Definition: A universal side divisor, is an element $s\in R\setminus R^\times$ such that for every $x\in R$ either $s\mid x$, or there is some unit $u\in R^\times$ such that $s\mid x+u$.
Fact: A Euclidean domain $R$ has universal side divisors.
Proof: Let $s$ be a non unit of minimal norm $N(s)$. Let $x\in R$ and write $$x=sq+r$$ Then either $r=0$, and so $s\mid x$, or $N(r)<N(s)$. But since $N(s)$ is minimal we must have $r$ a unit, so $s\mid x-r$. Thus $s$ is a universal side divisor.
I'm doing the following excercises:
Suppose $R=\Bbb Z[\alpha]$, where $\alpha=\frac{-1+\sqrt{-19}}{2}$. Prove that $R$ is not a Euclidean domain by showing that it has no universal side divisors.
a) Find the units of $R$.
b) Take the usual norm $N$ on $R$, $N(z)=z\bar z$. Show that the only elements of norm less than $5$ are $\{0,\pm 1,\pm 2\}$.
c) Taking $x=2$ in the definiton of a universal side divisor, show that any universal side divisor must be a non-unit divisor of $2$ or $3$.
d) Find all non-unit divisors of $2$ and $3$.
e) Now taking $x=\alpha$ in the definition of a universal side divisor, find a contradiction.
What I would like to know is:
- Is my proof correct?
- Why is the step b) needed?
Here we go:
a) Find the units of $R$.
Let $N(a+b\alpha)=a^2-ab+5b^2=1$, there are two cases to consider:
- $a\geq b$, then $1\geq 5b^2$, so $b=0$ and $a=1$
- $b\geq a$, then $1\geq a^2+4b^2$, so again $b=0$ and $a=-1$.
Thus the only units are $\pm 1$.
b) Take the usual norm $N$ on $R$, $N(z)=z\bar z$. Show that the only elements of norm less than $5$ are $\{0,\pm 1,\pm 2\}$.
Again, there are two cases to consider:
- $a\geq b$, then $5>a^2-ab+5b^2\geq 5b^2$, so $b=0$, and $a=0,1,2$
- $b\geq a$, then $5>a^2-ab+5b^2\geq a^2+4b^2$, so $b=0$, and $a=0,-1,-2$
Thus the elements $0,\pm 1, \pm 2$ is the only elements with norm less than $5$.
c) Taking $x=2$ in the definiton of a universal side divisor, show that any universal side divisor must be a non-unit divisor of $2$ or $3$.
Since $R^\times=\{\pm 1\}$ a universal side divisor $s$ must divide one of $2, (2+1), (2-1)$. Since $s$ is not a unit, we must have $s\mid 2$ or $s\mid 3$.
d) Find all non-unit divisors of $2$ and $3$.
Both $2,3$ are irreducible in $R$. To see this assume that $N(2)=N(x)N(y)=4$, and $N(3)=N(z)N(w)=9$. Is is then enough to show that $N(x),N(y)\neq 2$ and $N(z),N(w)\neq 3$. There are again two cases:
- If $a\geq b$, then $a^2+4b^2\geq 2,3\geq 5b^2$, and so $b=0$ and $a^2=2,3$, which is impossible.
- If $b\geq a$, then $5b^2\geq 2,3\geq a^2+4b^2$, and so $b=0$. Contradiction.
So both $2,3$ are irreducible.
e) Now taking $x=\alpha$ in the definition of a universal side divisor, find a contradiction.
If $R$ is a Euclidean domain, then there is a universal side divisor $s$ such that $s\mid 2$ or $s\mid 3$. As both $2,3$ are irreducible we have $s=\pm 2,\pm 3$. Therefore $N(s)=4,9$. Also $s$ divides one of $\alpha,\alpha-1,\alpha+1$. Computing norms we have:
- $N(\alpha)=5$
- $N(\alpha+1)=5$
- $N(\alpha-1)=6$
As $4$ or $9$ divides none of them, we must have that $s$ doesn't divide any of $\alpha,\alpha-1,\alpha+1$. Contradiction. Therefore $R$ is not a Euclidean domain.
