The combinatorial Nullstellensatz

The “combinatorial Nullstellensatz” is a relatively elementary statement due to Noga Alon (1999) whose name, while possibly frightening, really says what it is and what it is good for. (A freely available version is there.) Nullstellensatz is the classic name for a theorem of David Hilbert that relates loci in $F^n$ defined by polynomial equations and the ideals of the polynomial ring $F[T_1,\dots,T_n]$ generated by these equations, when $F$ is an algebraically closed field. The word itself is German and means “theorem about the location of zeroes”. The precise correspondence is slightly involved, and stating it here precisely would derail us from the goal of this post, so let us stay with these informal words for the moment. The adjective combinatorial in the title refers to the fact that Alon deduced many beautiful consequences of this theorem in the field of combinatorics, and for some of them, furnishes quite simple proofs. We'll give one of them below, the Cauchy—Davenport inequality, but my main motivation in writing this text was to discuss the proof of the combinatorial Nullstellensatz, in particular untold aspects of the proofs which took me some time to unravel.$\gdef\Card{\operatorname{Card}}$

The context is the following: $F$ is a field, $n$ is a positive integer, and one considers finite sets $S_1,\dots,S_n$ in $F$, and the product set $S=S_1\times \dots \times S_n$ in $F^n$. One considers polynomials $f\in F[T_1,\dots,T_n]$ in $n$ indeterminates and their values at the points $(s_1,\dots,s_n)$ of $S$. For every $i\in\{1,\dots,n\}$, set $g_i = \prod_{a\in S_i} (T_i-a)$; this is a polynomial of degree $\Card(S_i)$ in the indeterminate $T_i$.

Theorem 1. — If a polynomial $f\in F[T_1,\dots,T_n]$ vanishes at all points $s\in S$, there exist polynomials $h_1,\dots,h_n\in F[T_1,\dots,T_n]$ such that $f=g_1 h_1+\dots+g_n h_n$, and for each $i$, one has $\deg(g_i h_i)\leq \deg (f)$.

This theorem is really close to the classic Nullstellensatz, but the specific nature of the set $S$ allows to have a weaker hypothesis (the field $F$ is not assumed to be algebraically closed) and a stronger conclusion (the Nullstellensatz would assert that there exists some power $f^k$ of $f$ that is of this form, saying nothing of the degrees).

Its proof relies on a kind of Euclidean division algorithm. Assume that $f$ has some monomial $c_mT^m=c_mT_1^{m_1}\dots T_n^{m_n}$ where $m_i\geq \Card(S_i)$; then one can consider a Euclidean division (in the variable $T_i$), $T_i^{m_i}=p_i g_i + r_i$, where $\deg(r_i)<\Card(S_i)$. One can then replace this monomial $c_mT^m$ in $f$ by $c_m T^m/T_i^{m_i})r_i$, and, at the same time register $c_m T^m/T_i^{m_i} p_i$ to $h_i$. Since this amounts to subtract some multiple of $g_i$, the vanishing assumption still holds. Applying this method repeatedly, one reduces to the case where the degree of $f$ in the variable $T_i$ is $<\Card(S_i)$ for all $i$.
Then a variant of the theorem that says that a polynomial in one indeterminate has no more roots than its degree implies that $f\equiv 0$.

A weak point in the written exposition of this argument is the reason why the iterative construction would terminate. Intuitively, something has to decrease, but one needs a minute (or an hour, or a week) of reflexion to understand what precisely decreases. The problem is that if one works one monomial at a time, the degrees might stay the same. The simplest reason why this procedure indeed works belongs to the theory of Gröbner bases and to the proof that the division algorithm actually works: instead of the degrees with respect to all variables, or of the total degree, one should consider a degree with respect to some lexicographic ordering of the variables — the point is that it is a total ordering of the monomials, so that if one consider the leading term, given by the largest monomial, the degree will actually decrease.

The second theorem is in fact the one which is needed for the applications to combinatorics. Its statement is rather written in a contrapositive way — it will show that $f$ does not vanish at some point of $S$.

Theorem 2. — Let $f\in F[T_1,\dots,T_m]$ and assume that $f$ has a nonzero monomial $c_mT^m$, where $|m|$ is the total degree of $f$. If, moreover, $m_i<\Card(S_i)$ for all $i$, then there exists a point $s=(s_1,\dots,s_n)\in S$ such that $f(s)\neq 0$.

It is that theorem whose proof took me a hard time to understand. I finally managed to formalize it in Lean, hence I was pretty sure I had understood it. In fact, writing this blog post helped me simplify it further, removing a couple of useless arguments by contradiction! Anyway, I feel it is worth being told with a bit more detail than in the original paper.

We argue by contradiction, assuming that $f(s)=0$ for all $s\in S_1\times\dots \times S_n$. Applying theorem 1, we get an expression $f=\sum_{i=1}^n g_i h_i$ where $\deg(g_ih_i)\leq \deg(f)$ for all $i$. The coefficient of $T^m$ in $f$ is non zero, by assumption; it is also the sum of the coefficients of the coefficients of $T^m$ in $g_i h_i$, for $1\leq i\leq n$, and we are going to prove that all of them vanish — hence getting the desired contradiction. $\gdef\coeff{\operatorname{coeff}}$

Fix $i\in\{1,\dots, n\}$.  If $h_i=0$, then $\coeff_m(g_ih_i)=\coeff_m(0)=0$, hence we may assume that $h_i\neq0$. This implies that $\deg(g_i h_i)=\Card(S_i)+\deg(h_i) \leq \deg(f)=|m|$.
One then has
\[ \coeff_m(g_i h_i)=\sum_{p+q=m} \coeff_p (g_i) \coeff_q(h_i), \]
and it suffices to prove that all terms of this sum vanish. Fix $p,q$ such that $p+q=m$, assume that  $\coeff_p(g_i)\neq 0$, and let us prove that $\coeff_q(h_i)=0$.

By the very definition of $g_i$ as a product of $\Card(S_i)$ factors of the form $T_i-a$, this implies that $p_j=0$ for all $j\neq i$. Moreover, $p_i\leq p_i+q_i=m_i < \Card(S_i)$, by the assumption of the theorem, hence $p_i<\Card(S_i)$. This implies \[\Card(S_i)+\deg(h_i)\leq |m|= |p+q|=|p|+|q|=p_i+|q|<\Card(S_i)+|q|.\]
Subtracting $\Card(S_i)$ on both sides, one gets $\deg(h_i)<|q|$, hence $\coeff_q(h_i)=0$, as was to be shown.

To conclude, let us add the combinatorial application to the Cauchy—Davenport theorem.
$\gdef\F{\mathbf F}$

Theorem 3. — Let $p$ be a prime number and let $A$ and $B$ be nonempty subsets of $\F_p$, the field with $p$ elements. Denote by $A+B$ the set of all $a+b$, for $a\in A$ and $b\in B$. Then
\[ \Card(A+B) \geq \min (p, \Card(A)+\Card(B)-1).\]

First consider the case where $\Card(A)+\Card(B)>p$, so that $\min(p,\Card(A)+\Card(B)-1)=p$.  In this case, for every $c\in\F_p$, one has
\[\Card(A \cap (c-B))\cup \Card(A\cup (c-B))=\Card(A)+\Card(c-B)>\Card(A)+\Card(B)>p,\]
so that the sets $A$ and $c-B$ intersect. In particular, there exist $a\in A$ and $b\in B$ such that $a+b=c$ and $A+B=\F_p$, and the desired inequality holds.

Let us now treat the case where $\Card(A)+\Card(B)\leq p$, so that $\min(p,\Card(A)+\Card(B)-1)=\Card(A)+\Card(B)-1$. We will assume that the conclusion of the theorem does not hold, that is, $\Card(A+B)\leq \Card(A)+\Card(B)-2$, and derive a contradiction. Let us consider the polynomial $f\in \F_p[X,Y]$ defined by $f=\prod _{c\in A+B} (X+Y-c)$. The degree of $f$ is equal to $\Card(A+B)\leq \Card(A)+\Card(B)-2$. Choose natural integers $u$ and $v$ such that  $u+v=\Card(A+B)$, $u\leq\Card(A)-1$ and $v\leq \Card(B)-1$. The coefficient of $X^u Y^v$ is equal to the binomial coefficient $\binom{u+v}u$. Since $u+v\leq \Card(A)+\Card(B)-2\leq p-2$, this coefficient is nonzero in $\F_p$.  By theorem 2, there exists $(a,b)\in A\times B$ such that $f(a,b)\neq 0$. However, one has $a+b\in A+B$, hence $f(a,b)=0$ by the definition of $f$. This contradiction concludes the proof that $\Card(A+B)\geq \Card(A)+\Card(B)-1$. It also concludes this post.

