Divisibility in a Euclidean Domain

Question

Let $R$ be a Euclidean Domain. I am working on showing that $$ \text{If } a \, | \, bc \text{ with } a,b \neq 0 \text{ then } \frac{a}{(a,b)} \, \bigg| \, c. $$

Note that the first part of this problem is to show that it $(a,b) = 1$ and $a \, | \, bc$ then $a \, | \,c$. I had no problem with that, and believe it guides me towards the more generalized desired result. Letting $x = \frac{a}{(a,b)}$ I can show that $x \, | \, bc$. So if I can show that $$ (x,b) = 1, $$ then I'm done.

Edit: I realized my idea need not be true. Let $a = 8$ and $b=14$, then $(a,b) = 2$ but then $(4,14) \neq 1$. Okay back to the drawing board.

I am having a difficult time showing this final result. Any help is much appreciated!

Answer 1

Let $\mathrm{gcd}(a,b)=m$, i.e. $a=a_1m$ and $b=b_1m$, where $\mathrm{gcd}(a_1,b_1)=1$.

Then $\frac{bc}{a}=\frac{b_1c}{a_1}$, since $\mathrm{gcd}(a_1,b_1)=1$, we get that $a_1|c$. Let $\frac{c}{a_1}=x$.

Then \begin{equation} \frac{c}{\frac{a}{\mathrm{gcd}(a,b)}}=\frac{c}{\frac{a}{m}}=\frac{c}{a_1}=x. \end{equation}

Answer 2

Hint. Use the prime fatorization theorem: let $p$ be any prime element. Define $\alpha, \beta, \gamma \in \mathbb{N}$ so that $p^{\alpha} |a$, $p^{\alpha+1} \not| a$, $p^{\beta} | b$, $p^{\beta+1} \not | b$, $p^{\gamma} | b$, $p^{\gamma+1} \not | c$.

Your hypothesis now reads $\alpha \le \beta + \gamma$ and you want to prove $\alpha - \min\{\alpha, \beta\} \le \gamma$ ...

Distingis the cases $\alpha \le \beta$ and $\beta \le \alpha$: in the first case your goal becomes $\alpha \le \alpha +\gamma$ which is trivial, in the second case it becomes $\alpha \le \gamma+\beta$, which is your hypothesis.