5

I have been interested in whether we would be able to find a strict totally ordered commutative ring (thus an integral domain) that is not a GCD domain? I'm defining a strict totally ordered commutative ring as follows:

$(R,+,*,<,0,1)$ is a strict totally ordered commutative ring iff

  1. $(R,+,*,0,1)$ is a ring with multiplicative identity.
  2. $*$ is a commutative operation on $R$.
  3. $(R,<)$ is a strict toset (strict totally ordered set) so $<$ is irreflexive, transitive, asymmetric and connected (that is any two elements are comparable or equal).
  4. For all $x,y,z\in R$, $x<y$ implies $x+z<y+z$.
  5. For all $x,y\in R$, $0<x$ and $0<y$ implies $0<x*y$

We can prove from these axioms that the cancellation law for multiplication holds that For all $x,y,z \in R$ we have that $x*z=y*z$ implies $x=y$. So the Ordered Ring is also an Integral Domain. But does this tell us anything about whether the ring is also GCD domain? Are there any examples of Rings that can form strict totally ordered commutative rings but are not GCD domains?

Pymamba
  • 346

2 Answers2

5

You can take any non integrally closed subring of the reals. For example, you can take $A=\mathbb{Z}[\sqrt{d}]$, with $d>0$, $d\equiv 1 \ [4].$

It is not integrally closed since $\dfrac{1+\sqrt{d}}{2}$ is integral over $\mathbb{Z},$ hence over $A$, but does not lie in $A$.

UFD are integrally closed, so $A$ is not a UFD, and thus not a gcd domain since it is noetherian and has in particular ACCP (see the answer of Amateur_algebraist).

To make it explicit: take $d=5$, $a=4, b=2(1+\sqrt{5})$ .

Then, $a$ and $b$ has no gcd.

Indeed, assume that $d$ is such a gcd. Since $2$ divides $a$ and $b$, $2\mid d $ and $d=2c$.But $d$ divides $a$ and $b$, so $c$ divides $2$ and $1+\sqrt{5}$.

But these two elements are irreducible:

If $z=x+y\sqrt{5}$, set $N(z)=\vert x^2-5y^2\vert$.

$N(2)=N(1+\sqrt{5})=4$, so they are nonzero and non invertible ( a unit of $A$ has norm $1$).

A potential divisor of $2$ or $1+\sqrt{5}$ has norm $1,2,4$. In the first case, it is a unit, in the third one, the other factor is a unit, and the equation $N(z)=2$ has no solution, because $\pm 2$ is not a square mod $5$.

Note that $2$ and $1+\sqrt{5}$ are not associate (otherwise in particular, $b/a$ shoud be in $A$, which is not the case). Hence, they are coprime, meaning that $c$ is a unit. But now $1+\sqrt{5}$ is a common divisor, so it should divide $d=2c$, and thus it should divide $2$, since $c$ is a unit. This is not the case.

GreginGre
  • 16,641
3

Let $K = \Bbb Q(\sqrt{d})$ be a real quadratic number field with class number greater than $1$. (Wikipedia says you can take $d = 229$ or $257$.) Then its ring of integers $\mathcal O_K$ is:

  1. strict totally ordered, being a subring of $\Bbb R$;
  2. is not a PID or UFD, but has ACC on principal ideals: see here.

A commutative ring is a UFD $\Leftrightarrow$ it is a GCD domain and has ACC on principal ideals.

Amateur_Algebraist
  • 1,925