Combinatorics of partitions

Divided powers are an algebraic gadget that emulate, in an arbitrary ring, the functions $x\mapsto x^n/n!$ for all integers $n$, together with the natural functional equations that they satisfy. 

One of them is a nice binomial theorem without binomial coefficients : denoting $x^n/n!$ by $x^{[n]}$, one has $$ (x+y)^{[n]}=\sum_{k=0}^n x^{[n-k]} y^{[k]}. $$ 

Another formula looks at what happens when one iterates the construction: if everything happens nicely, one has $$ (x^{[n]})^{[m]}=\frac1{m!}\left(x^n/n!\right)^m = \frac1{(n!)^m m!} x^{nm} = \frac{(mn)!}{m! n!^m} x^{[mn]}. $$ 

 In the development of the theory, the remark that comes then is the necessary observation that the coefficient $(mn)!/{m! n!^m}$ is an integer, which is not obvious since it is written as a fraction. Two arguments are given to justify this fact. 

1. The formula $$ \frac{(mn)!}{m! n!^m} = \prod_{k=1}^{m} \binom{k n-1}{n-1} $$ which the authors claim can be proved by induction
2. The observation that $(mn)!/m!n!^m$ is the number of (unordered) partitions of a set with $mn$ elements into $m$ subsets with $n$ elements.

The latter fact is a particular case of a more general counting question: if $n=(n_1,\dots,n_r)$ are integers and $N=n_1+\dots+n_r$, then the number of (unordered) partitions of a set with $N$ elements into subsets of cardinalities $n_1,\dots,n_r$ is equal to $$\frac{N!}{\prod n_i! \prod_{k>0} m_k(n)!},$$ where $m_k(n)$ is the number of occurences of $k$ in the sequence $(n_1,\dots,n_r)$. 

The goal of this blog post is to present a combinatorial argument for the first equality, and an alternative expression of the second number as a product of integers. We also give a formal, group theoretical proof, that this quotient of factorials solves the given counting problem. 

 $\gdef\abs#1{\lvert#1\rvert}$ $\gdef\Card{\operatorname{Card}}$  

Counting partitions of given type 

 Let $S$ be a set with $N$ elements. A partition of $S$ is a subset $\{s_1,\dots,s_r\}$ of $\mathfrak P(S)$ consisting of nonempty, pairwise disjoint, subsets whose union is $S$. Its *type* is the “multiset” of integers given by their cardinalities. Because no $s_i$ is empty, the type is a multiset of nonzero integers. It is slightly easier to authorize zero in this type; then a partition has to be considered as a multiset of subsets of $S$, still assumed pairwise disjoint, so that only the empty subset can be repeated.

The group $G$ of permutations of $S$ acts on the set $\Pi_S$ of partitions of $S$. This action preserves the type of a partition, so that we get an action on the set $\Pi_{S,n}$ on the set of partitions of type $n$, of any multiset of integers $n$. It is nonempty if and only if $\abs n=N$.

This action is transitive: Given two partitions $(s_1,\dots,s_r)$ and $(t_1,\dots t_r)$ with the same type $(n_1,\dots,n_r)$, numbered so that $\abs{s_i}=\abs{t_i}={n_i}$ for all $i$, we just define a permutation of $S$ that maps the elements of $s_i$ to $t_i$.

By the orbit-stabilizer theorem, the number of partitions of type $n$ is equal to $\Card(G)/\Card(G_s)$, where $G_s$ is the stabilizer of any partition $s$ of type $n$. Since $\Card(S)=N$, one has $\Card(G)=N!=\abs n!$.

On the other hand, the stabilizer $G_s$ of a partition $(s_1,\dots,s_r)$ can be described as follows. By an element $g$ of this stabilizer, the elements of $s_i$ are mapped to some set $s_j$ which has the same cardinality as $s_i$. In other words, there is a morphism of groups from $G_s$ to the subgroup of the symmetric group $\mathfrak S_r$ consisting of permutations that preserve the cardinality. This subgroup is a product of symmetric groups $\mathfrak S_{m_k}$, for all $k> 0$, where $m_k$ is the number of occurrences of $k$ in $(n_1,\dots,n_r)$. Here, we omit the factor corresponding to $k=0$ because our morphism doesnt' see it.

This morphism is surjective, because it has a section. Given a permutation $\sigma$ of $\{1,\dots,r\}$ that preserves the cardinalities, we have essentially shown above how to construct a permutation of $S$ that induces $\sigma$, and this permutation belongs to $G_s$. More precisely, if we number the elements of $s_i$ as $(x_{i,1},\dots,x_{i,n_i})$, we lift $\sigma$ to the permutation that maps $x_{i,j}$ to $x_{\sigma(i),j}$ for all $i,j$. This makes sense since $n_{i}=n_{\sigma(i)}$ for all $i$.

The kernel of this morphism $G_s\to \prod_{k>0} \mathfrak S_{m_k}$ consists of permutations that fix each $s_i$. It is clearly equal to the product of the symmetric groups $\prod \mathfrak S_{n_i}$.

One thus has $\Card(G_s)=\prod n_i! \prod_{k>0} m_k!$, as was to be shown.

A combinatorial interpretation of the first relation

As we said above, that relation $$ \frac{(mn)!}{m^ n!^m} = \prod_{k=1}^{m-1} \binom{kn-1}{n-1} $$ can be proved by induction, just playing with factorials, but we want a combinatorial interpretation for it. (Here we assume that $n\geq 1$.)
By the preceding paragraph, the left hand side is the number of partitions of a set with $mn$ elements of type $(n,\dots,n)$. We may assume that this set is numbered $\{1,\dots,mn\}$.

Such a partition can be defined as follows. First, we choose the subset that contains $1$: this amounts to choosing $n-1$ elements among $\{2,\dots,mn\}$, and there are $\binom{mn-1}{n-1}$ possibilities. Now, we have to choose the subset that contains the smallest element which is not in the first chosen subset. There are $n-1$ elements to choose among the remaining $mn-n-1=(m-1)n-1$ ones, hence $\binom{(m-1)n-1}{n-1}$ possibilities. And we go on in the same way, until we have chosen the $m$ required subsets. (For the last one, there is course only one such choice, and notice that the last factor is $\binom{n-1}{n-1}=1$.)

An integral formula for the number of partitions of given type

Finally, we wish to give an alternating formula for the number of partitions of given type of a set $S$ that makes it clear that it is an integer. Again, we can assume that the set $S$ is equal to $\{1,\dots,N\}$ and that $n=(n_1,\dots,n_r)$ is a multiset of integers such that $\abs n=N$.

Let us write $n$ in an other way, $n=(0^{m_0},1^{m_1},\dots)$, meaning that $0$ is repeated $m_0$ times, $1$ is repeated $m_1$ times, etc. One has $N=\sum k m_k$. The idea is that we can first partition $S$ into a subsets of cardinalities $m_1$, $2m_2$,… and then subpartition these subsets: the first one into $m_1$ subsets with one element, the second into $m_2$ subsets with two elements, etc.

The number of ordered partitions with $m_1$, $2m_2$… elements is the multinomial number $$ \binom{N}{m_1,2m_2,\dots}.$$ And the other choice is the product of integers as given in the preceding section. This gives $$ \frac{\abs n!}{\prod_i n_i! \prod_{k>0}m_k(n)!} = \binom{N}{m_1,2m_2,\dots}\prod_{k>0} \frac{(km_k)!}{k!^{m_k} m_k(n)!}.$$ We can also write the fractions that appear in the product as integers: $$ \frac{\abs n!}{\prod_i n_i! \prod_{k>0}m_k(n)!} =\binom{N}{m_1,2m_2,\dots} \prod_{k>0} \prod_{m=1}^{m_k} \binom {km-1}{k-1}.$$

Number theory and finite automata

$\gdef\abs#1{\lvert#1\rvert}$

There are many parts of number theory I don't know of, and today's story is about one I learnt very recently, at the intersection of number theory and computer science. It is about the natural numbers that we can recognize using a finite automaton. 

Finite automata and automatic functions

Finite automata

Finite automata are very elementary computers. These machines can receive instructions of finitely many sorts, and can be in finitely many states; moreover, their behaviour is prescribed: when, in a given state, they receive one instruction, they just move to another state. Mathematically, one can say that there is a finite set $A$ of instructions, a finite set $S$ of states, and a mapping $\phi\colon A \times S\to S$. One state $\varepsilon$ is labeled as the initial state; wen reading a list of instructions $[a_1,\dots,a_m]$, the machine goes through the states $s_0=\varepsilon$, $s_1=\mu(a_1,s_0)$, $s_2=\mu(a_2,s_1)$, etc. until it reaches and outputs its the final state $s_n$. All in all, the automaton defines a function $\Phi$ from the set $A^*$ of lists in $A$ to the set $S$. Such functions are called automatic.

