## The annihilator of the maximal ideal in a local ring

I’ve sometimes been bothered by the notion that in  Noetherian local ring, the annihilator of the maximal ideal could be nonzero.  This is bothersome primarily because in practice, one never sees interesting examples of this occurrence.  I recently was thinking about this  and realized that an easy argument shows that there are, and can be, no interesting examples.

Let $R$ be a (commutative) Noetherian local ring, with maximal ideal $\mathfrak{m}$ and residue field $\Bbbk$.  Let $I$ be the annihilator of $\mathfrak{m}$.  Then the $R$-action on $I$ factors through $\Bbbk$; i.e., $I$ is a $\Bbbk$-vector space.  Moreover, since $I$ is finitely generated ($R$ being Noetherian), it is a finitely-generated vector space, and hence of finite length as an $R$-module.

There is a natural ring structure on $R/I \oplus I$ (this is essentially the associated graded construction, since $I = I/I^2$).  Originally, I had a paragraph here claiming that $R$ is necessarily isomorphic to $R/I \oplus I$; however, I had not really thought about it carefully, and I did not actually have an argument for this.  If this is true, it implies that $R$ can be reconstructed from $R/I$ and the length (= rank = dimension over $\Bbbk$) of $I$, which would give some credence to my claim that there can be “no interesting examples.”  Any comments from readers are welcome.

Posted in Mathematical | 5 Comments

## Natural selection of African-Americans

When I initially wrote this post, I designed it to be as rhetorically strong as possible, rather than to be either true or logically persuasive.  I also put a disclaimer at the beginning stating that this was what I was doing, and that the subsequent argument should not be taken too seriously.  However, when a colleague told me that he found the post offensive in spite of my disclaimer, I decided to moderate it to something closer to my actual beliefs.

So, here is the “more accurate” version: If you are going to do studies comparing intelligence of different “races” (which is already ethically questionable, since the results have the possibility of causing many people distress and very little chance of being useful for any good purpose), and then attribute the results to genetics (which is extremely questionable scientifically, since–among other things–it is virtually impossible to control for culture and upbringing), then you should distinguish whether your results about “black people” are for Africans or for African-Americans.  If you are going to argue that intelligence may have evolved differently on different continents, you should also consider that African-Americans were subject to a different selection process from Africans generally (who should, themselves, probably not be lumped together, but that’s another post).  Briefly, not all “black people” belong to the same group, even for the purposes of your study.

There is also one clarification of the original argument I would like to make. In a parenthetical aside, I observed that Africans who were sold into slavery were typically captured by other Africans.  That is simply my understanding of how the slave trade worked historically (see the book “Slaves in the Family,” by Edward Ball).  It is not relevant to the argument, and is not intended in any way to exonerate the European and American slavers who treated their captives, not as captives, but as disposable merchandise.

Posted in Non-mathematical | Tagged

## The local-to-global lemma

[Prerequisites: Point-set topology]

Proposition: (Local-to-Global Lemma) Let $X$ be a connected topological space.  Suppose $\sim$ is an equivalence relation on $X$ such that every point has a neighborhood of equivalent points.  Then any two points of $X$ are equivalent.

Proof: The equivalence classes of $\sim$ form a partition of $X$ into disjoint open sets.  Since $X$ is connected, there is only one non-empty equivalence class. QED

Exercises:

1. If $X$ is a connected open subset of $\mathbb{R}^n$, then $X$ is path-connected.

2. If $X$ is a connected open subset of $\mathbb{R}^n$, then any two points of $X$ can be connected by a piecewise-linear path in $X$.

3. If $X$ is a connected smooth manifold, then any two points of $X$ can be connected by a piecewise-smooth path.

4. If $X$ is a connected smooth manifold, then any two points of $X$ can be connected by a smooth path.  (Hint: Use bump functions to reparametrize the path from the previous exercise.)

5. Let $X$ be a connected smooth manifold.  For any two points $x, y \in X$, there exists a diffeomorphism $f \colon X \to X$, smoothly isotopic to the identity, such that $f(x) = y$. (Note: This is proved on pp. 23-24 of Milnor’s excellent book Topology from the differentiable viewpoint. In the process, Milnor proves the local-to-global lemma on the fly, without ever explicitly stating it.)

6. A topological space is called irreducible if every non-empty open subset is dense.  Suppose that $X$ is a connected topological space, such that every point has an irreducible neighborhood.  Show that $X$ is irreducible.

Thoughts: After proving that an open subset of Euclidean space is connected iff it is path-connected, point-set topology books typically generalize this by proving statements like “a locally path-connected space is connected iff it is path-connected.” However, the property of local path-connectedness (i.e., every point has arbitrarily small path-connected neighborhoods) is much stronger than the proof really requires.  If one attempts to get the weakest possible hypotheses, one ends up with definitions like the following:

Definition: A space $X$ is semi-locally path-connected if every point has a neighborhood $U$ such that for any two points $y, z \in U$, there is a path in $X$ from $y$ to $z$.

