If $R$ be a UFD.Let $a \in R$ be such that $a = {p_{1}}^{r_{1}}{p_{2}}^{r_{2}}...{p_{n}}^{r_{n}}$, where $p_{i}$'s are irreducible (or prime) element in $R$, for $i = 1,2,...,n$.Now if ${p_{i}}^{k_{i}}$ divides $a$ then can I say that ${p_{i}}^{k_{i}}|{p_{i}}^{r_{i}}$?If the answer is 'yes' then please tell me why?
Thank you in advance.
Let us first assume that $p_{i}$ is not an associate of $p_{j}$ for $i \neq j$.
Now,if ${p_{i}}^{k_{i}}|a$ then we claim that $k_{i} \leq r_{i}$.Because if ${p_{i}}^{k_{i}}|a$ then $\exists$ $b = {p_{1}}^{s_{1}}.{p_{2}}^{s_{2}}...{p_{n}}^{s_{n}}$ such that $a = {p_{i}}^{k_{i}}.b$. Simplifying we get
$a = {p_{1}}^{s_{1}}.{p_{2}}^{s_{2}}...{p_{i}}^{k_{i} + s_{i}}...{p_{n}}^{s_{n}}$. Now since $R$ is a UFD and $p_{i}$ is not associated to $p_{j}$ for $i \neq j$ so we must have $r_{i} = k_{i} + s_{i} \geq k_{i}$.Which proves the claim.
Now we always write ${p_{i}}^{r_{i}} = {p_{i}}^{k_{i}}.{p_{i}}^{r_{i} - k_{i}}$ , where ${p_{i}}^{r_{i} - k_{i}} \in R$ since $r_{i} \geq k_{i}$. This shows that ${p_{i}}^{k_{i}}|{p_{i}}^{r_{i}}$.This completes the proof.