In practice, some states could be labeled as admissible, in which case the machine unlocks the door, and the set of lists $\alpha\in A^*$ such that $\Phi(\alpha)$ is admissible is then called the language of the automaton.

The basic question of automata theory is to describe the (so-called “regular”) languages recognizable by some automaton, more generally the automatic functions. It appears that they are of a very restricted nature: due to the elementary nature of finite automata, they are much more restricted than the kind of languages that would be recognizable by a program in the programming language of your choice. The reason is that they have a fixed memory while “theoretical” computers, as modelled by Turing machines for example, have a potentially infinite memory, that is, as large as the computation may require.

Numbers

Number theory enters when you decide that the set $A$ of instructions are digits 0, 1, 2,…, 9 and the lists in $A^*$ represent a number. You then wish to understand what numbers can be recognizable, or what functions $\Phi\colon\mathbf N\to S$ are automatic, for a finite set $S$. This is what happens, for example, in the very elementary finite automata that stand at almost all Paris doors as electronic locks. They recognize exactly one 4-digit number (and any 4-digit number) puts them back in the initial state.

When it comes about digits, they can be the usual ones, 0, 1, 2,… , 9, writing number in base 10, but any other basis can be considered, and any set of digits that allows to write all numbers can be used. For example, one can consider writing numbers in base 3, with digits 0, 1, 2, 3, 4.

Let us thus fix some basis $b$. The first theorem is that the set of automatic functions does not depend on the set of digits which is considered. The reason is that one can build another kind of machine (called a transducer) that works like a finite automaton except that it output something at each step. Here, we can have a transducer that outputs the digits in some representation given the digits in another one. There could be carries to take care of, but when the basis is the same, their impact is controlled, and it can be controlled by finitely many of them. It is important for this argument that the basis stays the same, and we will see that in a more striking way later on. For this reason, we will not specify the set of digits in the sequel.

Examples

Here are some examples of recognizable sets of numbers.

• The empty set works. It suffices to have no terminal state.
• Singletons work. For example, to recognize “314”, one needs one initial state “$\varepsilon$”, three states labeled “3”, “31” and “314”, and a junk state “J”. The machine moves from “$\varepsilon$” to “3” if it receives “3”, and to “J” otherwise; from “3” to “31” if it receives “1”, etc.
• The union and the intersection of two recognizable sets are recognizable. Just make an automaton whose sets are pair of states of the two automata that recognize the given states, and simulate the moves of each of them. Similarly, the complement of a recognizable set is recognizable.
• By the previous examples, finite sets and complements of finite sets are recognizable.
• Here is a number theoretically more interesting example : arithmetic progressions are recognizable. To produce an automaton that, say, recognize all integers congruent to 1 modulo 4, it suffices to have a machine with 4 states that computes the euclidean reminder step by step.
• Another number theoretically interesting example: powers of $b$. Indeed their writing in base $b$ consists of one digit “1” followed by a series of “0”.
• A number theoretically annoying property of recognizable sets: since a finite automaton has finitely many states, the “pumping lemma” shows that for any large enough recognizable number, one can replicate at will some inner part of the writing and get another number. I suppose one can deduce from this property that the set of prime numbers cannot be recognizable. (This is exercise 5.8.12 in the book Automatic sequences, by J-P. Allouche and J. Shallit.)
• Numbers recognizable in base $b$ are also recognizable in base $b^2$ (or some other power) and conversely: it suffices to emulate the base $b^2$-writing of a number using base $b$-digits, or conversely.

Building from characteristic functions of admissible languages, the preceding results can be turned to examples of automatic functions. In particular, we obtain that functions which are ultimately periodic are automatic.

Cobham's theorem

Statement

Number theory can also enter a topic from two sides at a time, and here is the point of this blog post. Can one compare the sets of recognizable numbers from different bases? By the following theorem, they are genuinely distinct.

Theorem (Cobham, 1969). — Let $S$ be a finite set and let $f\colon \mathbf N\to S$ be a function which is automatic in two bases $b$ and $c$. If $b$ and $c$ are not powers of the same integer, then $f$ is ultimately periodic.
Since ultimately periodic functions are automatic in any basis, this theorem is definitely optimal.

The original proof of Cobham's theorem is said to be difficult, and this theorem has obtained various other simpler proofs, often partially flawed. I wish to describe here the recent proof by Thijmen Krebs (publisher version, arXiv:1801.06704) which calls itself “more reasonable”.

The assumption that $a$ and $b$ are not powers of the same integer will appear in different, equivalent, form in the proof: they do not have common powers.

Local periods

To prove that $f$ is ultimately periodic, Krebs proves that $f$ is periodic on large overlapping intervals.

Lemma 1. — Let $f: \mathbf N\to S$ be a function and let $I$ and $J$ be intervals of integers. We assume that $f$ has period $p$ on $I$ and period $q$ on $J$. If $\operatorname{Card}(I\cap J)\geq p+q$, then $f$ has period $p$ on the interval $I\cup J$.

The proof is elementary. Consider $x\in I\cup J$ such that $x+p\in I\cup J$. If both $x$ and $x+p$ belong to $I$, then $f(x)=f(x+p)$. Let us assume otherwise. If $x$ belongs to $I$, but not $x+p$, then $x+p\in J$; since $I\cap J$ has at least $p+q$ elements, $x$ must belongs to $J$. The other cases are similar and show that both $x, x+p$ belong to $J$. Using that $I\cap J$ is large, we pick up elements $y,y+p$ in $I\cap J$ such that $x\equiv y \pmod q$. Then $f(x)=f(y)=f(y+p)=f(x+p)$, the first and last equality using the $q$-periodicity on $J$, and the middle one the $p$-periodicity on $I$.

A diophantine approximation lemma

Lemma 2. — Let $a$ and $b$ be two real numbers such that $a,b>1$. For every $\varepsilon>0$, there exist nonzero natural numbers $m$ and $n$ such that $\abs{a^m-b^n}<\varepsilon b^n$.

The result is obvious if $a$ and $b$ have a common power, so we may assume that this is not the case. Then various proofs are possible. For example, one may consider the additive subgroup of $\mathbf R$ generated by $\log(a)$ and $\log(b)$; by assumption, it is dense hence there exists a small linear combination $m\log(a)-n\log(b)$; taking exponentials, $a^m/b^n$ is close to~$1$, as claimed.

Krebs gives a “pigeonhole-like” proof which is nice. For every integer $m\geq1$, consider the unique integer $n_m\geq 1$ such that $1\leq a^m/b^{n_m}<b$. There must be two integers $m$ and $p$ such that $m<p $ and such that $a^m/b^{n_m}$ and $a^p/b^{n_p}$ differ at most from $\varepsilon$, and this implies the result.

Some functional equations

Let $f\colon \mathbf N\to S$ be a function which is automatic in some basis $c$, computed by an automaton with set of states $S$. For every $s\in S$, let $L(s)$ be the set of integers $n$ such that $f(n)=s$. Note that these sets form a partition of $\mathbf N$.

The following lemma is a variant of the pumping lemma in the theory of automata; it is by it that automatic functions acquire their specific properties.

Lemma 3. — For every integers $x,y\in L(s)$ and any integer $z$ such that $0\leq z < c^n$, one has $f(x c ^ n+z)=f(yc ^n +z)$.

Indeed, when the automaton reads the integer $xc^n+z$, it first reads $x$, and goes to state $s$; similarly, when it reads $yc^n+z$, it first reads $y$ and reaches the state $s$ too. In both cases it then reads $z$ and ends up in the same final state.

Construction of local periods

We now consider a function $f$ which is automatic in two bases $a$ and $b$ which do not have common powers. We thus have two finite automata $A$ and $B$, with sets of states $S$ and $T$. For each state $s\in S$, we have a set $L(s)$ of integers as above; these sets form a partition of $\mathbf N$. Let $S^\infty$ be the set of states $s\in S$ such that $L(s)$ is infinite. For $t\in T$, we define $M(t)$ similarly, as well as $T^\infty$.

Let $s\in S^\infty$. Since the $M(t)$, for $t\in T$, form a finite partition of $\mathbf N$ and $L(s)$ is infinite, there exists a state $t(s)$ and two distinct integers $x(s)$ and $y(s)\in L(s)\cap M(t(s))$. Let $K$ be an integer strictly greater than all $x(s), y(s)$, for $s\in S^\infty$.

