# In $k[X,Y,Z,W]/(XW−YZ)$, there is no GCD of $xw$ and $xy$

I wanted to find an integral domain in which we do not have the property that any two elements have a greatest common divisor (gcd).

Somewhere on this forum someone claimed that the following quotient ring of the polynomial ring in four variables over a field $k$ is such a ring: $A:=k[X,Y,Z,W]/(XW−YZ)$. I would appreciate some help in proving this. We denote the classes of $X,Y,Z,W$ in $A$ by $x,y,z,w$.

1) $A$ is an integral domain
It suffices to prove that $XW−YZ$ is irreducible in $k[X,Y,Z,W]$. But this holds with the Eisenstein criterion if we consider $XW−YZ$ as an element in $k[Y,Z,W][X]$.

2) The elements $xw$ and $xy$ have no gcd in $A$.
Obviously the idea is to use that (because of the relation $xw=yz$), both $x$ and $y$ are common divisors of $xw$ and $xy$. But I'm having a hard time proving rigorously that there cannot exist a gcd.

• If $d=\gcd(xw,xy)$ then $d=xd'$ (since $x\mid d$) and therefore $d'=\gcd(w,y)$. But in this answer I showed that $x,y,z,w$ are irreducible, so $d'=1$. We get $d=x$. On the other side, $y\mid xw(=yz)$ and $y\mid xy$, so $y\mid x$, a contradiction. Jun 22 '17 at 21:26
• At OP: the argument sketched in the comment of @user26857 essentially applies a general form of Euclid's Lemma, viz $\ (y,w)=1,\ y\mid xw\,\Rightarrow\, y\mid x\ \ {\bf if}\ \ (xw,xy)\,$ exists, see this answer. As I explain there, failure of Euclid's Lemma yields such nonexistant gcds (this is a generic example of such failure). Jun 23 '17 at 20:34
• Many thanks for the arguments above! I will now try to put all the pieces together and then post it here Jun 24 '17 at 14:04

Here is a proof I put together using the above arguments of @user26857.

First I state all the lemmas we will use in the proof. We use the notation $(a,b)$ for $\gcd(a,b)$.

Lemma $1$: Let $R$ be an integral domain and $a,b\in R$ two irreducible elements that aren't associates. Then $(a,b)=1$.
Euclid's Lemma: Let $R$ be an integral domain and $a,b,c\in R$. If $(ac,bc)$ exists, then $(a,b) = (ac,bc)/c$.
Claim $2$: $y\nmid x$ in our Ring $A$. In particular, $y$ and $x$ aren't associates.
Claim $3$: $x,y,z,w$ are irreducible in our ring $A$.

Proofs for Euclid's Lemma and Claim $3$ were given by @Bill Dubuque and @user26857 in the comments to my question. Lemma $1$ is trivial.
Proof of Claim $2$:
Suppose $y\mid x$, i.e. that there exists a polynomial $f$ such that $y\cdot f(x,y,z,w)=x$. Then $Y\cdot f(X,Y,Z,W)-X\in (XW-YZ)$. By plugging in $0$ for $Y,Z,W$ we get $0\cdot f(X,0,0,0)-X=0$, a contradiction.

Finally, we prove that $d=(xw,xy)$ doesn't exist. If $d=(xw,xy)$ then $d=xd′$ (since $x \mid d$) and therefore by Euclid's Lemma $d′=(w,y)$. Now, $y$ and $w$ are both irreducible by Claim $3$ and not associates by Claim $2$ (just replace $x$ with $w$ in proof of Claim $2$), so by Lemma $1$ we get $d′=1$. This means $d=x$. On the other side, $y∣xw(=yz)$ and $y∣xy$, so $y∣x$, a contradiction to Claim $2$ and we are done.

• 1) I'm not sure if that result is called "Euclid's Lemma". 2) Irreducibility of $x,y,z,w$ is given in a linked thread, not in the comment. Jun 24 '17 at 22:20

I would suggest you denote the classes of $X,Y,Z,W$ by $x,y,z,w.$ Otherwise, it's difficult to know in which ring you perform computations.

I didn't check the details, but it seems that $A$ may be identified to a subring of $B=k(Y)[X,W]$ (namely $k[X,Y,XW/Y,W]$), by sending $Z$ to $XW/Y$. The latter ring is a UFD. If $xw$ and $xy$ have a gcd $d=P(x,y,z,w)$, then $D=P(X,Y,XW/Y,W)$ would be a gcd of $XW$ and $XY$ in $B$. Hence, $D$ would be of the form $F(Y)X$, and I'm sure you can derive a contradiction from here.

Otherwise, I can provide you another kind of example: take $A=\mathbb{Z}[i\sqrt{5}]$. This is an integral domain.

One can show that the common divisors of $9$ in $A$ are $\pm 1,\pm 3,\pm 9,\pm (2+i\sqrt{5}), \pm (2-i\sqrt{5})$.

(Hint: solve the equations $\vert z\vert^2=1,3$ or $9$ , for $z\in A.$)

I let you deduce from the latter that $9$ and $3(2+i\sqrt{5})$ have no gcd in $A.$

(HInt: find the possible common divisors among the list above, and check that none of them satisfy the definition of a gcd)

• I will try to work this out now and then report my progress. Jun 21 '17 at 13:41
• Ok. I'll try ot have a closer look to your example. If I find a simpler method, I'll get back to you. Jun 21 '17 at 13:48
• 1) The OP didn't ask for an alternative example. 2) I have serious doubts that your argument is correct. Basically you say that if $A\subset B$ is a ring extension, and $B$ a UFD then $\gcd(a_1,a_2)$ in $A$ is also a $\gcd$ in $B$. One can easily find counterexamples to this claim. Jun 22 '17 at 22:14