Let $R$ a ring with maximum common divisor. If $a,b,c \in R$ such that $a|bc$ and $(a,b)=1$ then $a|c$.

Question

Let $R$ a ring with maximum common divisor. Show that if $a,b,c \in R$ such that $a|bc$ and $(a,b)=1$ then $a|c$.

Comments: I tried to use the Bezout's theorem, but in my course we saw it only applies to principal domains, I tried to use the setting as well, but could not finish.

Answer 1

Hint $\ $ You can generalize the Bezout-based proof by replacing the Bezout linear combination $\,\color{#c00}{ja+kb = 1}\,$ by the gcd $\,\color{#0a0}{(a,b)=1}.\,$ Then the use of the integer distributive law is replaced by the use of the GCD Distributive Law, namely

$\qquad\qquad a\mid bc\,\Rightarrow\,a\mid ac,bc\,\Rightarrow\, a\mid jac\!+\!kbc = (\color{#c00}{ja+kb})c = c$

$\qquad\qquad a\mid bc\,\Rightarrow\, a\mid ac,bc\,\Rightarrow\, a\mid\, (ac,\ bc)\ =\ \color{#0a0}{(a,\ b)}\,c = c$