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 Noetherian domain of dimension two.

Let $\mathfrak{m}_1,\mathfrak{m}_2$ be two disctinct maximal ideals of height two. Are there always infinitely many prime ideals contained in $\mathfrak{m}_1\cap\mathfrak{m}_2$?

I have done the following case:

Let $k$ be an algebraically closed field. Let $R=k[x,y]$. Let $\mathfrak{m}_1, \mathfrak{m}_2,\ldots,\mathfrak{m}_r$ be $r$ disctinct maximal ideals. Then there are infinitely many prime ideals $\mathfrak{p}\subset \bigcap_{i=1}^r \mathfrak{m}_i$.

Proof: By Nullstellensatz, write $\mathfrak{m}_i=(x-a_i,y-b_i)$. If all $a_i$ are distinct. Then we may find a polynomial $F=y-f(x)$ which is irreducible (i.e. the interpolation polynomial). And we may add more points to get more different irreducible polynomials.

Since $\mathfrak{m}_i$ are distinct pairwise. We can always find a linear transformation $A\in GL_2$ so that we come back to above. Here it will use the field $k$ is infinite.

Corollary This is true for $\mathbb{A}^2_k$ for any field $k$.

It seems that it is difficult to go more far.

Let $A$ be a finitely generated $k$-algebra which is a domain of dimension two. Let $\mathfrak{m}_1\neq\mathfrak{m}_2$ be two maximal ideals. I am even unable to find a prime ideal $0\neq\mathfrak{p}\subset \mathfrak{m}_1\cap\mathfrak{m}_2$.


share|cite|improve this question
What do you mean by infinite prime? Googling the term didn't help. – Fredrik Meyer Apr 9 '12 at 17:08
@FredrikMeyer, I am sorry. – wxu Apr 9 '12 at 17:10
@wxu: I have undeleted your earlier question. I will transfer your update to it from this post to the other question. – Zev Chonoles Apr 9 '12 at 17:38
@FredrikMeyer: I believe wxu meant infinitely, not infinite. – Zev Chonoles Apr 9 '12 at 17:40
up vote 1 down vote accepted

For a variety of dimension $n$ over an algebraically closed field, and finitely many closed points, you can always find infinitely many irreducible positive dimensional subvarieties passing through all of them. In your situation, the closed points correspond to maximal ideals, and you can actually find infinitely primes in their intersection (i.e. irreducible subvarieties passing through those points) of any fixed height $c$ (i.e. codimension $c$), for $c=1, \ldots , n-1$. This is done by reducing to the case where your variety is projective (via Chow's lemma), blowing up the points in question, and using Bertini's theorem. This does it for the case where your ring is an affine domain, i.e. a finitely generated algebra over an algebraically closed field which is a domain.

share|cite|improve this answer

It is well known that if $A$ is a PID then the non-maximal prime ideals of $A[x]$ are principal.

Let $A$ be a DVR with uniformizing parameter $t$ and let $R = A[x]$ be the ring of polynomials with coefficients in $A$. For every non-negative integer $n$, the ideal $\mathfrak{m}_n = (t^n x -1)$ is maximal because $R / \mathfrak{m}_n$ is the field of fractions of $A$. Pick the product (i.e the intersection) $I$ of a finite number of $\mathfrak{m}_n$'s. $I$ is a principal ideal and $\{ 0 \}$ is the only prime ideal contained in $I$.

share|cite|improve this answer
From a geometric point of view this counterexample is not really satisfying (see the title), because the maximal ideals you consider are of height one. Actually they are the generic points of "horizontal curves" on $\mathrm{Spec}(A[x])$. – Hagen Knaf Apr 10 '12 at 11:39

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.