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.

The term "maximum common divisor" has a couple different denotations. Which is yours? — Bill Dubuque, Jun 23 '14 at 18:19
If $(a,b)=1$, I always worth the Bezout's theorem, R is not even PID? — Croos, Jun 23 '14 at 18:27

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

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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

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

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

edited Apr 13 '17 at 12:21

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answered Jun 23 '14 at 18:22

Bill Dubuque

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@Groos I added the Bezout proof to show the analogy. Note that you may need to prove the gcd distributive law (and possibly some other minor points) depending on what you already know. The above is just a sketch of the main ideas. – Bill Dubuque Jun 23 '14 at 18:35

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

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