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Let $S = \{ax + by: x,y \in \mathbb{Z}\}$ and let $e > 0$ be the smallest element in $S$. Prove that $e \mid a$, and hence prove that $e = \gcd(a,b)$

I'm afraid I can't provide much of my initial working here because I'm at a loss of how to start.

I suppose that to show $e\mid a$ I would have to show that $eg = a$ for some $g \in \mathbb{Z}$. But I have no idea how I would do this.

Martin Sleziak
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leanne
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3 Answers3

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When I first saw this problem, I was pretty surprised at how the proof went, so I'll just get you started on it.

By the Division Theorem, we have: $$a=qe+r \text{ for } 0 \leq r < e$$ Now, we need to show $r=0$. To do this, we can use the fact that $e=ax+by$ for some $x,y \in \Bbb{Z}$ and that $a=1a+0b$, so: $$1a+0b=q(ax+by)+r \implies a(1-qx)-bqy=r$$ Now we can combine the fact that $r=au+bv$ for $u,v \in \Bbb{Z}$ and that $0 \leq r < e$ to prove $r=0$.

Thomas M
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Noble Mushtak
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  • Wow that's really helpful, thanks! In your second to last line, should that be $a(1-q) - qby = r$? Also wondering - how do you know that $r = au + bv$? – leanne Jun 22 '16 at 03:48
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    @leanne Thanks for the typo catch! Since we have $r=a(1-q)-bqy=a(1-q)+b(-qy)$, we can set $u=1-q \in \Bbb{Z}$ and $v=-qy \in \Bbb{Z}$ so that $r=au+bv$. – Noble Mushtak Jun 22 '16 at 12:17
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    @leanne Note that the proof depends only on the fact that $S$ is closed under remainders.In fact this follows from the fact that $S$ is closed under differences, and one can present the proof in a way that emphasizes this fundamental innate (group / ideal) structure - see my answer. Such algebraic structure is ubiquitous in number theory and algebra, so it is essential to bring it to the fore. – Bill Dubuque Jun 22 '16 at 15:13
  • @BillDubuque "Closed under remainders" means that the Division Theorem is true for $\Bbb{Z}$ and that it is a Euclidean domain, correct? Just want to make sure. – Noble Mushtak Jun 22 '16 at 15:15
  • @Noble The proof employs the Division Algorithm, and it will work in any Euclidean domain ( i.e. one enjoying such division with "smaller" remainder) to show that all ideals are principal - generated by any nonzero element of least size. In fact there is a sort of generalization to arbitrary PIDs, see the Dedekind-Hasse criterion. – Bill Dubuque Jun 22 '16 at 15:29
  • @BillDubuque Thanks for telling me about the Dedekind-Hasse criterion! I didn't know about that before. – Noble Mushtak Jun 22 '16 at 15:45
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    @Noble Strangely, that pretty little result is not as widely known as it deserves to be. So I try to correct that by emphasizing it whenever it seems enlightening to do so. – Bill Dubuque Jun 22 '16 at 15:51
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Here is a conceptual way to prove Bezout's Identity for the gcd. The set $\rm\,S\,$ of integers of the form $\rm\,a x + b y,\ x,y\in \mathbb Z,\,$ is closed under subtraction so, by the Lemma below, every positive $\rm\,k\in S\,$ is divisible by $\rm\,d = $ least positive $\rm\in S.\,$ So $\rm\,a,b\in S$ $\,\Rightarrow\,$ $\rm d\mid a,b,\,$ i.e. $\rm\,d\,$ is a common divisor of $\rm\,a,b,\,$ necessarily the greatest such by $\rm\ c\mid a,b\,$ $\Rightarrow$ $\rm\,c\mid d = ax+by$ $\Rightarrow$ $\rm\,c\le d$.

