Let $R$ be integral domain and $r \not | a.$ If $r$ is prime and $r^k|ab,$ then $r^k|b ?$

Question

Let $R$ be an integral domain and $a,b,r \in R.$

Let $r$ be prime. Suppose there exists positive integer $k$ such that $r^k$ divides $ab$ and $r$ does not divide $a.$ Could anyone advise me on how to prove/disprove that $r^k$ divides $b \ ?$

Thank you.

Answer 1

I assume you need to prove this only using the axiom that

If $r$ is prime and $r|ab$ then either $r|a$ or $r|b$.

Use induction:

For $k=1$ this is clear: since $r$ is a prime which divides $ab$ it has to divide either $a$ or $b$. Since it doesn't divide $a$, it must divide $b$.

Now suppose you proved your claim already up to but not including $k$, then let's prove it for $k$: $$r^k=r\cdot r^{k-1}|ab$$

We know that $r^{k-1}|b$, so $b=r^{k-1}b^\prime$ for some $b^\prime\in R$. Then (why?)

$$r|ab^\prime$$

Again, since $r$ is a prime that doesn't divide $a$, it must divide $b^\prime$ and therefore $b=r^{k-1}rb^{\prime\prime}$ for some $b^{\prime\prime}\in R$. That is, $r^k | b$.

Answer 2

Products of primes $\,p\,$ (i.e. $\,p\mid ab\,\Rightarrow\,p\mid a\,$ or $\,p\mid b),\,$ always factor uniquely (into atoms), obey Euclids' Lemma, etc in any domain, by simply iterating said defining property. Indeed

$\,p_i\nmid a,\,\ p_1\!\cdots p_n\mid ab\,\Rightarrow\, p_1\mid b,\,$ so $\,p_2\cdots p_n\mid a(b/p_1)\,$ $\overset{\rm induct}\Rightarrow$ $\,p_2\cdots p_n\mid b/p_1\,$ $\Rightarrow$ $\,p_1\cdots p_n\mid b$