3

Is there a proof for the problem below?

$R$ is a commutative, integral domain with unity in which for each pair $a,b\in R$, g.c.d. $(a,b)$ exists. I want to show that if $(a,b)=1$ and if $a\mid c$ and $b\mid c$ then $ab \mid c$.

In a Bézout domain we can apply Bézout's identity as they do here.

Is there a way to proceed without using Bézout's identity?

Community
  • 1
Calvin Nesbitt
  • 361
  • Presumably, you want without unique factorization, too. It follows from unique factorization. – Thomas Andrews Oct 10 '15 at 00:24
  • Yes, $R$ isn't a UFD as stated – Calvin Nesbitt Oct 10 '15 at 00:25
  • Well, no, you didn't state that. There are plenty of UFDs without Bezout, for example. – Thomas Andrews Oct 10 '15 at 00:29
  • Apologies, I meant that in the question $R$ is defined in such a way that it isn't necessarily a UFD. – Calvin Nesbitt Oct 10 '15 at 00:31
  • Is there a reason you believe it is true in this conditions? Like, was it a problem in a book? – Thomas Andrews Oct 10 '15 at 00:33
  • Yes, it's from Bhattacharya, Jain, Nagpaul's "Basic Abstract Algebra". I should also note that the problem doesn't explicitly say that one can't use Bézout. Just the way that $R$ has been defined means we can't. – Calvin Nesbitt Oct 10 '15 at 00:35
  • Okay, it's just a weirdly stated problem, so I thought it might have been a conjecture from you. The problem specifically mentions without Bézout's, but the assumptions are much weaker than that - the ring only needs a gcd, so there are lots of theorems weaker than Bézout's that are being excluded, as well. – Thomas Andrews Oct 10 '15 at 00:51
  • I edited your problem statement to make it clear that by "without using Bézout" you are limiting the hypotheses, not the possible approaches to a proof (the first is easily quantifiable, but the second is not and often leads to confusion on the part of answerers) – Mario Carneiro Oct 10 '15 at 01:51

2 Answers2

2

As user26857's comment on Kishor J's answer makes clear, the important point is showing that for any $a,b,c$ we have $(ca,cb)=c(a,b)$.

Since we are in an integral domain, we have the following:

If $xy\mid xz$, then $y\mid z$

Now observe that $c(a,b)$ divides both $ca$ and $cb$. So $c(a,b)\mid (ca,cb)$. We want to show that the opposite holds as well.

First observe that we have $c\mid c(a,b)\mid (ca,cb)$. So we may write $(ca,cb)$ as $cg$ for some $g$. Now we have:

  • $c(a,b)\mid cg$ - so $(a,b)\mid g$.

It remains to show that $g\mid (a,b)$. We note that:

  • $gc=(ca,cb)\mid ca$ - so $g\mid a$
  • $gc=(ca,cb)\mid cb$ - so $g\mid b$

Therefore, $g\mid (a,b)$, and so $g=(a,b)$ (up to multiplication by units). We get that $(ca,cb)=c(a,b)$.

Now we need only observe that if $(a,b)=1$ then $(ca,cb)=c$, and the result is clear, since $ab\mid ca,cb$ if $a,b\mid c$.

John Gowers
  • 25,678
0

$ab \mid ca$ and $ab \mid cb$. Hence, $ab \mid (ca, cb)$. But $(ca, cb)$ is $c$ if $(a, b) = 1$. Therefore, $ab \mid c$.

user26857
  • 53,190
Kishor J
  • 1