Applying lemma 3 for $\varepsilon = 1/6K$, we obtain integer $m,n$ such that $\abs{a^m-b^n}<a^m/6K$. For $s\in S^\infty$, we $p(s)=\pm (x(s)-y(s))(a^m-b^n)$; since $a$ and $b$ have no common power, this is a nonzero integer, and we choose the sign so that it is strictly positive.

For each integer $x$, we define an interval $I_x$ of integers by $I_x=[(x+\frac13)a^m,(x+\frac53)a^m]\cap\mathbf N$.

Lemma 4. — For every state $s\in S_\infty$ and any integer $x\in L(s)$, the function $f$ has period $p(s)$ on $I_x$.

To that aim, we consider $z\in [\frac13a^m,\frac53a^m]$ and prove that $f(xa^m+z)=f(xa^m+z+p)$. First of all, we have the inequality $$ \abs{z-x(s)(a^m-b^n)-b^n} \leq \abs{z-a^m}+(x(s)+1) \abs{a^m-b^n} \leq \frac23 a^m + K\abs{a^m-b^n} \leq \frac56a^m\leq b^n $$ because $\abs{a^m-b^n}\leq \frac16 a^m$. Consequently, $z$ is written with at most $n$ digits in base $b$. Applying lemma 3, we have $ f(x a^m+z) = f(x(s)a^m+z$, because $x$ and $x(s)\in L(s)$. Applying lemma 3 again, this is equal to $f(x(s)a^m+z-x(s)(a^m-b^n))$, hence to $f(y(s)a^m+z-x(s)(a^m-b^n)) = f(y(s)a^m+z+p(s))$. Applying lemma 3 a third time, this is equal to $f(xa^m+z+p(s)$. This concludes the proof.

Conclusion

Since the sets $L(s)$, for $s\notin S_\infty$ are finite, the set $L(s)$ for $s\in S_\infty$ cover all integers larger than some integer, say $x$.

For $y\geq x$, there exists $s\in S_\infty$ such that $y\in L(s)$ and we have shown that $f$ is periodic with period $p(s)\leq \frac16a^m$ on an interval $I(x)=[(y+\frac13)a^m,(y+\frac53)a^m]$. By abuse of notation, write $p(x)=p(s)$.

Let $J(z)$ be the union of the intervals $I(y)$, for $x\leq y\leq z$. One has $J(x)=I(x)$; for $z>x$, the intersection $J(z)\cap J(z-1)$ is $[(z+\frac13)a^m;(z+\frac23)a^m]$ hence has at least $\frac13a^m$ points, thus at least $p(x)+p(z)$. Using lemma 1, $f$ has period $p(x)$ on $J(z)$, for all $z$.

This concludes the proof that $f$ is ultimately periodic

Evaluating the operator norms of matrices

Let $E$ and $F$ be normed vector spaces, over the real or complex numbers, and let $u\colon E\to F$ be a linear map. The continuity of $u$ is proved to be equivalent to the existence of a real number $c$ such that $\|u(x)\|\leq c \|x\|$ for every $x\in E$, and the least such real number is called the operator norm of $u$; we denote it by $\|u\|$. It defines a norm on the linear space $\mathscr L(E;F)$ of continuous linear maps from $E$ to $F$ and as such is quite important. When $E=F$, it is also related to the spectrum of $u$ and is implicitly at the heart of criteria for the Gershgorin criterion for localization of eigenvalues.

$\gdef\R{\mathbf R}\gdef\norm#1{\lVert#1\rVert}\gdef\Abs#1{\left|#1\right|}\gdef\abs#1{\lvert#1\rvert}$

However, even in the simplest cases of matrices, its explicit computation is not trivial at all, and as we'll see even less trivial than what is told in algebra classes, as I learned by browsing Wikipedia when I wanted to prepare a class on the topic.

Since I'm more a kind of abstract guy, I will use both languages, of normed spaces and matrices, for the first one allows to explain a few things at a more fundamental level. I'll make the translation, though. Also, to be specific, I'll work with real vector spaces.

So $E=\R^m$, $F=\R^n$, and linear maps in $\mathscr L(E;F)$ are represented by $n\times m$ matrices. There are plentiful interesting norms on $E$, but I will concentrate the discussion on the $\ell^p$-norm given by $\norm{(x_1,\dots,x_m)} = (|x_1|^p+\dots+|x_m|^p)^{1/p}$. Similarly, I will consider the $\ell^q$-norm on $F$ given by $\norm{(y_1,\dots,y_m)} = (|y_1|^q+\dots+|y_n|^q)^{1/q}$. Here, $p$ and $q$ are elements of $[1;+\infty\mathclose[$. It is also interesting to allow $p=\infty$ or $q=\infty$; in this case, the expression defining the norm is just replaced by $\sup(|x_1|,\dots,|x_m|)$ and $\sup(|y_1|,\dots,|y_n|)$ respectively.

Duality

Whatever norm is given on $E$, the dual space $E^*=\mathscr L(E;\mathbf R)$ is endowed with the dual norm, which is just the operator norm of that space: for $\phi\in E^*$, $\norm\phi$ is the least real number such that $|\phi(x)|\leq \norm\phi \norm x$ for all $x\in E$. And similarly for $F$. To emphasize duality, we will write $\langle x,\phi\rangle$ instead of $\phi(x)$.

Example. — The dual norm of the $\ell^p$ norm can be computed explicitly, thanks to the Young inequality \[ \Abs{ x_1 y_1 + \dots + x_n y_n } \leq (|x_1|^p+\dots + |x_n|^p)^{1/p} (|x_1|^q+\dots+|x_n|^q)^{1/q}\] if $p,q$ are related by the relation $\frac1p+\frac1q=1$. (When $p=1$, this gives $q=\infty$, and symmetrically $p=\infty$ if $q=1$.) Moreover, this inequality is optimal in the sense that for any $(x_1,\dots,x_n)$, one may find a nonzero $(y_1,\dots,y_n)$ for which the inequality is an equality. What this inequality says about norms/dual norms is that if one identifies $\R^n$ with its dual, via the duality bracket $\langle x,y\rangle=x_1y_1+\dots+x_n y_n$, the dual of the $\ell^p$-norm is the $\ell^q$-norm, for that relation $1/p+1/q=1$.

If $u\colon E\to F$ is a continuous linear map, it has an adjoint (or transpose) $u^*\colon F^*\to E^*$, which is defined by $u^*(\phi)= \phi\circ u$, for $\phi\in F^*$. In terms of the duality bracket, this rewrites as \[ \langle \phi, u(x)\rangle = \langle u^*(\phi),x\rangle\] for $x\in E$ and $\phi\in F^*$.

Proposition. — One has $\norm{u^*}=\norm u$.

For $\phi\in F^*$, $\norm{u^*(\phi)}$ is the least real number such that $|u^*(\phi)(x)|\leq \norm{u^*(\phi)} \norm x$ for all $x\in E$. Since one has \[ |u^*(\phi)(x)|= |\langle u^*(\phi),x\rangle|=|\langle\phi, u(x)\rangle\leq \norm\phi \norm{u(x)} \leq \norm\phi \norm u\norm x, \] we see that $\norm {u^*(\phi)}\leq \norm\phi\norm u$ for all $\phi$. As a consequence, $\norm{u^*}\leq \norm u$.
To get the other inequality, we wish to find a nonzero $\phi$ such that $\norm{u^*(\phi)}=\norm{u}\norm\phi$. This $\phi$ should thus be such that there exists $x$ such that $|\langle u^*(\phi),x\rangle|=\norm u\norm\phi\norm x$ which, by the preceding computation means that $|\langle\phi, u(x)\rangle=\norm\phi\norm u\norm x$. Such $\phi$ and $x$ must not exist in general, but we can find reasonable approximations. Start with a nonzero $x\in E$ such that $\norm{u(x)}$ is close to $\norm u\norm x$; then using the Hahn-Banach theorem, find a nonzero $\phi\in F^*$ such that $\norm\phi=1$ and $|\phi(u(x))|=\norm {u(x)}$. We see that $\langle\phi, u(x)\rangle$ is close to $\norm u\norm\phi\norm x$, and this concludes the proof.
In some cases, in particular in the finite dimension case, we can use biduality to get the other inequality. Indeed $E^{**}$ identifies with $E$, with its initial norm, and $u^{**}$ identifies with $u$. By the first case, we thus have $\norm{u^{**}}\leq \norm {u^*}$, hence $\norm u\leq\norm{u^*}$.

