Greatest Common Divisor multiplicative proof

Question

Given : gcd(m,n) = 1 , gcd(d₁,d₂) = 1.
Show : gcd(m,d₁)*gcd(n,d₂) = gcd(mn,d₁d₂).

Is my proof for above correct ?
If no , what have I missed ?
If yes , is there any better way to prove it ?

Proof :

p_i , q_i , e_i , f_i are prime numbers.

m = p₁^k₁p₂^k₂....p_s^k_s...p_w^k_w
d₁ = p₁^a₁p₂a₂....p_s^a_s e_s+1^b_s+1 ...e_x^a_x
n = q₁^l₁q₂^l₂....q_t^l_t q_t+1^b_t+1 ...q_y^l_y
d₂ = q₁^g₁q₂^g₂....q_t^g_t f_t+1^g_t+1 ...f_z^g_z

gcd(m,d₁) = p₁^{min(k₁,a₁)}p₂^{min(k₂,a₂)}...p_s^min(k_s,a_s)
gcd(n,d₂) = q₁^{min(l₁,g₁)}q₂^{min(l₂,g₂)}...q_t^min(l_t,g_t)
mn = p₁^k₁p₂^k₂....p_s^k_s...p_w^k_wq₁^l₁q₂^l₂....q_t^l_t q_t+1^b_t+1 ...q_y^l_y
d₁d₂ = p₁^a₁p₂a₂....p_s^a_s e_s+1^b_s+1 ...e_x^a_xq₁^g₁q₂^g₂....q_t^g_t f_t+1^g_t+1 ...f_z^g_z
gcd(mn , d₁d₂) = q₁^{min(l₁,g₁)}q₂^{min(l₂,g₂)}...q_t^min(l_t,g_t)p₁^{min(k₁,a₁)}p₂^{min(k₂,a₂)}...p_s^min(k_s,a_s)

gcd(mn , d₁d₂) = gcd(m, d₁)*gcd(n, d₂)

Please do confirm if it is correct.

Sorry for the title , not appropriate.

Answer 1

The claim is false so your proof cannot be correct, e.g. put $\, m> 1\,$ below

$$\begin{align} (m,d_1)(n,d_2) &= (m\cdot n,\, d_1\cdot d_2)\\[.3em] 1 = (m,1)\,(1,m)\, &\neq (m\cdot 1,\,1\cdot m) = m\end{align}\qquad$$

To debug your proof put $\,m=2\,$ above and find the first line in your proof that fails for these values.

Remark $\ $ Generally if $\, (m,n) = 1 = (d_1,d_2)\,$ then we have

$$(mn,d_1 d_2) = (m,d_1 d_2) (n,d_1d_2) = (m,d_1)\color{#c00}{(m,d_2) (n,d_1)}(n,d_2)\qquad $$

and the latter $= (m,d_1) (n,d_2) \iff \color{#c00}{(m,d_2) = 1 = (n,d_1)}$