Because such a definition is so clunky, it often gets replaced by the cleaner, but unnecessarily restrictive, definition that “every point has a path-connected neighborhood.”  However, if one remembers the local-to-global lemma, then it becomes easy, not only to remember the weakest possible definition, but to come up with similar definitions on the fly.  Moreover, it provides guidance for proofs in situations like that of exercise 5 above, where coming up with such a “semi-local” property does not really make sense, since it can be shown to hold in every instance for which the definition would be meaningful.

## An amusing variant of the “doomsday argument”

The premise of this argument is that the vast majority of human beings ever to be born will live in a future in which they do not live their own lives.  Instead, they spend their time inside an “experience machine” that allows them to relive the thoughts, sensations, and emotions of selected individuals from the past.  Of course, not just any average Joe is worthy of being relived.  If a life is being relived, it is because it was notable in some important way.

At this point, we bring in a probabilistic argument reminiscent of the doomsday argument.  If the vast majority of humans ever to be born live their lives in the experience machine, then chances are, you yourself are one of these “experience-machine” people, rather than the person you believe you are.  Moreover, there must have been some reason that the person whose life you are reliving was worthy of being relived.

Conclusion: You (or rather, the person whose life you are living) are going to be famous.

Thoughts:  This argument is, of course, not meant to be taken seriously.  But I think it is interesting, and thus fits within the purview of this blog.

Posted in Non-mathematical

## Elimination theory as elimination of quantifiers [not about logic]

[Note: In spite of the term "elimination of quantifiers" in the title, this post does NOT involve mathematical logic--or at least, no more so than, e.g., the distinction between continuity and uniform continuity.]

[Prerequisite: Familiarity with the main theorem of elimination theory, in the form that the projection

$\mathbb{P}^n \times \mathbb{A}^m \to \mathbb{A}^m$

takes Zariski-closed sets to Zariski-closed sets.  Related theorems, such as Chevalley's theorem, are also helpful.]

There are, roughly speaking, two ways in which elimination theory is typically described: a “theoretical” version, such as the statement above; and a more “applied” or “elementary” version, which is about systems of equations and “elimination of variables.”  Both statements are important.  However, I have found the “elimination of variables” statement confusing and difficult to remember, much less apply.

Recently, I happened upon a way of thinking about elimination theory that seems to make it much easier to “translate” between the “applied” description and the “theoretical” description.  The key is to think about elimination theory in terms of elimination of quantifiers, rather than elimination of variables.

Often, in algebraic geometry, we encounter conditions like the following condition on $f$:

$\exists h \text{ s.t. } P(h,f),$

where $P$ is an algebraic condition on $h$ and $f$.  We might have $h$ and $f$ ranging over finite-dimensional vector spaces $V$ and $W$, respectively.  ($V$ and $W$ might be more general varieties, also.) We would like to turn this into a purely algebraic condition on $f$.

Here is the key insight (which, like so many “key insights”, is logically trivial): Let $X \subset V \times W$ be the set of all ordered pairs $(f, g)$ such that $P(f,g)$ holds.  The statement that “$P$ is an algebraic condition” is, presumably, a statement that $X$ is a nice set from an algebro-geometric point of view (e.g., Zariski closed, or possibly Zariski open, or locally closed, or constructible).  The statement above is precisely equivalent to the statement that

$f$ lies in the image of $X$ under the projection $V \times W \to V$.

We are interested in knowing whether this image can be described by a purely algebraic condition; and this is precisely the sort of question that the “theoretical” version of elimination theory is good at answering.

I give three examples to illustrate this technique.  The first, showing that injectivity is an open condition on a space of linear maps, is very much a toy example: there is a more direct proof using determinants that is, I think, a better proof for any purpose other than illustrating the use of elimination theory.  The third example has a direct proof using resultants.  I don’t know a direct proof for the second example, but I would not be at all surprised to learn that there is one. (Incidentally, if you do know one, please let me know about it.)

Example 1: Let $V$ and $W$ be finite-dimensional vector spaces over an infinite field $\Bbbk$. Consider the set $U \subset \mathrm{Hom}_{\Bbbk}(V,W)$ consisting of those linear maps $V \to W$ which are injective.  I will use elimination theory to show that $U$ is a Zariski-open subset of the vector space $\mathrm{Hom}_{\Bbbk}(V,W)$.

Proof: Let $\phi \colon V \to W$ be an element of $\mathrm{Hom}_{\Bbbk}(V,W)$.  Then $\phi$ is injective iff

$\forall v \in V \smallsetminus \{0\},\,\phi(v) \neq 0$.

The first issue is that the quantifier here is $\forall$, whereas the quantifier we can handle with elimination theory is $\exists$.  We can solve this by replacing $\forall$ by the equivalent $\lnot \exists \lnot$, where $\lnot$ denotes “not”:

$\lnot \exists v \in V \smallsetminus \{0\} \text{ s.t. } \phi(v) = 0$.

Now, to make this a statement regarding projective space, note that for any $\lambda \in \Bbbk^*$, we have that $\phi(v) = 0$ if and only if $\phi(\lambda v) = 0$.  Thus, the above is equivalent to

$\lnot \exists [v] \in \mathbb{P}V \text{ s.t. } \phi(v) = 0$.

