0
$\begingroup$

Suppose $R$ is a UFD.Let $a \in R$.Then it can be represented as a finite product of irreducible elements as $a = {p_{1}}^{r_{1}}{p_{2}}^{r_{2}}...{p_{n}}^{r_{n}}$, where $p_{i}$ are irreducible.Now how can show that ${p_{i}}^{k_{i}}$ cannot divide $a$ for $k_{i} > r_{i}$ for $i = 1,2,...,n$.Please help me.

Thank you in advance.

  • 0
    otherwise you will have ${p_{i}}^{k_{i} - r_{i}}$ is an associate of $1$.Which in turn implies $p_{i}$ is an associate of $1$, but this means $1$ is irreducible.So...... – 2017-01-10
  • 0
    @A.Chattopadhyay: Note the "U" in UFD. – 2017-01-10
  • 0
    @Xam why are deleting your answer? – 2017-01-11
  • 0
    I have a question related to your deleted answer. – 2017-01-11
  • 0
    @A.Chattopadhyay I did because my answer wasn't totally right. – 2017-01-11
  • 0
    Why? Can you explain me please @Xam? – 2017-01-11
  • 0
    As far as I know the factorization in UFD is unique upto associates.Isn't it? – 2017-01-11

0 Answers 0