Doubt regarding the manipulation of Bézout's identity

Question

I had learnt the Bézout's identity in which it stated that

For nonzero integers $a$ and $b$, let $d$ be the greatest common divisor $d = gcd(a, b)$. Then there exists integers $x$ and $y$ such that $$ d = ax + by $$

Now we could transform this into

$$ d = ax + 1 \cdot (by) $$ which is $$ d = gcd(a, 1) $$ Which is wrong as this would mean all the $gcd$s would be equal to $1$.

So can anyone tell me the correct way to manipulate Bézout's identity?

Answer 1

It is true that, if $d=\gcd(a,b),$ then there are integers $x$ and $y$ such that $d=ax+by.$

The converse is not true. It is not true that, if there are integers $x$ and $y$ such that $d=ax+by,$ then $d=\gcd(a,b).$ Thus your assertion after "which is" is a non sequitur.

What is true: if there are integers $x$ and $y$ such that $d=ax+by,$ and $d$ divides $a$ and $d$ divides $b,$ then $d=\gcd(a,b).$

Answer 2

That's wrong - this means that $d$ is a multiple of $\gcd(a,1)$ which in this case is $1$, hence there are no problems.

To see why this holds we have that

$$d=ax+by$$

which implies

$$\lambda d=a\cdot(\lambda x)+b\cdot(\lambda y)$$

Hence every multiple of $\gcd(a,b)$ is a linear combination of $a$ and $b$. This doesn't mean however that $2d$ is the $\gcd$.

Answer 3

If you are given a relation of the type $au+bv=d$, all you can say in general is that $d$ is a multiple of $\operatorname{gcd}(a,b)$. The converse is also true. In particular, you get Bézout's theorem:

Two nonzero integers $a$ and $b$ are relatively prime if and only if there exists integers $u$ and $v$ such that $$au+bv=1$$

Answer 4

The ideal $(a,b)$ generated by $a$ and $b$ in any P.I.D. $R$ is generated by a g.c.d. $d$ of $a$ and $b$ (which is defined up a unit factor), so by definition $$\{ax+by\mid x,y\in R\}$$ is the set of multiples of $d$.

Answer 5

In fact, $d = \gcd(a,b)$ is the smallest positive integer that can be written in the form $ax+by$. So, yes $d = ax + (by)\cdot 1$, but the smallest positive linear combination of $ax$ and $1$ is $1 = 0(ax) + (1)1$.