I'm doing Gathmann's commutative algebra notes Exercise 2.23:

Let $I$ be an ideal in a ring $R$ with $I\neq R$. We say that a prime ideal $P\unlhd R$ is minimal over $I$ if $I ⊂ P$ and there is no prime ideal $Q$ with $I ⊂ Q \subsetneq P$.
(a) Prove that there is always a minimal prime ideal over $I$.
(b) Determine all minimal prime ideals over $(x^2y, xy^2)$ in $R[x, y]$.
If $R$ is the coordinate ring of a variety and $I$ the ideal of a subvariety, what is the geometric interpretation of a prime ideal minimal over $I$?

The solution to (a) is here.

For (b), we want to find prime ideals between $(x^2y,xy^2)$ and $R[x,y]$ as small as possible. I don't know how to do this in a systematic way, but the obvious chains which contain $(x^2y,xy^2)$ and stop at a prime ideal are:





so $(x)$, $(y)$ are two possible minimal prime ideals over $(x^2y,xy^2)$.

Questions: How do I show there are no prime ideals between $(x)$ and $(x^2y,xy^2)$? How do I know if there are no other chains? For the geometric interpretation, I think to find minimal prime ideals over $I$ is to find irreducible subvarieties such that their intersection contains $V(I)$, is it correct?

  • $\begingroup$ Minimal prime ideals are generic points of irreducible components. Has your course covered generic points yet? $\endgroup$
    – David Lui
    Mar 18 at 13:44
  • $\begingroup$ @DavidLui No, it hasn't been mentioned yet. $\endgroup$ Mar 18 at 13:54

1 Answer 1


First, note that when one is looking for minimal primes over an ideal $I$, one can work with the radical $\sqrt{I}$ instead: $\sqrt{I}=\bigcap_{P\supset I, \text{ P prime}} P$ (ref), so if $P$ is a prime containing $I$ then $P$ must also contain $\sqrt{I}$. This simplifies our problem considerably: $\sqrt{(x^2y,xy^2)}=(xy)$, which is a principal ideal in a UFD.

The minimal primes over a principal ideal $(a)$ in a UFD are exactly the ideals $(a_i)$, where $a_i$ is an irreducible factor of $a$. First, if $(a)\subset P$, then $a=u\prod a_i^{e_i}\in P$, meaning at least one of the $a_i$ is in $P$, so every prime over $(a)$ contains at least one of the $a_i$. Next, $(a_i)$ is actually prime: if $xy\in(a_i)$, then by factoring $x$, $y$, and $xy$ in to irreducibles we see that $a_i$ must divide either $x$ or $y$. Finally, any prime containing more than one is not minimal (it properly contains $(a_i)$, a prime ideal).

For the geometric interpretation, think about the inclusion-reversing properties of $V(-)$: if $I\subset P$ is a minimal prime, then $V(I)\supset V(P)$ and $V(P)$ is a maximal subvariety. This means that $V(P)$ is an irreducible component of $V(I)$.

  • $\begingroup$ In Gathmann's notes, he says that the fact $\sqrt I=\cap_{P\supset I} P$ corresponds to the fact $V(P)$ are irreducible subvarieties of $V(I)$, so I think it is something else here. $\endgroup$ Mar 18 at 18:25
  • $\begingroup$ What do you mean by "... it is something else here."? $\endgroup$
    – KReiser
    Mar 18 at 18:27
  • $\begingroup$ Sorry, I didn't read it carefully, I thought you were saying that in the last paragraph. $\endgroup$ Mar 18 at 18:32
  • $\begingroup$ I can't tell what you're specifically referring to - you're using "it" and "that" a lot without being clear what parts of my writing you're referencing. If you have a question, I'm happy to answer, but please, make it specific so that I can understand exactly what's troubling you and attempt to address it. $\endgroup$
    – KReiser
    Mar 18 at 18:33
  • 1
    $\begingroup$ Well, you know what a subvariety is, and you can talk about containment relationships between subvarieties, right? A maximal subvariety is one that's maximal for the containment relation on subvarieties. This is just a direct translation of $P$ a minimal prime means that if $P'\subset P$ for another prime $P'$, then $P'=P$ to the language of subvarieties by applying $V(-)$ to everything. $\endgroup$
    – KReiser
    Mar 18 at 18:49

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