1

For natural numbers $a$ and $b$, show that $a \Bbb Z + b \Bbb Z = \gcd(a, b)\Bbb Z $

I just basically said that the gcd of a and b { written as C } obviously divides aZ and bZ, therefore it can be rewritten as

CdZ + CeZ -> C[dZ + eZ] -> CZ[d+e] where Cd=a and Ce=b.

if a and b and C are natural numbers, then so are d and e. the sum of d and e is an integer, that when multiplied by an integer produces an integer. therefore

CZ[d+e] = CZ, so aZ + bZ = GCD(a,b)Z

{ Where C is the gcd(a,b) }

BCLC
  • 13,459
Max
  • 87

2 Answers2

3

You know that $a\Bbb Z+b\Bbb Z=\{ax+by:x,y\in\Bbb Z\}$. Let $d$ be the minimum positive integer in this set. By the Euclidean algorithm, we can write $a=qd+r$ and $b=q'd+r'$, with $r=0$ or $0<r<d$ and $r'=0$ or $0<r'<d$. We cannot have $0<r<d$ since else $r=a-qd$ is again in $a\Bbb Z+b\Bbb Z$ and is positive, which contradicts our choice of $d$ as the smallest positive integer in $a\Bbb Z+b\Bbb Z$. Thus $r=0$. By the same reasoning, $r'=0$. Thus $d\mid a,b$, so $d$ is a common divisor. But $d=ax'+by'$ for some integers $x',y'$, so if $f\mid a,b$ then $f\mid ax'+by'=d$. Thus $d$ is the greatest common divisor. Since $d\mid a,b$, it follows that $d\mid ax+by$ for any choice of $x,y$ integers, so every element in $a\Bbb Z+b\Bbb Z$ is a multiple of $d=(a,b)$, thus $a\Bbb Z+b\Bbb Z\subseteq(a,b)\Bbb Z$. But $a\Bbb Z+b\Bbb Z$ is closed under addition and subtraction, so all integer multiples of $(a,b)$ are in $a\Bbb Z+b\Bbb Z$, that is $a\Bbb Z+b\Bbb Z\supseteq (a,b)\Bbb Z$. Thus $a\Bbb Z+b\Bbb Z=(a,b)\Bbb Z$

Given any ideal $(x_1,\ldots,x_n)$ in $\Bbb Z$, there exists a least positive element in it, call it $d$. Dividing each $x_i$ by $d$ one gets $x_i=q_id+r_i$. Since $r_i=0$ or $r_i<d$, you get $r_i=0$ by $d$ being minimum. Thus $d\mid x_i$ for each $i$. This means $(d)\supseteq (x_1,\ldots,x_n)$. Since $(d)$ is trivially contained in $(x_1,\ldots,x_n)$, you get $(x_1,\ldots,x_n)=(d)$. It is not difficult to check that $d=\gcd(x_1,\ldots,x_n)$. We've already shown $d$ is a common divisor. But since $d=\sum a_ix_i$ for some integers $a_i$, any common divisor of the $x_i$ divides $d$. Thus $d$ is, being positive, the greatest common divisor of the $x_i$.

Pedro
  • 122,002
  • I can't follow this. I'm not really strong on ideals. Is my proof bad, then? – Max Dec 06 '13 at 05:22
  • @Max I'll add something. – Pedro Dec 06 '13 at 05:26
  • The notation $d\mid a,b,d\mid ax+by$ frightened me for a second. (I realized the second comma was not the same species as the first!) It might be possible to replace the phrase "$a\Bbb Z+b\Bbb Z$ is closed under multiplication" (which should specify ambient multiplication) with closed under addition (and additive inverses), since that might be simpler for ideal-shy readers. @Max One thing I notice about your proof is that both equalities $d\Bbb Z+e\Bbb Z=(d+e)\Bbb Z$ and $(d+e)\Bbb Z=\Bbb Z$ are false. (Although the desired equality $d\Bbb Z+e\Bbb Z=\Bbb Z$ is true - see Bezout's theorem.) – anon Dec 06 '13 at 05:38
  • Why is that false? the first should follow by distributive laws and the second because of what i stated above, right? – Max Dec 06 '13 at 05:47
  • @Max The distributive law holds for numbers, but we're talking about operations on sets. Notice that everything in $(d+e)\Bbb Z$ is a multiple of $d+e$. If for example $d=e=1$ then this means everything is even in $(d+e)\Bbb Z$. However $\Bbb Z+\Bbb Z$ is the set of things representable as the sum of two integers, which is just all of $\Bbb Z$. This includes odd numbers, e.g. $0+1\in\Bbb Z+\Bbb Z$. So $\Bbb Z+\Bbb Z\ne(1+1)\Bbb Z$. – anon Dec 06 '13 at 06:01
  • But d=ax′+by′ for some integers x′,y′, so if f∣a,b then f∣ax′+by′=d. Thus d is the greatest common divisor. can you elaborate? – Max Dec 06 '13 at 07:18
  • @Max What is troubling you? – Pedro Dec 06 '13 at 14:39
1

Well firstly, how do you define $\gcd(a,b)$? I'll define it by the converse of a proposition in Artin Algebra (Prop 2.3.5)

enter image description here

Namely, we'll prove

Converse of Prop 2.3.5: Let $a$ and $b$ be integers not both zero. Suppose (a) $d$ divides $a$ and $b$, (b) any integer $e$ that divides $a$ and $b$ also divides $d$ and (c) there are integers $r$ and $s$ s.t. $d=ra+sb$. Then $\mathbb Z d=\mathbb Z a + \mathbb Z b$.

where $d:=\gcd(a,b)$ is defined by the integer given by the assumptions of the above stated Converse of Prop 2.3.5.

-

Pf:

($\supseteq$)

Let $n \in \mathbb Za + \mathbb Zb$. Then there are integers $n_a, n_b$ s.t. $n=n_a a + n_b b$. We must show $n \in \mathbb Zd$, i.e. we must find an integer $n_d$ s.t. $n=n_d d$.

By $(a)$, there are integers $d_a, d_b$ s.t. $d_a d = a, d_b d = b$. Thus, $$n=n_a a + n_b b = n=n_a d_a d + n_b d_b d.$$

Thus, $n_d = n_a d_a + n_b d_b$.

($\subseteq$)

Let $n \in \mathbb Zd$. Then there is an integer $n_d$ s.t. $n=n_d d$. We must show $n \in \mathbb Za + \mathbb Zb$, i.e. we must find integers $n_a, n_b$ s.t. $n=n_a a + n_b b$.

By $(c)$, there are integers $d_1, d_2$ s.t. $d=d_1 a + d_2 b$. Thus, $$n=n_d d = d_1 a n_d + d_2 b n_d$$

Thus, $n_a = d_1 n_d, n_b = d_2 n_d$.

QED

Note: Observe $(c)$ is Bézout's identity.

BCLC
  • 13,459