Traces of nilpotent matrices

I needed the following well-known result in the course I’m teaching:

Lemma. Let k be a field and M \in M_n(k) a nilpotent matrix. Then \operatorname{tr} M = 0.

The classical proof uses the flag of subspaces

    \[0 \subseteq \ker M \subseteq \ker M^2 \subseteq \ldots \subseteq \ker M^n = k^n\]

to produce a basis for which M is upper triangular. Here is a slick basis-independent commutative algebra proof that shows something better:

Lemma. Let R be a commutative ring with nilradical \mathfrak r, and let M \in M_n(k) be a nilpotent matrix. Then the characteristic polynomial p_M(t) = \det(tI-M) satisfies

    \[p_M(t) \in t^n + \mathfrak r[t]_{< n}.\]

Here we write I[t]_{<n} \subseteq R for the polynomials in R of degree smaller than n whose coefficients lie in a given ideal I \subseteq R.

Note that the formulation should ring a bell: in the previous post we saw that R[t]^\times = R^\times + t\mathfrak r[t]. When R is a domain, this reduces to R[t]^\times = R^\times, and the lemma just says that p_M(t) = t^n.

This suggests that we shouldn’t work with \det(tI-M) but with its anadrome (or reciprocal) t^n\det(t^{-1}I-M) = \det(I-tM).

Proof of Lemma. We have to show that \det(I-tM) \in 1 + t\mathfrak r[t]. Since M is nilpotent, there exists r \in \mathbf N with M^{r+1} = 0, so (I-tM)(I+tM+\ldots+t^rM^r) = I. Thus \det(I-tM) \in R[t]^\times = R^\times + t\mathfrak r[t]. Evaluating at t=0 shows that the constant coefficient is 1. \qedsymbol

Invertible polynomials

Here is a classic little lemma on polynomials:

Lemma. Let R be a commutative ring with nilradical \mathfrak r. Then R[t]^\times = R^\times + t\mathfrak r[t].

That is, a polynomial a_nt^n + \ldots + a_1t + a_0 \in R[t] is invertible if and only if a_0 is invertible and a_1,\ldots,a_n are nilpotent. This is the formulation you find in Atiyah–MacDonald, exercise 1.2(i). Note that when R is a domain (or more generally a reduced ring), this just says that R[t]^\times = R^\times (via the constant polynomials).

Proof. First suppose f = \sum_{i=0}^n a_nt^n with a_0 \in R^\times and a_1,\ldots,a_n \in \mathfrak r. Multiplying by a_0^{-1} we may assume a_0 = 1; this does not change the nilpotence assumption on a_1,\ldots,a_n. Then g := 1-f = -a_1t-\ldots - a_nt^n is nilpotent since it is a sum of nilpotent elements -a_it^i in the commutative ring R[t]. Thus we get g^{r+1} = 0 for some r \in \mathbf N, and

    \[(1-g)(1+g+\ldots+g^r) = 1\]

shows that f \in R[t]^\times. Conversely, let f \in R[t]^\times, and suppose fg = 1 for some polynomial g = b_mt^m + \ldots + b_1t+b_0 \in R[t]. Without loss of generality we may assume a_n \neq 0 \neq b_m. If R is a domain, then the leading term of fg is a_nb_mt^{m+n} \neq 0, so we must have n=m=0 and f and g are constant. In the general case, we conclude that \phi(f) \in S^\times \subseteq S[t] for any ring homomorphism \phi \colon R \to S to a domain S. In particular, taking S = R/\mathfrak p for any prime ideal \mathfrak p \subseteq R shows a_1,\ldots,a_n \in \mathfrak p and a_0 \not\in\mathfrak p. Taking the intersection over all prime ideals gives

\qedsymbol   \begin{align*}a_1,\ldots,a_n &\in \bigcap_{\mathfrak p \text{ prime}} \mathfrak p = \mathfrak r,\\ a_0 &\in \bigcap_{\mathfrak p \text{ prime}} (R \setminus \mathfrak p) = R^\times. \end{align*}

The 14 closure operations

I’ve been getting complaints that my lemmas have not been so lovely (or little) lately, so let’s do something a bit more down to earth. This is a story I learned from the book Counterexamples in topology by Steen and Seebach [SS].

A topological space X comes equipped with various operations on its power set \mathcal P(X). For instance, there are the maps S \mapsto S^\circ (interior), S \mapsto \bar S (closure), and S \mapsto S^{\text{c}} (complement). These interact with each other in nontrivial ways; for instance (\bar S)^{\text{c}} = (S^{\text{c}})^\circ.

Consider the monoid M generated by the symbols a (interior), b (closure), and c (complement), where two words in a, b, and c are identified if they induce the same action on subsets of an arbitrary topological space X.

Lemma. The monoid M has 14 elements, and is the monoid given by generators and relations

    \[M = \left\langle a,b,c\ \left|\ \begin{array}{cc} a^2=a, & (ab)^2=ab, \\ c^2=1, & cac=b.\end{array}\right.\right\rangle.\]

Proof. The relations a^2 = a, c^2=1, and cac=b are clear, and conjugating the first by c shows that b^2=b is already implied by these. Note also that a and b are monotone, and a(T) \subseteq T \subseteq b(T) for all T \subseteq X. A straightforward induction shows that if v and w are words in a and b, then vaw(S) \subseteq vw(S) \subseteq vbw(S). We conclude that

    \[abab(S) \subseteq abb(S) = ab(S) = aab(S) \subseteq abab(S),\]

so (ab)^2=ab (this is the well-known fact that ab(S) is a regular open set). Conjugating by c also gives the relation (ba)^2 = ba (saying that ba(S) is a regular closed set).

