# Geometric interpretation of Minimal Prime Ideals

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:

$$(x^2y,xy^2)⊂(x^2y)⊂(x^2)⊂(x)$$,

$$(x^2y,xy^2)⊂(x^2y)⊂(y)$$,

$$(x^2y,xy^2)⊂(xy^2)⊂(x)$$,

$$(x^2y,xy^2)⊂(xy^2)⊂(y^2)⊂(y)$$,

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?

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

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 $$V(I)$$.
• 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. Mar 18 at 18:25
• 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. Mar 18 at 18:49