The case $p=1$

We compute $\norm{u}$ when $E=\mathbf R^m$ is endowed with the $\ell^1$-norm, and $F$ is arbitrary. The linear map $u\colon E\to F$ thus corresponds with $m$ vectors $u_1,\dots,u_m$ of $F$, and one has \[ u((x_1,\dots,x_m))=x_1 u_1+\dots+x_m u_m. \] By the triangular inequality, we have \[ \norm{u((x_1,\dots,x_m))} \leq |x_1| \norm{u_1}+\dots+\abs{x_m}\norm{u_m} \] hence \[ \norm{u((x_1,\dots,x_m))} \leq (\abs{x_1} +\dots+\abs{x_m}) \sup(\norm{u_1},\dots,\norm{u_m}). \] Consequently, \[ \norm{u} \leq \sup(\norm{u_1},\dots,\norm{u_m}). \] On the other hand, taking $x=(x_1,\dots,x_m)$ of the form $(0,\dots,1,0,\dots)$, where the $1$ is at place $k$ such that $\norm{u_k}$ is largest, we have $\norm{x}=1$ and $\norm{u(x)}=\norm{u_k}$. The preceding inequality is thus an equality.

In the matrix case, this shows that the $(\ell^1,\ell^q)$-norm of a $n\times m$ matrix $A$ is the supremum of the $\ell^q$-norms of the columns of $A$.

The case $q=\infty$

We compute $\norm{u}$ when $F=\mathbf R^n$ is endowed with the $\ell^\infty$-norm, and $E$ is arbitrary. A direct computation is possible in the matrix case, but it is not really illuminating, and I find it better to argue geometrically, using a duality argument.

Namely, we can use $u^*\colon F^*\to E^*$ to compute $\norm{u}$, since $\norm u=\norm{u^*}$. We have seen above that $F^*$ is $\mathbf R^n$, endowed with the $\ell^1$-norm, so that we have computed $\norm{u^*}$ in the preceding section. The basis $(e_1,\dots,e_n)$ of $F$ gives a dual basis $(\phi_1,\dots,\phi_n)$, and one has \[ \norm{u}=\norm{u^*} = \sup (\norm{u^*(\phi_1)},\dots,\norm{u^*(\phi_n)}). \]

In the matrix case, this shows that the $(\ell^p,\ell^\infty)$-norm of a $n\times m$ matrix $A$ is the supremum of the $\ell^p$-norms of the lines of $A$.

Relation with the Gershgorin circle theorem

I mentioned the Gershgorin circle theorem as being in the same spirit as the computation of an operator norm, because its proof relies on the same kind of estimations. In fact, no additional computation is necessary!

Theorem (Gershgorin “circles theorem”). — Let $A=(a_{ij})$ be an $n\times n$ matrix and let $\lambda$ be an eigenvalue of $A$. There exists an integer $i$ such that \[ \abs{\lambda-a_{ii}}\leq \sum_{j\neq i} \abs{a_{ij}}. \]

For the proof, one writes $A=D+N$ where $D$ is diagonal has zeroes on its diagonal, and writes $\lambda x=Ax=Dx+Nx$, hence $(\lambda I-D)x=Nx$. Endow $\R^n$ with the $\ell^\infty$-norm. We can assume that $\norm x=1$. Then the norm of the right hand side is bounded above by $\norm N$, while the norm of the left hand side is $\sup(\abs{\lambda-a_{ii}} |x_i|)\geq |\lambda-a_{ii}|$ if $i$ is chosen so that $|x_i|=\norm x=1$. Given the above formula for $\norm N$, this implies the theorem.

The case $p=q=2$

Since Euclidean spaces are very useful in applications, this may be a very important case to consider, and we will see that the answer is not at all straightforward from the coefficients of the matrix.

We have to bound from above $\norm{u(x)}$. Using the scalar product, we write \[ \norm{u(x)}^2 = \langle u(x),u(x)\rangle = \langle u^*u(x),x\rangle, \] where $u^*\colon F\to E$ now denotes the adjoint of $u$, which identifies with the transpose of $u$ if one identifies $E$ with $E^*$ and $F$ with $F^*$ by means of their scalar products. Using the Cauchy-Schwarz formula, we get that $\norm{u(x)}^2\leq \norm{u^*u(x)}\norm x\leq \norm{u^*u} \norm x^2$, hence $\norm{u} \leq \norm{u^*u}^{1/2}$. This inequality is remarkable because on the other hand, we have $\norm{u^*u}\leq \norm{u^*}\norm{u}=\norm{u}^2$. Consequently, $\norm{u}=\norm{u^*u}^{1/2}$.

This formula might not appear to be so useful, since it reduces the computation of the operator norm of $u$ to that of $u^*u$. However, the linear map $u^*u$ is a positive self-adjoint endomorphism of $E$ hence, (assuming that $E$ is finite dimensional here), it can be diagonalized in a orthonormal basis. We then see that $\norm{u^*u}$ is the largest eigenvalue of $u^*u$, which is also its spectral radius. The square roots of the eigenvalues of $u^*u$ are also called the singular values of $u$, hence $\norm u$ is the largest singular value of $u$.

One can play with duality as well, and we have $\norm{u}=\norm{uu^*}^{1/2}$.

Other cases?

There are general inequalities relating the various $\ell^p$-norms of a vector $x\in\R^m$, and these can be used to deduce inequalities for $\norm u$, when $E=\R^m$ has an $\ell^p$-norm and $F=\R^n$ has an $\ell^q$-norm. However, given the explicit value of $\norm u$ for $(p,q)=(2,2)$ and the fact that no closed form expression exists for the spectral radius, it is unlikely that there is a closed form expression in the remaining cases.

Worse: the exact computation of $\norm u$ in the cases $(\infty,1)$, $(\infty,2)$ or $(2,1)$ is known to be computationally NP-complete, and I try to explain this result below, following J. Rohn (2000) (“Computing the Norm ∥ A ∥∞,1 Is NP-Hard”, Linear and Multilinear Algebra 47 (3), p. 195‑204). I concentrate on the $(\infty, 1)$ case ; the $(\infty,2)$ case is supposed to be analogous (see Joel Tropp's thesis, top of page 48, quoted by Wikipedia, but no arguments are given), and the case $(2,1)$ would follow by symmetry.

A matrix from a graph

Let us consider a finite (undirected, simple, without loops) graph $G$ on the set $V=\{1,\dots,n\}$ of $n$ vertices, with set of edges $E$, and let us introduce the following $n\times n$ matrix $A=(a_{ij})$, a variant of the incidence matrix of the graph $G$ (actually $nI-E$, where $I$ is the identity matrix and $E$ is the incidence matrix of $G$):

• One has $a_{ii}=n$ for all $i$;
• If $i\neq j$ and vertices $i$ and $j$ are connected by an edge, then $a_{ij}=-1$;
• Otherwise, $a_{ij}=0$.

For any subset $S$ of $V$, the cut $c(S)$ of $S$ is the number of edges which have one endpoint in $S$ and the other outside of $S$.

Proposition. — The $(\ell^\infty,\ell^1)$-norm of $A$ is given by \[ 4 \sup_{S\subseteq V} c(S) - 2 \operatorname{Card}(E) + n^2. \]

The proof starts with the following observation, valid for more general matrices.

Lemma. — The $(\ell^\infty,\ell^1)$-norm of a symmetric positive $n\times n$ matrix $A$ is given by $\norm A = \sup_z \langle z, Az \rangle$ where $z$ runs among the set $Z$ of vectors in $\R^n$ with coordinates $\pm1$.

The vectors of $Z$ are the vertices of the polytope $[-1;1]^n$, which is the unit ball of $\R^n$ for the $\ell^\infty$-norm. Consequently, every vector of $[-1;1]^n$ is a convex combination of vectors of $Z$. Writing $x=\sum_{z\in Z} c_z z$, we have \[\norm {Ax} = \norm{\sum c_z Az} \leq \sum c_z \norm {Az}= \sup_{z\in Z} \norm{Az}. \] The other inequality being obvious, we already see that $\norm A=\sup_{z\in Z}\norm{Az}$. Note that this formula holds for any norm on the codomain.
If, for $z\in Z$, one writes $Az=(y_1,\dots,y_n)$, one has $\norm{Az}=|y_1|+\dots+|y_n|$, because the codomain is endowed with the $\ell^1$-norm, so that $\langle z, Az\rangle = \sum z_i y_i\leq \norm{Az}$. We thus the inequality $\sup_{z\in Z} \langle z,Az\rangle \leq \norm A$.
Let us now use the fact that $A$ is symmetric and positive. Fix $z\in Z$, set $Az=(y_1,\dots,y_n)$ as above, and define $x\in Z$ by $x_i=1$ if $y_i\geq0$ and $x_i=-1$ otherwise. One thus has $\langle x, Az\rangle=\sum |y_i|=\norm{Az}$. Since $A$ is symmetric and positive, one has $\langle x-z, A(x-z)\rangle\geq0$, and this implies \[2\norm{Az}= 2\langle x, Az\rangle \leq \langle x, Ax\rangle+\langle z, Az\rangle, \] so that, $\norm{Az}\leq \sup_{x\in Z} \langle x, Ax\rangle$. This concludes the proof.

