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$a,b\in \mathbb{Z}$

Factors of $a$: $a_1,a_2,...,a_n$.

Factors of $b$: $b_1,b_2,...,b_m$

Prove that if $\frac{a}{b}$ is irreducible, then $\frac{a_ia_j}{b_kb_l}$ is irreducible for all $i,j,k,l$.

I proved the case where $i=j, k=l$ using the fact that the square root of any non-perfect square is irrational.

I can't prove the more general case though.

EDIT: not just prime factors, just all the integer factors. E.g. 12: 1,2,3,4,6,12

Simplex1
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3 Answers3

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If a prime number $p$ divides both $a_i a_j$ and $b_k b_l$, then it divides $a_i$ or $a_j$ hence $a$, and it also divides $b_k$ or $b_l$ hence $b$. This is impossible hence there is no such prime number and $\frac{a_i a_j}{b_k b_l}$ is irreductible.

Gribouillis
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  • Does $p$ need to be prime though? – Simplex1 Aug 04 '20 at 11:47
  • @Simplex1 For the implication $p | (a_i a_j) \Longrightarrow p | a_i \text{ or } p | a_j$, then yes, $p$ needs to be prime. – Gribouillis Aug 04 '20 at 11:57
  • @Simplex1 For the logic behind Gribouillis' answer, yes, $p$ needs to be prime. For the logic behind my alternative (proof by contradiction) approach, no, I could have structured my answer where $d$ is a common factor of $\frac{a_i a_j}{b_k b_l}.$ I simply used prime $p$ as a common factor because I thought that the logic behind my approach would then be more easily seen to be ir-refutable. – user2661923 Aug 04 '20 at 12:02
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Alternative approach : proof by contradiction.

Suppose that $\frac{a_i a_j}{b_k b_l}$ is not irreducible.
Then $\exists \;$ prime $\;p \;\ni p|(a_i \times a_j)$ and $p|(b_k \times b_l) \Rightarrow$
$p|a$ and $p|b \Rightarrow \frac{a}{b}$ is not irreducible.

user2661923
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  • Is this due to the fundamental theorem of arithmetic? If it is irreducible, then $a_i a_j = nb_k b_l$. So there exists a prime p that divides $a_i a_j$ and $nb_k b_l$ . Reductively; p divides some $c$ and $nd$. But how do you go from p divides $nd$ to p divides $d$? I'm just wondering how you went from the preposition to "there exists a prime...". Sorry, I'm pretty new to number theory to put it lightly. – Simplex1 Aug 04 '20 at 11:38
  • @Simplex1 "Is this due to the fundamental theorem of arithmetic?" : yes. Further, given that $(a_i \times a_j)$ divides $a$, since (by assumption) $p$ divides $(a_i \times a_j), p$ must divide any multiple of $(a_i \times a_j).$ Assuming that I have not misunderstood the posting, $a$ is a multiple of $(a_i \times a_j).$ The logic behind concluding that $p$ divides $b$ is identical. – user2661923 Aug 04 '20 at 11:42
  • Oh right that makes sense. Neat – Simplex1 Aug 04 '20 at 12:03
  • @Simplex1 Actually it has nothing to do with FTA. Rather it follows immediately from the coprime hypothesis - see my comment on the question. – Bill Dubuque Jun 20 '21 at 06:08
  • @BillDubuque Your approach is certainly valid. However, my use of the FTA, as an alternative approach is also valid. By the FTA, if $p ~| ~(a_i \times a_j)$ then $p$ shows up in the prime decomposition of $(a_i \times a_j).$ Therefore, WLOG, $p$ shows in the prime decomposition of $a_i$. Therefore, since $a_i ~| ~a$, $p$ must show in the prime decomposition of $a$. Therefore, $p ~| ~a.$ – user2661923 Jun 20 '21 at 07:15
  • I didn't say it was not valid. You don't use the full power of FTA, only that every nonzero nonunit has a prime factor. But you don't need that. Just replace $p$ by any common factor $d$ and then you get the proof in my comment - which is much more general - it works in any domain (even non-UFDs, rings without primes, etc). – Bill Dubuque Jun 20 '21 at 07:27
  • And we don't need FTA to prove $,p\mid a_i a_j\mid a\Rightarrow p\mid a.,$ Rather, that follows by transitivity of divisibility. – Bill Dubuque Jun 20 '21 at 07:31
  • There is however an ambiguity in the question, i.e. is $,i=j,$ allowed, i.e. can we repeat factors in $,a_i a_j$ even if they were not repeated in $,a?,$ If so then we need to say a little more (e.g. use Euclid's Lemma), but we can still prove it using only gcd laws, without any use of FTA or primes. – Bill Dubuque Jun 20 '21 at 07:36
  • @BillDubuque I was about to make the same point. That is, I interpreted the question as allowing $(a) = 12, a_i = 4, a_j = 6.$ Then your transitivity of divisibility argument breaks down. – user2661923 Jun 20 '21 at 07:39
  • @BillDubuque Further, a subjective argument may be made that use of the prime numbers, as I did, provides an easier proof for the OP to understand. Note the OP's 2nd comment, following my answer. – user2661923 Jun 20 '21 at 07:42
  • @BillDubuque Now that I review my comment following the answer, I realize that I (also) may have misinterpreted the original question. But this begs the question: for a new student, what is the easiest argument to understand. – user2661923 Jun 20 '21 at 07:44
  • If the question doesn't allow repeated factors then the transitivity argument is both simpler and more general. If is does allow them then it is a comparison between using Euclid's lemma for prime vs. general divisors - not really much difference in complexity, but lots in generality - since the general form works in any domain. But we're beating a dead horse at this point. – Bill Dubuque Jun 20 '21 at 07:46
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If $\frac{a_ia_j}{b_kb_l}$ were reducible then there would exist a prime number $p\in\mathbb{N}$ such that

$a_ia_j=p\cdot\alpha\;$,

$b_kb_l=p\cdot\beta\;$.

So the prime number $p$ would be a factor of $a_i$ or $a_j$ and it would also be a factor of $b_k$ or $b_l$.

Consequently $p$ would be a factor of $a$ and a factor of $b$ too.

For this reason the fraction $\frac{a}{b}$ would not be irreducible and it is a contradiction.

So it is impossible that $\frac{a_ia_j}{b_kb_l}$ is reducible and it means that $\frac{a_ia_j}{b_kb_l}$ is irreducible.

Angelo
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