# What does “distinct” mean here? 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?

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

## 1 Answer

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 $i \neq j$.