Lemma $\ \ $ Let $\,\rm S\ne\emptyset \,$ be a set of integers $>0\,$ closed under subtraction $> 0,\,$ i.e. for all $\rm\,n,m\in S, \,$ $\rm\ n > m\ \Rightarrow\ n-m\, \in\, S.\,$ Then the least $\rm\:\ell\in S\,$ divides every element of $\,\rm S.$

Proof ${\bf\ 1}\,\ $ If not there is a least nonmultiple $\rm\,n\in S,\,$ contra $\rm\,n-\ell \in S\,$ is a nonmultiple of $\rm\,\ell.$

Proof ${\bf\ 2}\,\rm\,\ \ S\,$ closed under subtraction $\rm\,\Rightarrow\,S\,$ closed under remainder (mod), when it is $\ne 0,$ since mod is simply repeated subtraction, i.e. $\rm\ a\ mod\ b\, =\, a - k b\, =\, a\!-\!b\!-\!b\!-\cdots\! -\!b.\,$ Thus $\rm\,n\in S\,$ $\Rightarrow \rm\, (n\ mod\ \ell) = 0,\,$ else it is in $\,\rm S\,$ and smaller than $\rm\,\ell,\,$ contra minimality of $\rm\,\ell.$

Remark $\ $ In a nutshell, two inductions yield

$\rm S\ closed\ under\ {\bf subtraction} $
$\Rightarrow\:\rm S\ closed\ under\ {\bf mod} = remainder = repeated\ subtraction $
$\Rightarrow\:\rm S\ closed\ under\ {\bf gcd} = repeated\ mod\ (Euclid's\ algorithm)$

Viewed constructively this yields the extended Euclidean algorithm for the gcd.

The conceptual structure will be clarified when one studies ideals of rings, where the above proof generalizes to show that Euclidean domains are PIDs.

Bill Dubuque
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  • This is the most intuitive proof I have seen thus far. The standard proof of Bezout's identity uses the division algorithm and is not something you would naturally think of (some authors may skip it entirely). It leads to nightmarish questions. I did have to sit down with a piece of paper and look at some examples of positive integer sets closed under subtraction e.g. ${2,4,6,8}$ . – john Nov 01 '22 at 09:19
  • One thing I could mention, you seem to switch between two sets. To be explicit let $S = { ax + by : x , y \in \mathbb Z }$ and $S' = {ax+by > 0 : x, y \in \mathbb Z }$. You showed that if we define $ d = \min( S ') $ , then $ d \mid a $ and $d \mid b $. But two issues. How do we know $a,b$ are in $S ' $, since we did not assume $a,b$ are positive. I suppose if $ a < 0 $ then $a(-1) + b(0) > 0 $ demonstrates $-a \in S ' $ for integers $ x = -1, y = 0 $. And if $d | -a $ it follows that $d | a $. But how do you show $d$ divides any element in the set $S$. Also $a,b$ must be nonzero. – john Nov 01 '22 at 10:34
  • @john Yes, generally if nonempty $S$ is closed under subtraction then it is also closed under negation $(s\in S\Rightarrow s-s = 0\in S\Rightarrow 0-s = -s\in S),,$ so $,a,b\in S\Rightarrow |a|,|b|\in S,,$ and $a\Bbb Z + b\Bbb Z = |a|\Bbb Z + |b|\Bbb Z,,$ so wlog we may assume $a,b\ge 0,$ (and not both are $0).,$ You may find of interest a proof of Euclid's Lemma by this method. – Bill Dubuque Nov 01 '22 at 12:10
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One must define $e$ as the smallest positive element of $S$.

Since $e\in \{ax + by: x,y \in \mathbb{Z}\}$, there are some numbers $x_1,y_1$ such that $e = ax_1+by_1$.

Suppose some positive member $f$ of $S = \{ax + by: x,y \in \mathbb{Z}\}$ were not a multiple of $e$. Then we have some numbers $x_2,y_2$ such that $f = ax_2+by_2$.

Since $e$ is the smallest, we must have $e<f$. Let $r$ be the remainder when $f$ is divided by $e$. Then using the relations \begin{align} f & = ax_2 + by_2 \\ e & = ax_1 + by_1 \end{align} you can show that $r \in S$. But $0<r$ since $f$ is not a multiple of $e$ and $r<e$ since $r$ was a remainder upon division by $e$.

That means there is a positive member of $S$, namely $r$ that is smaller than $e$. That can't happen since $e$ is the smallest.

Hence every member of $S$ is a multiple of $e$. Therefore $e$ is a common divisor of $a\in S$ and $b\in S$.

Michael Hardy
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  • You showed that every positive $k \in S$ is a multiple of $e $ where $e$ stands for the least positive element in $S $. Then how do you go from there to show every member of $S$ is a multiple of $e$? Specifically how do you show that $e$ is a common divisor of $a \in S$ or $b\in S$, since $a,b$ may not even be positive. – john Nov 01 '22 at 10:05