Posted by: Alexander Ellis | November 15, 2008

## A Silly Prime Avoidance Lemma

The other day I read this post over at Notational Notions. The main result is the following.

Proposition: Let $k$ be an infinite field and $V$ any $k$-vector space. Then $V$ cannot be written as a union of finitely many proper subspaces.

$V$ can be either finite- or infinite-dimensional in the above. The very basic idea here is that in an algebraic setting, unions of “algebraically nice” subsets need not be “algebraically nice.” An immediate corollary of this Proposition is the following (somewhat silly) version of the Prime Avoidance Lemma.

Silly Prime Avoidance (SPA) Lemma: Let $R$ be a ring containing an infinite field. Let $J,I_1,\ldots,I_n$ be ideals of $R$ such that $J\subset\bigcup_jI_j$. Then $J\subset I_j$ for some $j$.

The goal of this post is to discuss the SPA Lemma and to see what it means in terms of “geometry over a finite field.”

Proof: Considering $R$ as a $k$-vector space, every ideal of $R$ is a subspace. Now apply the Proposition with $V=J$, and the subspaces in question being $I_j\cap J$. Since there are finitely many $I_j$, we can only have $J\subset\bigcup_j I_j$ if $I_j\cap J=I_j$ for some $j$. Q.E.D.

(Recall that the full Prime Avoidance Lemma allows the following modification of the hypotheses: $R$ an arbitrary ring, as long as at most two of the $I_j$ are non-prime. While neither of the two sets of hypotheses are stronger than the other, the “Silly” version seems a bit less “serious” to me. See Chapter 3 of Eisenbud’s Commutative Algebra with a View Toward Algebraic Geometry for a thorough discussion.)

Can we relax the condition that $k$ be infinite? Certainly not, since if $k$ is finite, then $R$ can be a finite set, which is easy to exhaust by a finite union: Consider $\mathbb{F}_2[x,y]/(x^2,xy,y^2)$. Then the ideal $(x,y)$ is in the union of the ideals $(x)$, $(y)$, and $(x+y)$. but is not contained in any of them individually. (Note that none of the ideals $(x)$, $(y)$, or $(x+y)$ are prime.) So really, in order to ask whether or not we can strengthen the SPA Lemma, we should address the following problem.

Problem: Let $k$ be a finite field and let $R$ be a unital $k$-algebra. Find an ideals $J,I_1,\ldots,I_n$ of $R$ such that $J\subset\bigcup_jI_j$ but $J$ not contained in $I_j$ for all $j$. Let’s ask that $R$ be Noetherian, too. (Hints (highlight to view): 1. We will need at least three $I_j$ which are not prime, by the full Prime Avoidance Lemma. 2. Clearly, $J$ cannot be a principal ideal; in particular, $R$ cannot be a PID.)

While you think about the Problem, let’s consider what this means for geometry over finite fields. Let $R=k[x_1,\ldots,x_m]/(F_1,\ldots,F_k)$ be a finitely presented $k$-algebra and $J=(f_1,\ldots,f_s)$ an ideal of $R$. Suppose $Z_1, \ldots, Z_n$ are closed subsets of $\text{Spec}(R)$ with corresponding ideals $I_j$. The SPA Lemma says that if on each particular $Z_j$ not all the $f_i$ vanish, then there is some $h = \sum g_jf_j$ (with $g_j\in R$) which vanishes on none of the $I_j$. (When we ask if $f$ vanishes at a point of $\text{Spec}(R)$, we are asking whether the image of f in the residue field at the corresponding prime is zero; see Chapter 1 of Eisenbud and Harris’s The Geometry of Schemes.).

What, then, would a solution to the Problem be?  It would be an affine scheme $X=\text{Spec}(R)$ (or affine algebraic set, if you don’t like schemes and don’t mind taking R to be nilpotent-free) over $\mathbb{F}_q$ and a finite set of polynomials $f_1, \ldots, f_s$ on $\text{Spec}(R)$, along with closed subsets $Z_1, \ldots, Z_n$ such that every combination $\sum f_ig_i$ (with $g_i \in R$) vanishes on one of the $Z_j$.  One consequence of this is that we have no hope of “simultaneously separating” the $Z_j$ using a function from the (non-unital) subalgebra $J = (f_1, \ldots, f_s) \subset R$: i.e., there cannot be a function $h \in J$ which takes different values on each of the $Z_n$.  One possible intuitive explanation for this is that the residue fields are extensions of $\mathbb{F}_q$; in nice cases, then, they will all be finite themselves.  And as long as they don’t vary too wildly (e.g., if there is a $q'$ such that all the residue fields are $\mathbb{F}_{q'}$ at the largest), then the set of possible values of functions is a finite set; by the Pigeonhole Principle, if we choose $n$ too large, then we’ll never separate them all simultaneously.  Our result is stronger: not only will some value repeat, but at least one of the values will always be zero.

Does anyone know a better geometric intuition for this, or a more precise statement of what sort of “non-separation” this entails?  I would love to understand this more clearly.  More generally, I’d like to understand how well we can separated closed subsets of arbitrary schemes.  Anyway, we conclude with a solution to our problem.

Solution: (Highlight to view.)  Let $R = \mathbb{F}_2[x,y]$ and $J = (x,y)$.  Let $I_1' = (x)$, $I_2' = (y)$, and $I_3' = (x+y)$; note that these may be prime.  For $j=1,2,3$, let $I_j$ be the ideal generated by $I_j'$ and all polynomials of degree 2 and higher with no linear or constant term.  Since R is Noetherian, these must be finitely generated (even though we don’t need this to be true, it’s nice to know it is).  It is easily verified that this is a solution to the problem.

Remark 1: (Highlight to view.)  This solution shows that the hypothesis that “no more than two of the $I_j$ are non-prime” is sharp.

Remark 2: (Highlight to view.)  Two easy generalizations of the solution: you can do this over any $\mathbb{F}_q$ and for any number of variables by just adding another ideal of the form $I_j$ for each possible homogeneous linear polynomial.  And the quotient of this by any ideal generated by functions with no linear or constant term will still be an example; in particular, this yields an example of Krull dimension one, e.g. $\mathbb{F}_2[x,y]/(y^2-x^3)$.

1. Updated: a slight correction made to the definition of $I_j$ in the solution (only adjoining higher-degree generators with no linear part).