In an ideal, pairwise non-coprime implies globally non-coprime? 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 $I$ ?
 A: 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 $I\subset (d)$.
A: Suppose that the ground field $k$ is infinite.
Let $f_1, \dots, f_m$ be a system of generators of $I$ with $f_1\ne 0$.  For each irreducible factor $p$ of $f_1$, consider the sub-vector space of $k^{m-1}$
$$V_p=\{ (t_2, \dots, t_m) \in k^{m-1}\mid p \text{ divides  } t_2f_2+\dots+t_m f_m \}.$$ 
Then $k^{m-1}$ is the union of the various $V_p$. As Ewan explained  in the comments, this is because for all $(t_2,\dots, t_m)\in k^{m-1}$, the sum $t_2f_2+\dots+t_m f_m\in I$ has a commun irreducible factor $p$ with $f_1$ and then $(t_2,\dots, t_m)\in V_p$. As $f_1$ has only finitely many irreducible factors and $k$ is infinite, $k^{m-1}=V_p$ for some irreducible factor $p$ of $f_1$. This implies in particular that $p$ divides $f_i$ for all $i\le m$, hence $p$ divides all elements of $I$. 
Remark If $k$ is finite, the above proof doesn't work. We could consider instead of constants $t_2,\dots, t_m$ polynomials of some bounded degrees $d$. But I don't have a clear idea of whether this will work or not.
