Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $R$ be a Euclidean domain, and let $r_{1}$, $r_{2}$, $r_{3}, \ldots,r_{n}$ be (distinct) elements of $R$. Prove that there are elements $a_{1}$, $a_{2}$, $a_{3},\ldots,a_{n}$ such that $d = a_{1}r_{1} + a_{2}r_{2} + a_{3}r_{3} + \cdots + a_{n}r_{n}$ is the greatest common divisor of $r_{1}$, $r_{2}$, $r_{3},\ldots, r_{n}$.

Okay so I know that a Euclidean domain is an integral domain where if $a,b \neq 0$, $\deg(a) \leq d(ab)$ and if $a \in R$ and $b \neq 0$, $a=qb+r$ for some $q,r \in R$, and $\deg(r) < \deg(b)$ or $r=0$. I do not know how to use this information to solve the problem. I would appreciate help very much. Thanks!

share|cite|improve this question
An Euclidean domain is a PID. – wxu May 1 '12 at 7:16
Can you do the case $n=2$? It's essentially the Euclidean algorithm from elementary number theory. Then you can go for induction. – Gerry Myerson May 1 '12 at 7:34

Hint $\:$ The set $\rm\: I = \{ a_1\:r_1\! + \cdots + a_n\:r_n,\: a_i \in R\}\:$ is an ideal, i.e. it is closed under subtraction and scaling by elements of $\rm R.\:$ Just as for the classical extended Euclidean algorithm, show that any element $\rm\:r\in I\:$ of least Euclidean value divides all elements of $\rm\: I\:$ (else if $\rm\ r\nmid i\:$ then its remainder $\rm\: i\ mod\ r = i - q\:\!r\in I\:$ has Euclidean value smaller than $\rm\!\:r\Rightarrow\Leftarrow).$ Further, $\rm\:r\:$ is divisible by every common divisor of the $\rm\:r_i\:$ since it is an $\rm R$-linear combination of the $\rm\:r_i.\:$ Therefore, since $\rm\:r\:$ is a common divisor of the $\rm\:r_i\:$ that is divisible by every common divisor, we conclude $\rm\:r = gcd(r_i).$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.