Thus, in any reduced word in M, no two consecutive letters agree because of the relations a^2=a, b^2=b, and c^2=1. Moreover, we may assume all occurrences of c are at the start of the word, using ac=cb and bc=ca. In particular, there is at most one c in the word, and removing that if necessary gives a word containing only a and b. But reduced words in a and b have length at most 3, as the letters have to alternate and no string abab or baba can occur. We conclude that M is covered by the 14 elements

    \[\begin{array}{ccccccc}1, & a, & b, & ab, & ba, & aba, & bab,\\c, & ca, & cb, & cab, & cba, & caba, & cbab. \end{array}\]

To show that all 14 differ, one has to construct, for any i\neq j in the list above, a set S_{i,j} \subseteq X_{i,j} in some topological space X_{i,j} such that i(S_{i,j}) \neq j(S_{i,j}). In fact we will construct a single set S \subseteq X in some topological space X where all 14 sets i(S) differ.

Call the sets i(S) for i \in \langle a,b\rangle the noncomplementary sets obtained from S, and their complements the complementary sets. By the arguments above, the noncomplementary sets always satisfy the following inclusions:
Inclusions between different setsIt suffices to show that the noncomplementary sets are pairwise distinct: this forces a(S) \neq \varnothing (otherwise a(S) = ba(S)), so each noncomplementary set contains a(S) and therefore cannot agree with a complementary set.

For our counterexample, consider the 5 element poset P given by

    \[\begin{array}{ccccc}p\!\!\! & & & & \!\!\!q \\ \uparrow\!\!\! & & & & \!\!\!\uparrow \\ r\!\!\! & & & & \!\!\!s \\ & \!\!\!\nwarrow\!\!\! & & \!\!\!\nearrow\!\!\! & \\ & & \!\!\!t,\!\!\!\!\! & & \end{array}\]

and let X be the disjoint union of P^{\operatorname{Alex}} (the Alexandroff topology on P, see the previous post) with a two-point indiscrete space \{u,v\}. Recall that the open sets in P^{\operatorname{Alex}} are the upwards closed ones and the closed sets the downward closed ones. Let S = \{p,s,u\}. Then the diagram of inclusions becomes Lattice of inclusions for the exampleWe see that all 7 noncomplementary sets defined by S are pairwise distinct. \qedsymbol

Remark. Steen and Seebach [SS, Example 32(9)] give a different example where all 14 differ, namely a suitable subset S \subseteq \mathbf R. That is probably a more familiar type of example than the one I gave above.

On the other hand, my example is minimal: the diagram of inclusions above shows that for all inclusions to be strict, the space X needs at least 6 points. We claim that 6 is not possible either:

Lemma. Let X be a finite T_0 topological space, and S \subseteq X any subset. Then aba(S) = ab(S) and bab(S) = ba(S).

But in a 6-element counterexample X, the diagram of inclusions shows that any point occurs as the difference i(S) \setminus j(S) for some i,j \in \langle a,b\rangle\setminus 1. Since each of i(S) and j(S) is either open or closed, we see that the naive constructible topology on X is discrete, so X is T_0 by the remark from the previous post. So our lemma shows that X cannot be a counterexample.

Proof of Lemma. We will show that ab(S) \subseteq aba(S); the reverse implication was already shown, and the result for bab(S) follows by replacing S with its complement.

In the previous post, we saw that X is the Alexandroff topology on some finite poset (X,\leq). If T \subseteq X is any nonempty subset, it contains a maximal element t, and since X is a poset this means u \geq t \Rightarrow u=t for all u \in T. (In a preorder, you would only get u \geq t \Rightarrow u \leq t.)

The closure of a subset T \subseteq X is the lower set \bigcup_{t \in T} X_{\leq t}, and the interior is the upper set \{t \in T\ |\ X_{\geq t} \subseteq T\}. So we need to show that if x \in ab(S) and y \geq x, then y \in ba(S).

By definition of ab(S), we get y \in b(S), i.e. there exists z \in S with z \geq y. Choose a maximal z with this property; then we claim that z is a maximal element in X. Indeed, if u \geq z, then u \geq x so u \in b(S), meaning that there exists v \in S with v \geq u. Then v \geq z \geq y and v \in S, which by definition of z means v=z. Thus z is maximal in X, hence z \in a(S) since z \in S and X_{\geq z} = \{z\}. From z \geq y we conclude that y \in ba(S), finishing the proof. \qedsymbol

The lemma fails for non-T_0 spaces, as we saw in the example above. More succinctly, if X = \{x,y\} is indiscrete and S = \{x\}, then aba(S) = \varnothing and ab(S) = X. The problem is that x is maximal, but X_{\geq x} \supsetneq \{x\} and x \not\in a(S).


[SS] L.A. Steen, J.A. Seebach, Counterexamples in Topology. Reprint of the second (1978) edition. Dover Publications, Inc., Mineola, NY, 1995.