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I would appreciate hints to this. I've done part (a) but am unconfident.
Wondering how I could approach part (b)

Question's comment -- The aim of this question is to use the Division Algorithm and the definition of greatest common divisor (gcd) to show that $d_0 = \gcd(a,b)$.

Question:

Let $a,b$ be integers, not both zero, let $S$ be the set of integers defined by
$$S = \{ ax + by | x,y \in \mathbb{Z}\}$$ and let $d_0$ be the smallest positive integer in the set $S$.
Prove:
a. If $s\in S$, then $d_0$ is a divisor of $s$.
b. $d_0$ is a divisor of both $a$ and $b$.
c. If $d$ is a divisor of both $a$ and $b$, then $d$ is a divisor of $d_0$.
d. $d_0 =\gcd(a,b),$ and hence there exists integers $x,y$ such that $ax + by = \gcd(a,b)$.

Attempt:

Note I don't expect full (or any hints to the problem(s) as there are quite a bit, but I would just appreciate hints.
a. Let $s \in S$. Then $s = d_0 q + r$ for some $q,r \in \mathbb{Z}$.
By definition, $d_0 = ax_0 + by_0$ for some $x_0,y_0 \in \mathbb{Z}$.
$s = ax_s + by_s$ for some $x_s,y_s \in \mathbb{Z}$.
Hence, $$ax_s + by_s = q(ax_0 + by_0) + r\\ a(x_s - qx_0) + b(y_s - qy_0) = r.$$ Define $x_r:= x_s - qx_0 \in \mathbb{Z}$ and $y_r:= y_s - qy_0 \in \mathbb{Z}$.
Then this shows that $r \in S$ which implies $d_0 | s$.

Twenty-six colours
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    Hint : Bézout's identity – Mohbenay Aug 06 '17 at 10:01
  • I believe that's what we're trying to prove in part (d). – Twenty-six colours Aug 06 '17 at 10:02
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    This is correct, but has one drawback, in my opinion: it relies on Euclidean division. However the same result is true even for P.I.D.s which have no Euclidean division algorithm, and it can be proved in a very simple way, using the minimality of $d_0$. – Bernard Aug 06 '17 at 10:05
  • Note that you also need to prove that $d_0$ is well-defined, i.e. that $S$ has at least one positive element. The PID case mentioned by @Bernard is known as the Dedekind-Hasse PID test, e.g. see this answer. – Bill Dubuque Aug 06 '17 at 16:03

3 Answers3

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a. I'm not quite sure how you can just conclude from $r\in S$ that $d_0|s$ holds. But you can easily do it. Due to the euclidean division you get either $r=0$, which would imply $d_0|s$, or $0<r<d_0$, which is a contradiction to the minimality of $d_0$.

b. Use the last statement and $a,b\in S$.

c. Because $d_0\in S$ there exist integers $x,y$ with $d_0=ax+by$. So if $d|a,b$, then also $d|ax+by=d_0.$

d. We already showed $d_0|a,b$, which impliess $d_0|gcd(a,b)$, by the definition of the gcd. But $gcd(a,b)|a,b$, which impliess using c. that $gcd(a,b)|d_0$. So we have two positive numbers and each divides the other one, meaning they are equal, i.e. $d_0=gcd(a,b).$

Frieder Jäckel
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  • +1 ... I was in the middle of typing a more extended answer, but you was faster! Well done ;) – rtybase Aug 06 '17 at 10:26
  • Thanks. For d. Why does $d_0|a,b$ imply $d_0|\gcd(a,b)$?
    I'm not sure what definition is used, (or how to prove that).     – Twenty-six colours Aug 06 '17 at 10:30
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    Because every number which divides two numbers also divides their gcd. If that is not entirely clear write $a=\prod_i p_i^{\alpha_i}$ and $b=\prod_i p_i^{\beta_i}$. Then every divisor $d=\prod_i p_i^{\gamma_i}$ must satisfy $\gamma_i\leq \min{\alpha_i,\beta_i}$ and for the gcd equality is attained for all $i.$ – Frieder Jäckel Aug 06 '17 at 10:50
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a Your proof is correct, but incomplete. When applying Euclidean division, $0 \leq r < d_0$. You show that $r \in S$, is positive and smaller than $d_0$ (which is smallest by definition) thus a contradiction.

rtybase
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  • Thank you, I can't see why applying Euclidean division would show that $0 \leq r\leq d_0$. – Twenty-six colours Aug 06 '17 at 10:31
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    When $s=d_0q+r$, you are applying Euclidean division, which is a theorem, which states that $0\leq r < |d_0|$. But $d_0$ is positive, thus according to Euclidean division $0\leq r < d_0$. – rtybase Aug 06 '17 at 10:37
  • $q$ should also be positive right by definition (or convention)? – Twenty-six colours Aug 06 '17 at 10:43
  • Not necessarily, Euclidean division states nothing about $q$ (apart from it's unique) and it can't since, particularly for this case, $s \in S$ can also be negative. From $s=d_0q+r < 0$ we will deduct that $q <0$ since everything else ($d_0,r$) is positive. But this doesn't matter, since you show that $r \in S$ too. – rtybase Aug 06 '17 at 10:47
  • There are no restrictions on the sign of $q.$ – Frieder Jäckel Aug 06 '17 at 10:52
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$(a)\ $ $\,d_0\nmid s\,$ $\,\Rightarrow\, 0 < s\bmod d_0 < d_0\,$ is a smaller positive element of $S$, contra minimality of $d_0\,$ (notice $\,s\bmod d_0 = s - k\, d_0\,$ for some integer $k\,$ so it lies in $S,\,$ i.e. $S$ is $\rm\color{#0a0}{closed\ under\ mod}$).

$(b)\ $ Applying prior to $\,s = a,b\in S\,$ yields $\,d_0\mid a,b,\,$ i.e. $\,d_0$ is a common divisor of $\,a,b$

$(c)\ \ d_0\in S\,\Rightarrow\, d_0 = \color{#c00}aj+\color{#c00}bk\,$ so $\,d\mid \color{#c00}{a,b}\,\Rightarrow\, d\mid d_0$

$(d)\ \ d_0\,$ is a common divisor of $\,a,b$ by $(b),\,$ necessarily the greatest common divisor since by $(c),\,$ $\,d\mid a,b\,\Rightarrow\,d\mid d_0\,\Rightarrow\, d\le d_0.\,$ Therefore, by definition, $\,d_0 = \gcd(a,b).\,$ Furthermore, we deduce from $\,d_0\in S\,$ said Bezout linear representation $\,\gcd(a,b) = ax+by.$

Remark $\ $ Said more concisely $\, a\Bbb Z + b\Bbb Z = d_0\Bbb Z\,\Rightarrow\, d_0 = \gcd(a,b).\,$ This viewpoint is clarified when one learns ideal theory. But one can understand this more conceptual viewpoint even without knowledge of ideals (or groups), see this answer (which greatly clarifies the essence of the matter).

Bill Dubuque
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