To prove the theorem, we will apply the preceding lemma. We observe that $A$ is symmetric, by construction. It is also positive, since for every $x\in\R^n$, one has \[\langle x,Ax\rangle=\sum a_{ij}x_ix_j \geq n \sum x_i^2 -\sum_{i\neq j} x_i x_j = (n+1)\sum x_i^2- (\sum x_i)^2\geq \sum x_i^2 \] using the Cauchy-Schwarz inequality $(\sum x_i)^2\leq n\sum x_i^2$. By the preceding lemma, we thus have \[ \norm A = \sup_{z\in\{\pm1\}^n} \langle z, Az\rangle. \] The $2^n$ vectors $z\in Z$ are in bijection with the $2^n$ subsets of $V=\{1,\dots,n\}$, by associating with $z\in Z$ the subset $S$ of $V$ consisting of vertices $i$ such that $z_i=1$. Then, one can compute \[ \langle z, Az\rangle = \sum_{i,j} a_{ij} z_iz_j = 4c(S) - 2\operatorname{Card}(E) + n^2. \] It follows that $\norm A $ is equal to the indicated expression.

The last step of the proof is an application of the “simple max-cut” NP-hardness theorem of Garey, Johnson and Stockmeyer (1976), itself a strenghtening of Karp (1973)'s seminal result that “max-cut” is NP-complete. I won't explain the proofs of these results here, but let me explain what they mean and how they relate to the present discussion. First of all, computer scientists categorize problems according to the time that is required to solve them, in terms of the size of the entries. This notion depends on the actual computer that is used, but the theory of Turing machines allows to single out two classes, P and EXP, consisting of problems which can be solved in polynomial, respectively exponential, time in term of the size of the entries. A second notion, introduced by Karp, is that of NP problems, problems which can be solved in polynomial time by a “non deterministic Turing machine” — “nondeterministic” means the computer can parallelize itself at will when it needs to consider various possibilities. This class belongs to EXP (because one can simulate in exponential time a polynomial time nondeterministic algorithm) and also corresponds to the class of problems whose solution can be checked in polynomial time.

Our problem is to find a subset $S$ of $\{1,\dots,n\}$ that maximizes $c(S)$. This is a restriction of the “general max-cut” problem where, given an integer valued function $w$ on the set of edges, on wishes to find subset that maximizes $c(S;w)$, the sum of the weights of the edges which have one endpoint in $S$ and the other outside of $S$. Karp (1973) observed that the existence of $S$ such that $c(S;w)\geq m$ is an NP problem (if one is provided $S$, it takes polynomial time to compute $c(S;w)$ and to decide that it is at least $m$), and the naïve search algorithm is in EXP, since there are $2^n$ such subsets. Moreover, Karp proves that any NP problem can be reduced to it in polynomial time. This is what is meant by the assertion that it is NP-complete. Consequently, determining $\sup_S c(S;w)$ is NP-hard: if you can solve that problem, then you can solve the “max-cut” problem in polynomial time, hence any other NP-problem. A subsequent theorem by Garey, Johnson and Stockmeyer (1976) established that restricting the max-cut problems to $\pm1$ weights is still NP-hard, and this completes the proof of Rohn's theorem.

(Aside, to insist that signs matter: a theorem of Edmonds and Karp (1972), one can solve the “min-cup” problem in polynomial time, which consists in deciding, for some given integer $m$, whether there exist $S$ such that $c(S;w)\leq m$.)

The topology on the ring of polynomials and the continuity of the evaluation map

Polynomials are an algebraic gadget, and one is rarely led to think about the topology a ring of polynomials should carry. That happened to me, though, more or less by accident, when María Inés de Frutos Fernández and I worked on implementing in Lean the evaluation of power series. So let's start with them. To simplify the discussion, I only consider the case of one inderminate. When there are finitely many of them, the situation is the same; in the case of infinitely many indeterminates, there might be some additional subtleties, but I have not thought about it.

$\gdef\lbra{[\![}\gdef\rbra{]\!]} \gdef\lpar{(\!(}\gdef\rpar{)\!)} \gdef\bN{\mathbf N} \gdef\coeff{\operatorname{coeff}} \gdef\eval{\operatorname{eval}} \gdef\colim{\operatorname{colim}}$

Power series

A power series over a ring $R$ is just an expression $\sum a_nT^n$, where $(a_0,a_1, \dots)$ is a family of elements of $R$ indexed by the integers. After all, this is just what is meant by “formal series”: coefficients and nothing else.

Defining a topology on the ring \(R\lbra T\rbra\) should allow to say what it means for a sequence $(f_m)$ of power series to converge to a power series $f$, and the most natural thing to require is that for every $n$, the coefficient $a_{m,n}$ of \(T^n\) in $f_m$ converges to the corresponding coeffient $a_m$ of $T^n$ in \(f\). In other words, we endow \(R\lbra T\rbra \) with the product topology when it is identified with the product set \(R^{\bN}\). The explicit definition may look complicated, but the important point for us is the following characterization of this topology: Let \(X\) be a topological space and let \(f\colon X \to R\lbra T\rbra\) be a map; for \(f\) to be continuous, it is necessary and sufficient that all maps \(f_n\colon X \to R\) are continuous, where, for any \(x\in X\), \(f_n(x)\) is the \(n\)th coefficient of \(f(x)\). In particular, the coeffient maps \(R\lbra T\rbra\to R\) are continuous.

What can we do with that topology, then? The first thing, maybe, is to observe its adequacy wrt the ring structure on \(R\lbra T\rbra\).

Proposition. — If addition and multiplication on \(R\) are continuous, then addition and multiplication on \(R\lbra T\rbra\) are continuous.

Let's start with addition. We need to prove that \(s\colon R\lbra T\rbra \times R\lbra T\rbra\to R\lbra T\rbra\) is continuous. By the characterization, it is enough to prove that all coordinate functions \(s_n\colon R\lbra T\rbra \times R\lbra T\rbra\to R\), \( (f,g)\mapsto \coeff_n(f+g) \), are continuous. But these functions factor through the \(n\)th coefficient maps: \(\coeff_n(f+g) = \coeff_n(f)+\coeff_n(g)\), which is continuous, since addition, coefficients and projections are continuous. This is similar, but slightly more complicated for multiplication: if the multiplication map is denoted by \(m\), we have to prove that the maps \(m_n\) defined by $m_n(f,g)=\coeff_n(f\cdot g)$ are continuous. However, they can be written as \[ m_n(f,g)=\coeff_n(f\cdot g) = \sum_{p=0}^n \coeff_p(f)\coeff_{n-p}(g). \] Since the projections and the coefficient maps are continuous, it is sufficient to prove that the maps from \(R^{n+1} \times R^{n+1}\) to \(R\) given by \[((a_0,\dots,a_n),(b_0,\dots,b_n))\mapsto \sum_{p=0}^n a_p b_{n-p} \] are continuous, and this follows from continuity and commutativity of addition on \(R\), because it is a polynomial expression.

Polynomials

At this point, let's go back to our initial question of endowing polynomials with a natural topology.

An obvious candidate is the induced topology. This looks correct; in any case, it is such that addition and multiplication on \(R[T]\) are continuous. However, it lacks an interesting property with respect to evaluation.

Recall that for every \(a\in R\), there is an evaluation map \(\eval_a\colon R[T]\to R\), defined by \(f\mapsto f(a)\), and even, if one wishes, the two-variable evaluation map \(R[T]\times R\to R\).
The first claim is that this map is not continuous.

An example will serve of proof. I take \(R\) to be the real numbers, \(f_n=T^n\) and \(a=1\). Then \(f_n\) converges to zero, because for each integer \(m\), the real numbers \(\coeff_m(f_n)\) are zero for \(n>m\). On the other hand, \(f_n(a)=f_n(1)=1\) for all \(n\), and this does not converge to zero!

