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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 $I$ ?

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Just a curiosity: how did you come up with this question? –  user26857 Nov 25 '12 at 8:50
    
@YACP : the way I come up with every question here : trying to solve another question, in this case math.stackexchange.com/questions/236736/… . –  Ewan Delanoy Nov 26 '12 at 4:37

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up vote 3 down vote accepted

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 $I\subset (d)$.

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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.

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It would be nice to have a commutative algebra style proof. If we can show that $I$ has height $1$, then it is contained in a prime ideal $\mathfrak p$ of height $1$. In a UFD, such a prime ideal is principal. Then a generator of $\mathfrak p$ divides $I$. –  user18119 Nov 24 '12 at 9:42
    
what is the height of an ideal? The minimal number of generators (in a Noetherian context)? –  Ewan Delanoy Nov 24 '12 at 16:36
    
@EwanDelanoy: it is the minimum of the heights of the prime ideals containing the ideal. For a prime ideal $\mathfrak p$, the height is the maximum of the lengths of strictly increasing chains of prime ideals contained in $\mathfrak p$. –  user18119 Nov 24 '12 at 18:07
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@YACP : let $\overrightarrow{t}=(t_2, \ldots ,t_m) \in k^{m-1}$, we must show that $\overrightarrow{t}$ is in at least one of the $V_p$. We know that the gcd $G$ of $f_1$ and $t_2f_2+ \ldots +t_mf_m$ is non-constant. Since it divides $f_1$, there is at least one $p$ dividing $G$, qed. –  Ewan Delanoy Nov 25 '12 at 7:08

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