TITLE: In an ideal, pairwise non-coprime implies globally non-coprime?
QUESTION [6 upvotes]: Let $R$ be a polynomial ring $R=k[X_1,X_2, \ldots ,X_n]$. Let $I$ be an ideal of $R$ such that any two elements of $I$ have a non-constant gcd. Does it follow that there is a non-constant $D$ dividing all the elements of $I$ ?

REPLY [3 votes]: The polynomial ring $R=k[X_1,\ldots,X_n]$ is, in particular, a Krull domain. For Krull domains we have the following result (a generalization of the well known result for Dedekind domains which says that every ideal is $2$-generated):
Theorem. Let $R$ be a Krull domain and $I$ a fractionary ideal of $R$. Then there exist $a,b\in I$ such that $R:(R:I)=R:(R:(a,b))$.
Proof. See R. Fossum, The Divisor Class Group of a Krull Domain, Proposition 5.11.
In our case set $d=\gcd(a,b)$. Then $(a,b)\subset (d)$ and therefore $R:(R:(a,b))\subset (d)$. It is obvious that $I\subset R:(R:I)$, so $I\subset (d)$.