So we have to change the topology on polynomials if we want that this map be continuous, and we now give the correct definition. The ring of polynomials is the increasing union of subsets \(R[T]_n\), indexed by integers \(n\), consisting of all polynomials of degree less than \(n\). Each of these subsets is given the product topology, as above, but we endow their union with the “inductive limit” topology. Explicitly, if \(Y\) is a topological space and \(u\colon R[T]\to Y\) is a map, then \(u\) is continuous if and only if, for each integer \(n\), its restriction to \(R[T]_n\) is continuous.

The inclusion map \(R[T]\to R\lbra T\rbra\) is continuous, hence the topology on polynomials is finer than the topology induced by the topology on power series. As the following property indicates, it is usually strictly finer.

We can also observe that addition and multiplication on \(R[T]\) are still continuous. The same proof as above works, once we observe that the coefficient maps are continuous. (On the other hand, one may be tempted to compare the product topology of the inductive topologies, with the inductive topology of the product topologies, a thing which is not obvious in the direction that we need.)

Proposition. — Assume that addition and multiplication on \(R\) are continuous. Then the evaluation maps \(\eval_a \colon R[T]\to R\) are continuous.

We have We have to prove that for every integer \(n\), the evaluation map \(\eval_a\) induced a continuous map from \(R[T]_n\) to \(R\). Now, this map factors as a projection map \(R[T]\to R^{n+1}\) composed with a polynomial map \((c_0,\dots,c_n)\mapsto c_0+c_1a+\dots+c_n a^n\). It is therefore continuous.

Laurent series

We can upgrade the preceding discussion and define a natural topology on the ring \(R\lpar T\rpar\) of Laurent series, which are the power series with possibly negative exponents. For this, for all integers \(d\), we set \(R\lpar T\rpar_d\) to be the set of power series of the form \( f=\sum_{n=-d}^\infty c_n T^n\), we endow that set with the product topology, and take the corresponding inductive limit topology. We leave to the reader to check that this is a ring topology, but that the naïve product topology on \(R\lpar T\rpar\) wouldn't be in general.

Back to the continuity of evaluation

The continuity of the evaluation maps $f\mapsto f(a)$ were an important guide to the topology of the ring of polynomials. This suggests a more general question, for which I don't have a full answer, whether the two-variable evaluation map, \((f,a)\mapsto f(a)\), is continuous. On each subspace $R[T]_d\times R$, the evaluation map is given by a polynomial map ($(c_0,\dots,c_d,a)\mapsto c_0 +c_1a+\dots+c_d a^d$), hence is continuous, but that does not imply the desired continuity, because that only tells us about $R[T]\times R$ with the topology $\colim_d (R[T]_d\times R)$, while we are interested in the topology $(\colim_d R[T]_d)\times R$. To compare these topologies, note that the natural bijection $\colim_d (R[T]_d\times R) \to (\colim_d R[T]_d)\times R$ is continuous (because it is continuous at each level $d$), but the continuity of its inverse is not so clear.

I find it amusing, then, to observe that sequential continuity holds in the important case where $R$ is a field. This relies on the following proposition.

Proposition. — Assume that $R$ is a field. Then, for every converging sequence $(f_n)$ in $R[T]$, the degrees $\deg(f_n)$ are bounded.

Otherwise, we can assume that $(f_n)$ converges to $0$ and that $\deg(f_{n+1})>\deg(f_n)$ for all $n$. We construct a continuous linear form $\phi$ on $R[T]$ such that $\phi(f_n)$ does not converge to $0$. This linear form is given by a formal power series $\phi(f)=\sum a_d c_d$ for $f=\sum c_dT^d$, and we choose the coefficients $(a_n)$ by induction so that $\phi(f_n)=1$ for all $n$. Indeed, if the coefficients are chosen up to $\deg(f_n)$, then we fix $a_d=0$ for $\deg(f_n)<d<\deg(f_{n+1})$ and choose $a_{\deg(f_{n+1})}$ so that $\phi(f_{n+1})=1$. This linear form is continuous because its restriction to any $R[T]_d$ is given by a polynomial, hence is continuous.

Corollary. — If $R$ is a topological ring which is a field, then the evaluation map $R[T]\times R\to R$ is sequentially continuous.

Consider sequences $(f_n)$ in $R[T]$ and $(a_n)$ in $R$ that converge to $f$ and $a$ respectively. By the proposition, there is an integer $d$ such that $\deg(f_n)\leq d$ for all $n$, and $\deg(f)\leq d$. Since evaluation is continuous on $R[T]_d\times R$, one has $f_n(a_n)\to f(a)$, as claimed.

Remark. — The previous proposition does not hold on rings. In fact, if $R=\mathbf Z_p$ is the ring of $p$-adic integers, then $\phi(p^nT^n)=p^n \phi(T^n)$ converges to $0$ for every continuous linear form $\phi$ on $R[T]$. More is true since in that case, evaluation is continuous! The point is that in $\mathbf Z_p$, the ideals $(p^n)$ form a basis of neighborhoods of the origin.

Proposition. — If the topology of $R$ is linear, namely the origin of $R$ has a basis of neighborhoods consisting of ideals, then the evaluation map $R[T]\times R\to R$ is continuous.

By translation, one reduces to showing continuity at $(0,0)$. Let $V$ be a neighborhood of $0$ in $R$ and let $I$ be an ideal of $R$ such that $I\subset V$. Since it is an subgroup of the additive group of $R$, the ideal $I$ is open. Then the set $I\cdot R[T]$ is open because for every $d$, its trace on $R[T]_d$, is equal to $I\cdot R[T]_d$, hence is open. Then, for $f\in I\cdot R[T]$ and $a\in R$, one has $f(a)\in I$, hence $f(a)\in V$.

Here is one case where I can prove that evaluation is continuous.

Proposition. — If the topology of $R$ is given by a family of absolute values, then the evaluation map $(f,a)\mapsto f(a)$ is continuous.

I just treat the case where the topology of $R$ is given by one absolute value. By translation and linearity, it suffices to prove continuity at $(0,0)$. Consider the norm $\|\cdot\|_1$ on $R[T]$ defined by $\|f\|_1=\sum |c_n|$ if $f=\sum c_nT^n$. By the triangular inequality, one has $|f(a)|\leq \|f\|_1 $ for any $a\in R$ such that $|a|\leq 1$. For every $r>0$, the set $V_r$ of polynomials $f\in R[T]$ such that $\|f\|_1<r$ is an open neighborhood of the origin since, for every integer $d$, its intersection with $R[T]_d$ is an open neighborhood of the origin in $R[T]_d$. Let also $W$ be the set of $a\in R$ such that $|a|\leq 1$. Then $V_r\times W$ is a neighborhood of $(0,0)$ in $R[T]\times R$ such that $|f(a)|<r$ for every $(f,a)\in V_r\times W$. This implies the desired continuity.

Flatness and projectivity: when is the localization of a ring a projective module?

Projective modules and flat modules are two important concepts in algebra, because they characterize those modules for which a general functorial construction (Hom module and tensor product, respectively) behave better than what is the case for general modules.

This blog post came out of reading a confusion on a student's exam: projective modules are flat, but not all flat modules are projective. Since localization gives flat modules, it is easy to obtain a an example of a flat module which is not projective (see below, $\mathbf Q$ works, as a $\mathbf Z$-module), but my question was to understand when the localization of a commutative ring is a projective module.

$\gdef\Hom{\operatorname{Hom}}\gdef\Spec{\operatorname{Spec}}\gdef\id{\mathrm{id}}$

Let me first recall the definitions. Let $R$ be a ring and let $M$ be a (right)$R$-module.

The $\Hom_R(M,\bullet)$-functor associates with a right $R$-module $X$ the abelian group $\Hom_R(M,X)$. By composition, any linear map $f\colon X\to Y$ induces an additive map $\Hom_R(M,f)\colon \Hom_R(M,X)\to \Hom_R(M,X)$: it maps $u\colon M\to X$ to $\phi\circ u$. When $R$ is commutative, these are even $R$-modules and morphisms of $R$-modules. If $f$ is injective, $\Hom_R(M,f)$ is injective as well, but if $f$ is surjective, it is not always the case that $\Hom_R(M,f)$ is surjective, and one says that the $R$-module $M$ is projective if $\Hom_R(M,f)$ is surjective for all surjective linear maps $f$.

The $\otimes_R$-functor associates with a left $R$-module $X$ the abelian group $M\otimes_R X$, and with any linear map $f\colon X\to Y$, the additive map $M\otimes_R X\to M\otimes_R Y$ that maps a split tensor $m\otimes x$ to $m\otimes f(x)$. When $R$ is commutative, these are even $R$-modules and morphisms of $R$-modules. If $f$ is surjective, then $M\otimes_R f$ is surjective, but if $f$ is injective, it is not always the case that $M\otimes_R f$ is injective. One says that $M$ is flat if $M\otimes_R f$ is injective for all injective linear maps $f$.

