Greatest Common Divisors in integral domains

Question

Assume that $R$ is an integral domain. I want to prove that every finite subset $B$ has a greatest common divisor if and only if $B$ has a least common multiple. (As the (helpful) comment below shows, this is not immediately obvious.) I tried to use mathematical induction on the number of elements in $B$. If $B$ has 2 elements, say $b_1$, and $b_2$ then the identity $${\rm gcd}(b_1, b_2){\rm lcm}(b_1, b_2)={b_1}{b_2}$$ helps. The above identity can be proved using elementary divisibility arguments, and my checking it through seems to show that holds in an arbitrary integral domain. But even if $B$ has $3$ elements, the identity linking the product to the greatest common divisor, and least common multiple becomes complicated. Not sure how to set up the induction step. Thanks.

Answer 1

but I am trying to answer the question in a general integral domain

Not possible, since there are examples of domains that don't admit GCDs. The special ones that do are called gcd domains and that is indeed equivalent to having lcms.

If two elements in an integral domain are coprime, can one write 1 as an R -linear combination of these elements?

No, for example, try to write $1$ as a combination of $x$ and $y$ in $\mathbb Z[x,y]$.

I tried to prove this by induction on the number of elements in B

I think all you need induction for, maybe, is to reassure yourself that the gcd condition you have works for all such $B$ iff it works for all pairs of elements. The same goes for lcms, then you can work with the much simpler pairwise conditions.

If the gcd of any subset B in an integral domain R can be written as a finite linear combination of the elements in B (with R coefficients), then R must be a PID. No idea how to prove this!

Neither do I, because this is also not true. A Bézout domain is a GCD domain in which you can write GCDs as a linear combination, but Bezout domains are not PIDs in general.

Answer 2

We argue by mathematical induction. If $B$ is a singleton $\{b\}$, then $b$ is both its own greatest common divisor, and its own least common multiple. So the existence of $b$ ensures that the existence of the gcd is equivalent to the existence of the lcm. If $n=2$, and $B$ is an arbitrary subset $B=\{b_1, b_2\}$, then the identity $${\rm gcd}(b_1, b_2){\rm lcm}(b_1, b_2) = {b_1}{b_2}$$ proves that the existence of the gcd is equivalent to the existence of the lcm. To treat the case with more than two elements, I first established the following identities by mathematical induction $${d_n}:={\rm gcd}(b_1,\cdots ,b_n) = {\rm gcd}({\rm gcd}(b_1,\cdots ,b_{n-1}), b_n),$$ and $${m_n}:={\rm lcm}(b_1,\cdots ,b_n) = {\rm lcm}({\rm lcm}(b_1,\cdots ,b_{n-1}), b_n).$$ Then assume (as our induction hypothesis) that the existence of the gcd is equivalent to the existence the lcm for any subset consisting of $k$ elements for any $2\leq k\leq n$, and assume that $B$ is an arbitrary subset of $R$ consisting of $n+1$ elements. If $d_{n+1}$ exists, since $$d_{n+1}={\rm gcd}(d_n,b_{n+1}),$$ the existence of $d_n$ implies the existence of $m_n$, (by our induction hypothesis), and by the result for $n =2$, we conclude that $${\rm lcm}(m_n, b_{n+1}) = {\rm lcm}(b_1,\cdots ,b_{n+1})=m_{n+1}$$ must exist. The converse implication is proved with a very similar argument, and the assertion follows by mathematical induction.