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Let $A$ be a UFD. Assume that $a,b \in A$ are relatively prime, $c \in A$ and $a | bc$. To prove that $a|c$, is the following approach correct (or do you have to use some type of prime factorization argument)?

By the relatively prime assumption, $\gcd(a,b)$ is a unit. Call that unit $u \in A^{\times}$. Then we can write $u = ax + by$ for some $x,y \in A$. But then, $c = c \cdot u = c(ax+by)=cax+cby$. However, it is clear that $a|cax$ and the fact that $a|bc$ implies $a|cby$. Thus, $a|cax+cby=c$.

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I edited the title to make it more descriptive. Please feel free modify it as you wish. –  Bill Dubuque Dec 7 '12 at 3:19

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up vote 2 down vote accepted

No, e.g. in $\rm\:\Bbb Q[x,y],\:$ $\rm\:gcd(x,y) = 1\:$ but $\rm\: x r + y s = 1\:\Rightarrow\: 0 = 1\:$ by evaluating at $\rm\,x,y=0.$

Here are $\,2\,$ proofs, using $(1)$ existence and uniqueness of prime factorizations, and $(2)$ gcds.

$(1)$ $\rm\ a\mid bc\:$ so $\rm\:ad = bc\:$ for some $\rm\:d.\:$ By existence, we can factor all four terms into prime factors. By uniqueness, the same multiset of primes occurs on both sides. So all of the primes in the factorization of $\rm\:a\:$ must also occur on the RHS, necessarily in the factorization of $\rm\:c,\:$ since, by hypothesis, $\rm\:a,b\:$ are coprime, hence share no prime factors. Therefore $\rm\:a\mid c\:$ since all of its prime factors (counting multiplicity) occur in $\rm\:c.\:$ Note that this inference can be expressed purely multiset-theoretically: $\rm\: A\cap B = \{\,\},\ \ A \cup D\,=\, B\cup C\:$ $\:\Rightarrow\:$ $\rm\:A\subset C,\:$ where $\rm\:C =$ multiset of primes in the unique prime factorization of $\rm\:c,\:$ and similarly for $\rm\:A,B,D.$

$(2)\ $ Using gcds: $\rm\ \ a\mid ac,bc\:\Rightarrow\:a\mid \color{#0A0}{(ac,bc)=(a,b)c} = c\:$ by $\rm\:(a,b)=1,\:$ and by the following

Theorem $\rm\ \ \ (a,b)\ =\ (ac,bc)/c\quad \rm\color{#0A0}{(\,gcd\ distributive\ law)}$

Proof $\rm\quad\ d\mid a,b\ \iff\ dc\mid ac,bc\ \iff\ dc\mid (ac,bc)\ \iff\ d\mid(ac,bc)/c$

The theorem holds more generally in any domain as long as $\rm\ (ac,bc)\ $ exists. See my post here for further discussion of this property and its relationship with Euclid's Lemma.

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so how would you go about proving it then? –  afedder Dec 7 '12 at 1:53
    
Since $A$ is a UFD, we can let $a=p_1p_2...p_n$ for some primes $p_1,p_2,...,p_n \in A$. Then, for all $i$, $p_i$ does not divide b. But I don't know how to proceed. –  afedder Dec 7 '12 at 1:55
    
So you don't need a prime factorization argument? –  afedder Dec 7 '12 at 1:57
1  
No, factorization into primes isn't needed, one needs only the existence of gcds, as the second proof above shows. The gcd argument is more general since there are domains which have no primes yet gcds exist, e.g. the ring of all algebraic integers, which has no irreducibles (so no primes) since every element factors, e.g. $\rm\: a = \sqrt{a}\cdot \sqrt{a}\ \ $ –  Bill Dubuque Dec 7 '12 at 2:32

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