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Let $D$ be a PID, $E$ a domain containing $D$ as a subring. Is it true that if $d$ is a gcd of $a$ and $b$ in $D$, then $d$ is also a gcd of $a$ and $b$ in $E$?

Zev Chonoles
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gottigen
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2 Answers2

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In any domain $D$, for $a,b \in D \setminus \{0\}$, if the ideal $\langle a,b \rangle_D = \{xa + yb \ | \ x,y \in D\}$ of $D$ is principal, then any generator $d$ of the ideal is a gcd of $a$ and $b$. (Note that in general gcd's are unique precisely up to units, i.e., the corresponding principal ideal is unique.)

So if $D$ is a PID, there is $d \in D$ such that $\langle a,b \rangle_D = dD$. Now push forward these ideals to $E$:

$\langle a,b \rangle_E = \langle a,b \rangle_D E = (d D) E = dE$.

Thus the ideal of $E$ generated by $a$ and $b$ is still principal and still generated by the same element $d$ (now well-determined up to a unit of $E$; note that the unit group of $E$ could be larger than the unit group of $D$).

So in summary: yes.

Pete L. Clark
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  • Lots of thanks to point out my mistake. – gottigen Feb 02 '11 at 15:43
  • Sorry to bother, but how do you see $\langle a,b\rangle_E\subset\langle a,b\rangle_D E$? If the elements of $\langle a,b\rangle_E$ are of form $xa+yb$ for $x,y\in E$, how can they be put in form $(ua+vb)e$ for $u,v\in D$ and $e\in E$? – Son Bi Jul 17 '12 at 22:23
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    @Son Bi: The ideal $\langle a,b \rangle_D E$ is the set of all sums of elements $x_i e_i$ with $x_i \in \langle a,b \rangle_D$ and $e_i \in E$. Because $\langle a,b \rangle_E$ and $\langle a,b \rangle_D E$ are both ideals of $E$, to show the first is contained in the second it is enough to show that $a,b \in \langle a,b \rangle_D E$, which is clear. – Pete L. Clark Jul 18 '12 at 07:39
  • @PeteL.Clark Hi how do i ensure that the ideal of E generated by a and b is maximal in E such that d is also the gcd of a, b in E. From what I see your proof only shows that d is a common divisor of a, b. you do state that it is. But I can't see why. Thanks – user10024395 Apr 08 '15 at 02:21
  • @user136266: Write $d = xa+yb$. Then if $e$ divides $a$ and $e$ divides $b$, then $e$ divides $d$. – Pete L. Clark Apr 08 '15 at 05:37
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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, 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

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

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 $\rm\:u,v\in D.$

Bill Dubuque
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  • Hi, can I know which part of your proof uses the fact that D is a PID? It seems that PID is not used. – user10024395 Apr 08 '15 at 03:17
  • @user136266 It assumes $,\rm (a,b) = (c),$ is principal to derive the equational specification for the gcd. – Bill Dubuque Apr 08 '15 at 03:34
  • so if it is not a pid, $(a,b) \neq (c)$? What does gcd mean then in that case? I have always thought that (a,b)=(c) even when it is not a PID – user10024395 Apr 08 '15 at 04:03
  • @user136266 Generally a gcd is a common divisor that is divisible by every common divisor; equivalently $\ c\mid a,b\iff c\mid \gcd(a,b)., $ gcds exist in any UFD, which needn't be a PID, e.g. $,\Bbb Q[x,y],,$ where $,\gcd(x,y)=1,$ but $,(x,y)\neq (1),,$ since $, xf + y g = 1,$ $\Rightarrow$ $,0 = 1,$ by evaluating at $, x = 0 = y.\ \ $ – Bill Dubuque Apr 08 '15 at 13:33