Justifying expansion: $\gcd(a,c) \cdot \gcd(b,c) = \gcd(ab,bc,ac,cc)$

Solution 1:

For $p$ a prime, and $n$ an integer, write $n_p$ for the power of $p$ dividing $n$, that is, for the nonnegative integer $\nu$ such that $p^{\nu}$ divides $n$ but $p^{\nu+1}$ doesn't.

The power of $p$ dividing $\gcd(a,c)$ is $\min(a_p,c_p)$; similarly for $\gcd(b,c)$; so the power of $p$ dividing $\gcd(a,c)\gcd(b,c)$ is $\min(a_p,c_p)+\min(b_p,c_p)$, which is $\min(a_p+b_p,a_p+c_p,b_p+c_p,2c_p)$, which is $\min((ab)_p,(bc)_p,(ac)_p,(c^2)_p)$, QED.

Solution 2:

It is an immediate consequence of expanding using the GCD distributive and associative $\rm\color{#c00}{laws}$, analogous to how we expand $\,(a\!+\!c)(b\!+\!c)$ using the same laws for integer arithmetic, i.e.

$$\begin{align}\color{#0a0}{(a,c)}(b,c) &\,=(\color{#0a0}{(a,c)}b,\ \ \color{#0a0}{(a,c)}c)\ \ \ \text{via Distributive Law}\\[.1em] &\,= ((ab,cb),(ac,cc))\ \ \text{via Distributive Law}\\[.1em] {\rm so}\,\ \ (a,\,\ c)(b,\,\ c)&\,= (\,ab,\ bc,\ \ \,ac,cc)\ \ \ \ \text{via Associative Law}\\[.2em] \text{It's a gcd analog of } \ (a\!+\!c)(b\!+\!c) &\,= \ \,ab\!+\!bc\!+\!ac\!+\!cc,\ \text{by both satisfy said }\rm\color{#c00}{laws}\\[.3em] \text{It is exactly that }\:\!\rm (A\!+\!C)(B\!+\!C)&\rm \,=AB\!+\!BC\!+\!AC\!+\!CC\ \text{ as ideals: } A=(a)\rm\ etc \end{align}\quad$$

In summary, the GCD operation behaves just like addition in integer arithmetic - it is associative and commutative and multiplication distributes over it, so we can perform GCD arithmetic analogously to integer arithmetic. To better facilitate the analogy it may help to denote the gcd operation by an infix addition symbol, e.g. $\oplus$ as in this answer, so the OP identity becomes

$$(a\oplus c)(b \oplus c)\, =\, ab \oplus bc \oplus ac \oplus cc\qquad\ \ \ \ $$

Other proofs using Bezout are not only more complex but also less general since they don't apply in more general gcd domains where Bezout fails, e.g. polynomial rings like $\,\Bbb Z[x]\,$ and $\,\Bbb Q[x,y]$.

Ditto for proofs using (unique) prime factorization, since there are gcd domains that are not UFDs, e.g. the ring of all algebraic integers is Bezout so a gcd domain, but it has no irreducibles so no primes since e.g. $\,\alpha = \sqrt \alpha \sqrt \alpha$.

A nice exercise using the above is the proof of the "Freshman's Dream" GCD Binomial Theorem

$$\begin{align} (a\oplus b)^n &=\ a^n\oplus\, b^n\\[.3em] {\rm i.e}\ \ \ \gcd(a,b)^n &= \gcd(a^n,b^n)\end{align}\qquad\qquad\qquad\ $$