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Proposition (Dummit&Foote - p.287)

Let $a,b$ be two nonzero elements of a unique factorization domain $R$ and suppose $a=up_1^{e_1}\cdots p_n^{e_n}$ and $b=vp_1^{f_1}\cdots p_n^{f_n}$ are prime factorizations for $a$ and $b$, where $u,v$ are units, the primes $p_1,...,p_n$ are distinct and the exponents $e_i$ and $f_i$ are $\geq 0$. Then $p_1^{min\{e_1,f_1\}} ... p_n^{min\{e_n,f_n\}}$ is a g.c.d of $a$ and $b$.

What does "distinct" mean here?

If this just means that $p_i\neq p_j$, then I think this proposition is false.

And below is why I think this is false.

Need a verification of this

( Suppose $p_1,...,p_n$ are primes such that each of them does not associate to another. Let $p_{n+1}$ be a prime which associates to $p_n$, but $p_n\neq p_{n+1}$. Then $a,b$ can have two different representation. One consists of $p_1,...,p_n$ and another consists of $p_1,...,p_{n+1}$. Since $min\{e_n,f_n\}+min\{e_{n+1},f_{n+1}\}\neq min\{e_n+e_{n+1},f_n+f_{n+1}\}$, $a,b$ can have two distinct g.c.ds each of which does not associate to another. This is impossible.)

So, does the hypothesis of this proposition mean that $p_i$'s don't associate to others?

  • $\begingroup$ Probably:$i\neq j\implies p_i$ and $p_j$ are not associated. $\endgroup$ – drhab Feb 20 '15 at 12:00
  • $\begingroup$ @drhab so does this proposition assume $p_i$ and $p_j$ to be not associated? And is my argument correct? $\endgroup$ – Rubertos Feb 20 '15 at 12:01
  • 1
    $\begingroup$ Yes, your argument is correct. $\endgroup$ – Crostul Feb 20 '15 at 12:02
  • $\begingroup$ @Crostul Thank you for checking. $\endgroup$ – Rubertos Feb 20 '15 at 12:07

This community wiki solution is intended to clear the question from the unanswered queue.

By distinct he means they are not associates, i.e., $p_i \nsim p_j$ if $i \neq j$.


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