$a\mid bc$ where $a$, $b$ are relatively prime implies $a\mid c$ in a unique factorization domain

Question

I'm asked to solve

If $R$ is a unique factorization domain and for $a$, $b$ two elements that are relatively prime in $R$ and $a$ divides $bc$, then $a$ divides $c$.

While trying to prove this, I've proved that $a$ and $c$ are associate. And this is more than what is required, so I'm afraid I've done something wrong. Can anybody help me?

First of all, if $a$ was irreducible, then it is prime and we are done. If it wasn't then it can be written as a finite product of irreducible elements, $b$ and $c$ can be written similarly.

$a$ divides $bc$ implies that $bc = ka$ for some $k \in R$. Now since the irreducible elements need to be associate and $a$ doesn't divide $b$. the irreducible factors of $a$ are associate to the ones of $c$. Which implies that $a$ is associate to $c$.

Answer 1

There is an easy proof that does not require unique factorisation, just the existence of least common multiples. Since $\gcd(a,b)=1$ and a least common multiple of $a,b$ exists, that least common multiple must be (the association class of) $ab$. Now $bc$ is by hypothesis a multiple of $a$ and it is clearly a multiple of $b$, so by the defining property of least common multiples it is a multiple of $ab$. We have $ab\mid bc$, and one can simplify that by$~b$ to give $a\mid c$.

Answer 2

The same proof that works for $\mathbb Z$ works for every UFD.

For each prime $p$ in $R$, let $v_p(x)$ be the exponent of $p$ in the factorization of $x$. Then $a$ divides $bc$ iff $v_p(a) \le v_p(bc)$ for all $p$. Now, $ v_p(bc) = v_p(b) + v_p(c )$ and so $v_p(a) \le v_p(b) + v_p(c )$. Since $a$ and $b$ are relatively prime, $v_p(b)=0$ whenever $v_p(a)>0$. Hence $v_p(a)>0$ implies $v_p(a) \le v_p(c )$ for all $p$. This means that $a$ divides $c$.

If the proof above is too abstract, here it is made concrete:

Since $a$ divides $bc$, we can write $ad=bc$, for some $d\in R$. Now consider the factorization into irreducibles (or primes): $$ u_a p_1^{\alpha_1}\cdots p_n^{\alpha_n}\ u_d p_1^{\delta_1}\cdots p_n^{\delta_n} = u_b p_1^{\beta_1}\cdots p_n^{\beta_n}\ u_c p_1^{\gamma_1}\cdots p_n^{\gamma_n} $$ for some units $u_*$. Here I have used the same set of primes for $a$, $b$, $c$, and $d$. Just use a zero exponent if a prime does not divide one of those. It does no harm.

All this means that $\alpha_i+\delta_i = \beta_i+\gamma_i$. Since $a$ and $b$ are relatively prime, $\beta_i=0$ whenever $\alpha_i>0$ and so in this case $\alpha_i\le\alpha_i+\delta_i = 0+\gamma_i=\gamma_i$. This means that $a$ divides $c$.

Answer 3

Your suspicion is correct: you can't deduce that $a$ and $c$ are associate. Think to the case $a=6$, $b=5$ and $c=12$ in the integers: $a$ and $b$ are coprime, $a$ divides $bc=60$, but $a$ and $c$ are certainly not associate.

Here's a possible proof by induction on the number of terms in a factorization of $a$, let's call it the length of $a$.

If the length of $a$ is $1$, then $a$ is irreducible (and so prime); since $a$ does not divide $b$, but divides $bc$, it must divide $c$.

Now suppose the assertion is true for elements that have length $n$ and assume $a$ has length $n+1$. Then we can write $a=pA$, where the length of $A$ is $n$ (fill in details) and $p$ is prime. Note that $A$ and $b$ are coprime (fill in details).

Since $p$ divides $a$ and so divides $bc$, we get that $p$ divides $bc$. But $p$ does not divide $b$, because $a$ and $b$ are coprime (fill in details). Therefore $p$ divides $c$ and we can write $c=pC$. So we have that $a=pA$ divides $bc=pbC$ and so $A$ divides $bC$. Since $A$ has length $n$ and is coprime with $b$, the induction hypothesis tells us that $A$ divides $C$: $C=AD$. Therefore $$ c=pC=pAD=aD $$ and so $a$ divides $c$.

Fill in the details and the proof is complete.