How to prove that if $\gcd(n,k)=1$ then $\gcd(f,nk)=\gcd(f,n)·\gcd(f,k)$ by primes

Question

$$\gcd(f,nk)=\gcd(f,n)·\gcd(f,k)$$ when $(n,k)=1$ by the fundamental theorem of arithmetic.

I tried many times to prove this equation, but I did not get it.

Answer 1

Let $s$ be big enough and $p_1,\cdots ,p_s$ an enumeration of prime numbers then $$f = p^{a_1}_1\cdots p^{a_s}_s,n =p^{b_1}_1\cdots p^{b_s}_l,k = p^{c_1}_1\cdots p^{c_s}_s, $$ and we know that $(n,k)=1$ implies $\min \{b_i,c_i\}=0$ for all $i$.
Also, $(f,nk)=(p^{a_1}_1\cdots p^{a_s}_s,p^{b_1}_1\cdots p^{b_s}_l* p^{c_1}_1\cdots p^{c_s}_s)=(p^{a_1}_1\cdots p^{a_s}_s, p^{b_1+c_1}_1\cdots p^{b_s+c_s}_s)=p^{\min \{a_1,b_1+c_1\}}_1\cdots p^{\min \{a_s,b_s+c_s\}}_s,$ on the other hand $$(f,n) = p^{\min \{a_1,b_1\}}_1\cdots p^{\min \{a_s,b_s\}}_s,$$ $$(f,k) = p^{\min \{a_1,c_1\}}_1\cdots p^{\min \{a_s,c_s\}}_s,$$ hence $(f,n)(f,k)=p^{\min \{a_1,b_1\}+\min \{a_1,c_1\}}_1\cdots p^{\min \{a_s,b_s\}+\min \{a_s,c_s\}}_s$
So can you prove that $\min \{a,b+c\}=\min \{a,b\}+\min \{a,c\}$ given that $\min \{b,c\}=0$?