Let $R$ be subring of integral domain $S.$ Suppose $R$ is $\text{PID}.$ Let $a\in R$ be a greatest common divisor of $r_1,r_2$ in $R$. ($r_1,r_2 \in R$, not both zero). Could anyone advise me on how to prove $a$ is greatest common divisor of $r_1,r_2$ in $S?$

Hints will suffice, thank you.

  • $\begingroup$ Hint: there exists $s_1,s_2\in R$ such that $a=r_1s_1+r_2s_2$ and also notice that $a|r_1,r_2$ in $R$. $\endgroup$ – user119882 Mar 15 '14 at 15:49
  • $\begingroup$ Thank you. This means $(a) = (r_1) + (r_2)$ in $R$ and hence $(a) = (r_1) +(r_2)$ in $S.$ Let $(z) \subseteq S$ such that $r_1,r_2 \in (z).$ For any $x,y \in S, \ r_1x+r_2y \in (z),$ so $(a) \subseteq (z)$ in $S.$ Now, if $R$ is $\text{UFD}$ but not $\text{PID},$ how do I disprove the statement? $\endgroup$ – Alexy Vincenzo Mar 15 '14 at 16:02
  • $\begingroup$ I'm sorry, how you define $k[r_1, r_2]?$ $\endgroup$ – Alexy Vincenzo Mar 15 '14 at 16:14
  • $\begingroup$ So you need a counterexample, $k[xy,xz]\subset k[x,y,z]$($k$ is a field, and $k[x,y,z]$ is the polynomial ring over $k$ in three variables $x,y,z$, and $k[xy,xz]$ is the subring of $k[x,y,z]$ generated by $xy,xz$ and the field $k$). Now set $r_1=xy,r_2=xz$. $\endgroup$ – user119882 Mar 15 '14 at 16:20
  • $\begingroup$ $k[r_1,r_2]=\{f(r_1,r_2)| f(t_1,t_2)\in k[t_1,t_2]\}$ where $k[t_1,t_2]$ is the polynomial ring in two indeterminates over $k$. $\endgroup$ – user119882 Mar 15 '14 at 16:22

Hint $ $ gcds in PIDs 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, i.e. roots in D remain roots in rings $\rm\,R \supset D.\:$ More precisely, the Bezout identity for the gcd yields the following ring-theoretic equational specification for the gcd

$$\begin{eqnarray} \rm\gcd(a,b) = c &\iff&\rm (a,b) = (c)\\ &\iff&\rm a = c\:\color{#C00}x ,\ b= c\:\color{#C00} y ,\,\ a\:\color{#C00} u + b\: \color{#C00}v = c\ \ has\ roots\ \ \color{#C00}{x,y,u,v}\in D\end{eqnarray}$$

Proof $\ (\Leftarrow)\:$ In any ring $\rm R,\:$ $\rm\:a\: x = c,\ b\: y = c\:$ have roots $\rm\:x,y\in R$ $\iff$ $\rm c\ |\ a,b\:$ in $\rm R.$ Further if $\rm\:c = a\: u + b\: v\:$ has roots $\rm\:u,v\in R\:$ then $\rm\:d\ |\ a,b$ $\:\Rightarrow\:$ $\rm\:d\ |\ a\:u+b\:v = c\:$ in $\rm\: R.\:$ Hence we infer $\rm\:c = gcd(a,b)\:$ in $\rm\: R,\:$ being a common divisor divisible by every common divisor. $\ (\Rightarrow)\ $ If $\rm\:c = gcd(a,b)\:$ in D then the Bezout identity implies the existence of such roots $\rm\:u,v\in D.\ $ QED

Rings with such linearly representable gcds are known as Bezout rings. As above, gcds in such rings always persist in extension rings. In particular, coprime elements remain coprime in extension rings (with same $1$). This need not be true without such Bezout linear representations of the gcd. For example, $\rm\:\gcd(2,x) = 1\:$ in $\rm\:\mathbb Z[x]\:$ but the gcd is the nonunit $\:2\:$ in $\rm\:\mathbb Z[x/2]\subset \mathbb Q[x]$.

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  • $\begingroup$ Note: the gcd "equalities" should really by "associate to" but I'd like the answer to stay accessible to beginners so I leave that simple translation for more advanced readers. $\endgroup$ – Gone Sep 1 '19 at 0:30

Here is another proof.

Let $()_S$ and $()_R$ denote the ideal generated by the thing inside the parentheses in $S$ and $R$ respectively. As $a$ is a GCD of $r_1$ and $r_2$ in $R$, $(a)_R = (r_1,r_2)_R$. Now if $e\mid r_1$ and $e\mid r_2$ in $S$, then $(r_1,r_2)_S \subset (e)_S$. Obviously, $(r_1,r_2)_R \subset (r_1,r_2)_S$, so we get $(a)_R \subset (e)_S$, thus $a \in (e)_S$ so $e \mid a$ in $S$. And of course $a\mid r_1$ and $a\mid r_2$ in $S$. Therefore $a$ is a GCD in $S$ as well.

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