$, we have $$r=(1-2x)+ i(1-2y)\sqrt{5}.$$ Note that $1-2x$ and $1-2y$ are both odd, so each has absolute value $\ge 1$. It follows that the norm of the right-hand side is at least $(1)^2+(1)^2(5)=6$, and in particular cannot be $<4$.

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Hint: nonunit $a$ has min norm $\Rightarrow a$ is prime, by remainders mod $a$ have norm $0$ or $1$, so are $0$ or units, so $R/a$ is a field. But nonunit $2$ has min norm, and $2$ isn't prime in $R$ by your first equation.

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