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I'm studying unique factorization domains and I was able to prove that in a UFD if $gcd(a,b)=1$ and $a|bc$ then $a|c$ with the help of the experts here Prove general Euclid's Lemma in a UFD using prime factorization

Now since I'm taking a number theory class too, I've learned this lemma: if $gcd(a,b)=1$ and $a|c$ and $b|c$ then $ab|c$.

My question: Does this lemma hold in a UFD too?? If not can you explain why? It sounds natural to work but I don't know how to prove it....

Hassuni
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1 Answers1

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Yes, this holds in any UFD as well. Write down

$$a=p_1\cdots p_n$$ $$b=q_1\cdots q_m$$ $$c=z_1\cdots z_t$$

where each element on the right side is prime/irreducible.

Then $a|c$ implies that for each $p_i$ there is $z_j$ such that $p_i\equiv z_j$ (i.e. equal up to invertible). And so we have an injective function $A:\{1,\ldots,n\}\to\{1,\ldots, t\}$ with the property that $p_i\equiv z_{A(i)}$.

The same works for $b|c$ relationship, i.e. for any $q_i$ there is $z_j$ such that $q_i\equiv z_j$, giving us an injective function $B:\{1,\ldots, m\}\to\{1,\ldots,t\}$ with the property $q_i\equiv z_{B(i)}$.

Now $gcd(a,b)=1$ implies that $im(A)\cap im(B)=\emptyset$. Indeed, otherwise, if $A(i)=B(j)$ then $p_i\equiv z_{A(i)}\equiv z_{B(j)}\equiv q_{j}$ would be a common divisor of both $a$ and $b$. Therefore we have an injective function

$$A\sqcup B:\{1,\ldots,n,n+1\ldots,n+m\}\to\{1,\ldots, t\}$$ $$A\sqcup B(k)=\begin{cases} A(k) &\text{if }k\leq n \\ B(k-n) &\text{otherwise} \end{cases}$$

Which gives us that $z_i\equiv p_{A\sqcup B(i)}$ when $i\leq n$ and $z_{i-n}\equiv q_{A\sqcup B(i-n)}$ when $i> n$. This means that the decomposition of $a$ and the decomposition of $b$ can be found (disjointly) in the decomposition of $c$. Meaning $ab|c$.

freakish
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  • Hello! And thank you for the proof. I have 2 questions. First if we're factoring to primes why shouldn't it be $a=p_1^{α_1}...p_n^{α_n}$? And I don't think I understand the logic of ...Then $a|c$ implies that for each $p_i$ there is $z_j$ such that $p_i\equiv z_j$ (i.e. equal up to invertible). I am sorry if this sounds dumb but I don't get it :( – Hassuni Dec 18 '20 at 14:22
  • I've never said that $p_i\neq p_j$ when $i\neq j$. For example in naturals your representation of $12$ is $2^2\cdot 3$ while mine is $2\cdot 2\cdot 3$. These are equivalent, but for my usage the representation without powers is simpler. – freakish Dec 18 '20 at 14:46
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    For the second question $a|c$ is equivalent to $c=ar$ for some $r$. By applying the unique factorization to both sides we conclude that the decomposition of $c$ has to contain decomposition of $a$, up to invertible (as with everything in UFD). – freakish Dec 18 '20 at 14:47