GCD in a subring is GCD in a bigger ring?

Question

Let $R$ be a UFD which is a subring of an integral domain $S$. If $r_1$ and $r_2$ are two nonzero elements of $R$ with GCD $d$, is it true that $d$ is also a GCD of $r_1$ and $r_2$ in $S$?

I know this is true if $R$ is a PID.

Answer 1

No, e.g. $\rm\:\gcd(2,x) = 1\:$ in $\rm\:\mathbb Z[x]\:$ but the gcd is the nonunit $\:2\:$ in $\rm\:\mathbb Z[x/2]\subset \mathbb Q[x]$.

But gcds in a PID D persist in extension rings because the gcd may be specified by the solvability of (linear) equations over D and such solutions always persist in extension rings, see here.

Answer 2

Let $K$ be a field. First note, that $K[X,Y] \cong K[X,XY]$, via $X \mapsto X, \, Y \mapsto XY$, so $\gcd_{K[X,XY]}(X,XY) = 1$. Now consider the canonical inclusion $\iota \colon K[X,XY] \hookrightarrow K(Y)[X]$. Obviously $\gcd_{K(Y)[X]}(X,XY) = X \not \sim_{K(Y)[X]} 1$. Note that $K[X,XY]$ is a UFD and $K(Y)[X]$ is even a PID.

Answer 3

Warning: you must specify the subring (which may be the whole ring) relative to which the divisibility is defined.

Suppose that $R$ is a subring of an integral domain $S$, and let $R^*$, $S^*$ denote the sets of nonzero elements of $R$ resp. $S$. $\newcommand{\divides}{\mid}$ We say that $s\in S^*$ divides $s'\in S^*$ relative to $R$, and write $s\divides_R s'$, if there exists $r\in R^*$ such that $sr=s'$. If $s,s'\in S^*$, then $s\divides_R s'$ and $s'\divides_R s$ if and only if $s'=su$ for some unit $u$ of $R$; we write this (equivalence) relation as $s\sim_R s'$. $\newcommand{\set}[1]{{\{#1\}}}$ For any $s_1,s_2\in S^*$ we denote by $\gcd_R(s_1,s_2)$ a greatest lower bound (provided it exists) of $\set{s_1,s_2}$ in the preordered set $(S^*,{\divides_R})$; $\gcd_R(s_1,s_2)$ is determined only up to a factor in $R^\times$, the group of units of $R$.

If the divisibility in your question means the divisibility relative to the whole ring, both for $R$ and $S$, then the answer is no. $\newcommand{\ZZ}{\mathbb{Z}}$ $\newcommand{\QQ}{\mathbb{Q}}$ For example, let $R=\ZZ$ and $S=\QQ$. Then $\gcd_\ZZ(4,6)\sim_\ZZ 2\nsim_\ZZ 1$ in $(\ZZ,{\divides_\ZZ})$ and $\gcd_\QQ(4,6)\sim_\QQ 1$ in $(\QQ,{\divides_\QQ})$; but we do have $\gcd_\ZZ(4,6)\sim_\ZZ 2\nsim_\ZZ 1$ in $(\QQ,{\divides_\ZZ})$. (Added next day: actually this is not a counterexample -- se below.)

If the divisibility in your question is relative to $R$, in $R$ as well as in $S$, then the answer is yes. This fact is best proved in a more general context of GCD monoids, since the additive structure in rings is immaterial for your question -- actually it is a nuisance, always under our legs when we neither want it nor need it. I will give you the proof tomorrow.

Continued. $~$WHOA! I gave a non-counterexample, since also $\gcd(4,6)_\QQ\sim_\QQ 2$ because $2\sim_\QQ 1$ in $(\QQ,{\divides_\QQ})$. Such is a life of a mathematician: doing mathematics I am wasting great deal of time getting myself into dead ends and then backing out of them, and part of the time I am making mistakes, some of them unbelievably stupid. I will let the mistake stay, it has a didactic value.

Now let me see if I understand your question. We have a UFD $R$ which is a subring of an integral domain $S$. If $r_1$, $r_2$ are any nonzero elements of $R$ with a GCD $d$ relatively to $R$, then the question is whether $d$ is also a GCD of $r_1,r_2\in S$ relatively to $S$. Am I right? Is this the question? Let us assume it is.

In the UFD $R$ every pair of elements has a GCD (even when one of them, or both, are zero, since $\gcd_R(r,0)=r$ for every $r\in R$), that is, $R$ is a GCD domain. Nothing is said about the existence of GCDs for all pairs of elements of $S$; however, the question does not ask this, it just queries whether any two nonzero elements $r_1,r_2\in R\subseteq S$, which we know have a GCD $d$ in $R$, have this same $d$ as a GCD in $S$. This is indeed true if $R$ is a PID, for in this case $d=r_1r_1'+r_2r_2'$ for some $r_1',r_2'\in R$; if $c\in S$ is a common divisor of $r_1$, $r_2$ in $S$, then $r_1=cs_1$ and $r_2=cs_2$ for some $s_1,s_2\in S$, hence $d=c(s_1r_1'+s_2r_2')$, which shows that $c$ divides $d$ in $S$. The additive structure of rings does have
a role in this case.

Here comes a genuine counterexample for an $R$ that is only a UFD, not a PID. Let $X$, $Y$, $Z$ be formal variables, and consider the rings $R:=\ZZ[X,Y]$ and $S:=\ZZ[X,Y,Z,XZ^{-1},YZ^{-1}]$;
$R$ is a UFD, $S$ is an integral domain because it is a subring of the field $\QQ(X,Y,Z)$, and $R$ is a subring of $S$. We have $\gcd_R(X,Y)\sim_R 1$. The elements $X$ and $Y$ of the ring $S$ have a common divisor $Z$ in $S$; but $Z$ does not divide $1$ in $S$, thus $1$ is not a GCD of $X$ and $Y$ in $S$. At this point it is already clear that we have a counterexample, but we ask a related question: do $X$ and $Y$ have a GCD in $S$? In principle it might happen that $\gcd_S(X,Y)$ would not exist; but in our case $\gcd_S(X,Y)$ exists and is equal to (that is, it is $\sim_S$ to) $Z$ (you may enjoy proving this).

Ha! We do have a stronger question: do there exist a UFD $R$ and an integral domain $S$ containing $R$ as a subring, such that there are nonzero $r_1$, $r_2$ in $R$ that do not have a GCD in $S$ (this GCD is meant relatively to $S$)? Right now I cannot see how I would construct an example.

About the promised proof in the context of GCD monoids: this proof answers a question that you did not ask (but which I mistakenly understood to be the question you are asking), so I will not bother you with it.

(Added a little later.) Here is an example for the related stronger question; it is an evident extension of the counterexample for the original question. Let $X$, $Y$, $U$, $V$ be formal variables, and set $R:=\ZZ[X,Y]$, $S:=\ZZ[X,Y,U,V,X/U,Y/U,X/V,Y/V]$. $~$If $e\in S$ is a common divisor of
$X$ and $Y$ in $S$, then $e\sim_S 1$ or $e\sim_S U$ or $e\sim_S V$, and the set $\set{1,U,V}$ does not have a greatest element in the preordered set $(S,{\divides_S})$.