In other words, the complement of $U$ is the image of $\{(\phi, [v]) \mid \phi(v) = 0\}$ under the projection $\mathrm{Hom}_{\Bbbk}(V, W) \times \mathbb{P}V \to \mathrm{Hom}_{\Bbbk}(V,W)$.  By the main theorem of elimination theory, this image is closed.  Hence, its complement, $U$, is open.

Example 2: Let $S$ denote a finitely generated, graded $\Bbbk$-algebra, with the degree-zero part being precisely $\Bbbk$.  Let $0 < n < m$.  Then $\{(f, g) \in S_n \times S_m \mid f | g \}$ is a Zariski-closed subset of the $\Bbbk$-vector space $S_n \times S_m$.

Proof: The statement $f | g$ is equivalent to the statement that

$\exists h \in S_{m-n} \text{ s.t. } fh=g$.

This, in turn, is equivalent to

$\exists h \in S_{m-n} \text{ s.t. } fh \sim g$,

where we write $a \sim b$ if $a = \lambda b$ for some $\lambda \in \Bbbk^*$.  The condition $f h \sim g$ is closed in all variables and homogeneous in $h$, so by the main theorem of elimination theory, the set of all pairs $(f,g)$ for which such an $h$ exists is closed.

Example 3: With the same notation as above, but allowing $n \leq m$, the set $C := \{(f,g) \in S_n \times S_m \mid f, g \text{ have a common factor}\}$ is closed.

Proof: Let

$C_i = \{(f,g) \in S_n \times S_m \mid f, g \text{ have a common factor in } S_i\}$.

Clearly, $C = \bigcup_{1 \leq i \leq n} C_i$. Thus, it suffices to show that $C_i$ is closed.  Now, $(f,g) \in C_i$ if and only if

$\exists h \in S_i \text{ s.t. } h | f \text{ and } h | g$.

We just showed, in Example 2, that $h | f$ and $h | g$ are closed conditions; hence, their conjunction is also closed.  Furthermore, this condition is clearly homogeneous in $h$.  Hence, by the main theorem of elimination theory, $C_i$ is closed.

Thoughts:  In these examples, I have focused specifically on the use of the main theorem of elimination theory.  However, there are other, related theorems that can also be useful.  For instance, if $X$ and $Y$ are schemes of finite type over $\Bbbk$, then the projection $X \times_{\Bbbk} Y \to Y$ is necessarily open.  This follows from the fact that flat, finite-type morphisms of Noetherian schemes are open, although there is probably a more elementary proof.

Chevalley’s Theorem, stating that the image of a constructible set under a finite-type morphism is constructible (assuming all schemes Noetherian), can also be useful here.  The “elimination-theoretic” version states that if you write out any condition like the ones above (using things like $\forall$, $\exists$, $\lnot$, “and”, “or”, and “implies” at will), then as long as the “innermost conditions” are open or closed, the resulting condition will be constructible.  Since the hypotheses are quite weak, the conclusion in this case is not terribly strong; however, it can still be useful.  For instance, suppose you can write a set as the intersection of an infinite collection of open sets.  If you can describe the same set using finitely many “constructible” conditions, then the set is necessarily open.  [Proof: an arbitrary intersection of open sets is stable under generization.  If a constructible set is closed under generization, then it is open.]

## An immediate proof that algebraically closed fields are infinite

Lemma: A field $\Bbbk$ is algebraically closed if and only if every positive-degree polynomial map $\Bbbk \to \Bbbk$ is surjective.

Proof: If every positive-degree polynomial map is surjective, then in particular, every positive-degree polynomial sends something to zero; hence, $\Bbbk$ is algebraically closed.

Suppose that $\Bbbk$ is algebraically closed and $f \in \Bbbk[x]$ is a positive-degree polynomial.  For every $c \in \Bbbk$, the polynomial $f - c$ has a zero; hence, $f$ maps something to $c$.  QED

Theorem: Every algebraically closed field is infinite.

Proof: Let $\Bbbk$ be a finite field.  Then a map $\Bbbk \to \Bbbk$ is surjective iff it is injective.  Since the polynomial $x(x-1)$ maps both $0$ and $1$ to zero, it is not injective.  QED

Thoughts

When I was an undergraduate, there were essentially only two nontrivial facts I ever needed to know about finite fields.  The first was that every finite field has cardinality equal to a power of a prime.  This has a self-contained proof that was given in each of the first two algebra courses I took.

The second was that every algebraically closed field is infinite.  The usual justification for this was something like, “When you classify all the finite fields, you see that none of them is algebraically closed.”  While the classification of finite fields is beautiful and not terribly difficult, I think the proof above has an enormous advantage in brevity.  I would also note that it does not even require knowing that the order of a finite field is a power of a prime, unlike any proof relying on, e.g., the Frobenius map.

A final note is that a version of the Lemma actually holds for holomorphic functions over $\mathbb{C}$: Every non-constant holomorphic function defined on the entire complex plain is surjective, except possibly at one point.  For instance, the exponential function maps onto $\mathbb{C} \smallsetminus \{0\}$.

Posted in Mathematical | 9 Comments