We know that if $A$ is a PID, then we can guarantee the existence of a $\gcd$ between two elements. My question is, does the converse hold? If I know that every two elements have a $\gcd$, do I know that the ring is a PID? My intuition says no, in that case I would like a counter-example.
Thanks.
PIDs are precisely the UFDs with dimension $\le 1\,$ (i.e. every prime ideal $\ne 0$ is maximal). Hence any higher dimensional UFD provides a counterexample (because UFDs are always gcd domains), for example $\,\Bbb Z[x],\, \Bbb Q[x,y].\,$ More generally we have
Theorem $\rm\ \ \ TFAE\ $ for a $\rm UFD\ D$
$(1)\ \ $ prime ideals are maximal if nonzero, $ $ i.e. $\rm\ dim\,\ D \le 1$
$(2)\ \ $ prime ideals are principal
$(3)\ \ $ maximal ideals are principal
$(4)\ \ \rm\ gcd(a,b) = 1\, \Rightarrow\, (a,b) = 1, $ i.e. $ $ coprime $\Rightarrow$ comaximal
$(5)\ \ $ $\rm D$ is Bezout
$(6)\ \ $ $\rm D$ is a $\rm PID$
Proof $\ $ (sketch of $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1)\ $ where $\rm\,p_i,\,P\,$ denote primes $\neq 0$
$(1\Rightarrow 2)$ $\rm\ \ p_1^{e_1}\cdots p_n^{e_n}\in P\,\Rightarrow\,$ some $\rm\,p_j\in P\,$ so $\rm\,P\supseteq (p_j)\, \Rightarrow\, P = (p_j)\:$ by dim $\le1$
$(2\Rightarrow 3)^{\phantom{|^i}}\!\!\!$ $ \ $ max ideals are prime, so principal by $(2)$
$(3\Rightarrow 4)^{\phantom{|^i}}\!\!\!$ $\ \rm \gcd(a,b)=1\,\Rightarrow\,(a,b) \subsetneq (p) $ for all max $\rm\,(p),\,$ so $\rm\ (a,b) = 1$
$(4\Rightarrow 5)^{\phantom{|^|}}\!\!\!$ $\ \ \rm c = \gcd(a,b)\, \Rightarrow\, (a,b) = c\ (a/c,b/c) = (c)$
$(5\Rightarrow 6)^{\phantom{|^|}}\!\!\!$ $\ $ Ideals $\neq 0$ in Bezout UFDs are generated by an elt with least #prime factors
$(6\Rightarrow 1)^{\phantom{|^|}}\!\!\!$ $\ \ \rm (d) \supsetneq (p)$ properly $\rm\Rightarrow\,d\mid p\,$ properly $\rm\,\Rightarrow\,d\,$ unit $\,\rm\Rightarrow\,(d)=(1),\,$ so $\rm\,(p)\,$ is max
