Proof explanation: suppose $a\mid bc$ and $\gcd(a, b) = 1$. Then $a\mid c$.

Question

I have been given a proof, but I do not understand the "why" behind it. If someone could explain me each of its steps with great detail that would be amazing!

The proof I was given is the following one:

Since $\gcd(a,b)=1$, we have that $$ ax+by=1.\label{1}\tag{1} $$ We need see the relation between $c$ and $a$, so we multiply both sides of \eqref{1} by $c$ and get $$ c= acx+bcy.\label{2}\tag{2} $$ Now since $c \mid cax+cby$, $a\mid bc$ and $a \mid bcy$, we conclude $a\mid cax$ and this proves that $a \mid c$.

I do not get how $a$ has been proved to divide $c$ in this last step.

Answer 1

In the proof you were given, you have that $c= acx+bcy$ for some integers $x$ and $y$ (note Bézout's identity proves there exists these integers $x$ and $y$ which satisfy the original equation of $ax + by = 1$). Also, you know that $a\mid bc$, so $a$ divides the second term of $bcy$. In addition, $a \mid acx$, so $a$ divides the first term. Since $a$ divides both terms on the right, it must divide their sum (more generally, $a$ divides any linear combination of those terms), i.e., $c$, so $a\mid c$.

Answer 2

John explained the algebra. Below I explain one way to "understand the 'why' behind the proof".

Notice that the set $D$ of integers $\,d\,$ such that $\,a\mid d c\,$ is closed under subtraction by

$$a\mid dc,d'c\,\Rightarrow\, a\mid dc\!-\!d'c = (d\!-\!d')c\,\Rightarrow\, d-d'\in D\qquad$$

so a basic Theorem $\Rightarrow D\,$ is also closed under gcd, so $\,a,b\in D\,\Rightarrow\, (a,b)\!=\!1\in D,\,$ so $\,a\mid 1c=c$

The quoted proof is a special case of this proof, since the the gcd (and its Bezout identity) can be obtained by repeated subtraction: the Euclidean algorithm in subtractive (vs. remainder) form.

More conceptually: $ $ nonzero $d\in D$ are all the possible denominators for $\,c/a,\,$ because

$\,d\in D\!\!\overset{\rm def\!\!}\iff\! a\mid dc\!$ $\iff\!\! \exists\, j\!:\, aj = dc\!$ $\iff\! \!\exists\, j\!:\, \large \frac{c}a = \frac{j}d\!$ $\iff\! \large \frac{c}a\,$ is writable with denom $\,d$

The above closure property says such denominators are closed under subtraction so they're also closed under gcd. So $\,a,b\,$ denoms for $\,c/a\Rightarrow$ so too is $\,\gcd(a,b)\! =\! 1,\,$ i.e. $\,c/a = j/1\,$ so $\,a\mid c$.

Or, directly by Bezout $\, f= c/a\,$ and $\, af,bf\in\Bbb Z\,\Rightarrow f = (ax\!+\!by)f = x(af)\!+\!y(bf)\in\Bbb Z,\,$ which is a direct fractional view of the proof cited in the OP (but is still less conceptual, since it doesn't explicitly emphasize the fundamental innate structure - that denominators are closed under gcd).

Of course gcd-closure of denoms is also obvious from the well-known result that denominator set $D = \ell\Bbb\,\! Z\,$ is the set of multiples of the least denominator $\ell,\,$ so if $\,j\ell, k\ell\, $ are denoms then their gcd $\,(j\ell,k\ell) = (j,k)\ell$ is also a denom (being a multiple of $\ell)$. This well-known result is provable as above, see unique fractionization.

The algebraic (ring) essence of the matter is clarified by studying denominator (and order) ideals.