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 $E$?
Extending ideals from principal ideal domains
2 Answers
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.
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0Lots of thanks to point out my mistake. – 2011-02-02
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0Sorry 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 $e\in E$? – 2012-07-17
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1@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. – 2012-07-18
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0@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 – 2015-04-08
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0@user136266: Write $d = xa+yb$. Then if $e$ divides $a$ and $e$ divides $b$, then $e$ divides $d$. – 2015-04-08
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 $\rm\:u,v\in D.$
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0Hi, can I know which part of your proof uses the fact that D is a PID? It seems that PID is not used. – 2015-04-08
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0@user136266 It assumes $\,\rm (a,b) = (c)\,$ is principal to derive the equational specification for the gcd. – 2015-04-08
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0so 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 – 2015-04-08
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0@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.\ \ $ – 2015-04-08