Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $I$ ?

share|cite|improve this question
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… . – Ewan Delanoy Nov 26 '12 at 4:37
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 $I\subset (d)$.

share|cite|improve this answer

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.

share|cite|improve this answer
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
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.