These notions are quite abstract, and the development of homological algebra made them prevalent in modern algebra.

Example. — Free modules are projective and flat.

Proposition. — An $R$-module $M$ is projective if and only if there exists an $R$-module $N$ such that $M\oplus N$ is free.
Indeed, taking a generating family of $M$, we construct a free module $L$ and a surjective linear map $u\colon L\to M$. Since $M$ is projective, the map $\Hom_R(M,u)$ is surjective and there exists $v\colon M\to L$ such that $u\circ v=\id_M$. Then $v$ is an isomorphism from $M$ to $u(M)$, and one can check that $L=u(M)\oplus \ker(v)$.

Corollary. — Projective modules are flat.

Theorem (Kaplansky). — If $R$ is a local ring, then a projective $R$-module is free.

The theorem has a reasonably easy proof for a finitely generated $R$-module $M$ over a commutative local ring. Let $J$ be the maximal ideal of $R$ and let $k=R/J$ be the residue field. Then $M/JM$ is a finite dimensional $k$-vector space; let us consider a family $(e_1,\dots,e_n)$ in $M$ whose images form a basis of $M/JM$. Now, one has $\langle e_1,\dots,e_n\rangle + J M = M$, hence Nakayama's lemma implies that $M=\langle e_1,\dots,e_n\rangle$. Let then $u\colon R^n\to M$ be the morphism given by $u(a_1,\dots,a_n)=\sum a_i e_i$; by what precedes, it is surjective, and we let $N$ be its kernel. Since $M$ is projective, the morphism $\Hom_R(M,u)$ is surjective, and there exists $v\colon M\to R^n$ such that $u\circ v=\id_M$. We then have an isomorphism $M\oplus N\simeq R^n$, where $N=\ker(v)$. Moding out by $J$, we get $M/JM \oplus N/JN \simeq k^n$. Necessarily, $N/JN=0$, hence $N=JN$; since $N$ is a direct summand of $R^n$, it is finitely generated, and Nakayama's lemma implies that $N=0$.

Example. — Let $R$ be a commutative ring and let $S$ be a multiplicative subset of $R$. Then the fraction ring $S^{-1}R$ is a flat $R$-module.
Let $u\colon X\to Y$ be an injective morphism of $R$-modules. First of all, one identifies the morphism $S^{-1}R\otimes_R u\colon S^{-1}R\otimes_R X\to S^{-1}R\otimes_R Y$ to the morphism $S^{-1}u\colon S^{-1}X\to S^{-1}Y$ induced by $u$ on fraction modules. Then, it is easy to see that $S^{-1}u$ is injective. Let indeed $x/s\in S^{-1}X$ be an element that maps to $0$; one then has $u(x)/s=0$, hence there exists $t\in S$ such that $tu(x)=0$. Consequently, $u(tx)=0$, hence $tx=0$ because $u$ is injective. This implies $x/s=0$.

Theorem. — Let $R$ be a commutative ring. If $M$ is a finitely presented $R$-module, then $M$ is locally free: there exists a finite family $(f_1,\dots,f_n)$ in $R$ such that $R=\langle f_1,\dots,f_n\rangle$ and such that for every $i$, $M_{f_i}$ is a free $R_{f_i}$-module.
The proof is a variant of the case of local rings. Starting from a point $p\in\Spec(R)$, we know that $M_p$ is a finitely presented flat $R_p$-module. As above, we get a surjective morphism $u\colon R^n\to M$ which induces an isomorphism $\kappa(p)^n\to \kappa(p)\otimes M$, and we let $N$ be its kernel. By flatness of $M$ (and an argument involving the snake lemma), the exact sequence $0\to N\to R_p\to M\to 0$ induces an exact sequence $0\to \kappa(p)\otimes N\to \kappa(p)^n\to \kappa(p)\otimes M\to 0$. And since the last sequence is an isomorphism, we have $\kappa(p)\otimes N$. Since $M$ is finitely presented, the module $N$ is finitely generated, and Nakayama's lemma implies that $N_p=0$; moreover, there exists $f\not\in p$ such that $N_f=0$, so that $u_f\colon R_f^n\to M_f$ is an isomorphism. One concludes by using the quasicompactness of $\Spec(R)$.

However, not all flat modules are projective. The most basic example is the following one.

Example. — The $\mathbf Z$-module $\mathbf Q$ is flat, but is not projective.
It is flat because it is the total fraction ring of $\mathbf Z$. To show that it is not projective, we consider the free module $L={\mathbf Z}^{(\mathbf N)}$ with basis $(e_n)$ and the morphism $u\colon L\to\mathbf Q$ that maps $e_n$ to $1/n$ (if $n>0$, say). This morphism is surjective. If $\mathbf Q$ were projective, there would exist a morphism $v\colon \mathbf Q\to L$ such that $u\circ v=\id_{\mathbf Q}$. Consider a fraction $a/b\in\mathbf Q$; one has $b\cdot 1/b=1$, hence $b v(1/b)=v(1)$. We thus see that all coeffiencients of $v(1)$ are divisible by $b$, for any integer $b$; they must be zero, hence $v(1)=0$ and $1=u(v(1))=0$, a contradiction.
The proof generalizes. For example, if $R$ is a domain and $S$ does not consist of units, and does not contain $0$, then $S^{-1}R$ is not projective. (With analogous notation, take a nonzero coefficient $a$ of $v(1)$ and set $b=as$, where $s\in S$ is not $0$; then $as$ divides $a$, hence $s$ divides $1$ and $s$ is a unit.)

These recollections are meant to motivate the forthcoming question: When is it the case that a localization $S^{-1}R$ is a projective $R$-module?

Example. — Let $e$ be an idempotent of $R$, so that the ring $R$ decomposes as a product ot two rings $R\simeq eR \times (1-e)R$, and both factors are projective submodules of $R$ since their direct sum is the free $R$-module $R$. Now, one can observe that $R_e= eR$. Consequently, $R_e$ is projective. Geometrically, $\Spec(R)$ decomposes as a disjoint union of two closed subsets $\mathrm V(e)$ and $\mathrm V(1-e)$; the first one can be viewed as the open subset $\Spec(R_{1-e})$ and the second one as the open subset $\Spec(R_e)$.

The question was to decide whether this geometric condition furnishes the basic conditions for a localization $S^{-1}R$ to be projective. With the above notation, we recall that $\Spec(S^{-1}R)$ is homeomorphic to a the subset of $\Spec(R)$ consisting of prime ideals $p$ such that $p\cap S=\emptyset$. The preceding example corresponds to the case where $\Spec(S^{-1}R)$ is open and closed in $\Spec(R)$. In this case, we view $S^{-1}R$ as a quasicoherent sheaf on $\Spec(R)$, it is free of rank one on the open subset $\Spec(S^{-1}R)$, and zero on the complementary open subset. It is therefore locally free, hence the $R$-module $S^{-1}R$ is projective.

Observation. — The set $\Spec(S^{-1}R)$ is stable under generization. If $S^{-1}R$ is a projective $R$-module, then it is open.
The first part is obvious: if $p$ and $q$ are prime ideals of $R$ such that $p\subseteq q$ and $q\cap S=\emptyset$, then $p\cap S=\emptyset$. The second part follows from the observation that the support of $S^{-1}R$ is exactly $\Spec(S^{-1}R)$, combined with the following proposition.

Proposition. — The support of a projective module is open.
I learnt this result in the paper by Vasconcelos (1969), “On Projective Modules of Finite Rank” (Proceedings of the American Mathematical Society 22 (2): 430‑33). The proof relies on the trace ideal $\tau_R(M)$ of a module: this is the image of the canonical morphism $t\colon M^\vee \otimes_R M\to R$. (It is called the trace ideal, because when $M$ is free, $M^\vee\otimes_R M$ can also be identified with the module of endomorphisms of finite rank of $M$, a split tensor $\phi\otimes m$ corresponds with the endomorhism $x\mapsto \phi(x)m$, and then $t(\phi \otimes m)=\phi(m)$ is its trace.) Now, if $p$ belongs to the support of $M$, then $\tau_R(M)_p=R_p$, while if $p$ does not belong to the support of $M$, one has $M_p=0$, hence $\tau_R(M)_p=0$. In other words, the support of $M$ is the complement of the closed locus $\mathrm V(\tau_R(M))$ of $\Spec(R)$.

On the other hand, one should remember the following basic property of the support of a module.

Proposition. — The support of a module is stable under specialization. The support of a finitely generated module is closed.
Indeed, for every $m\in M$ and $p\in \Spec(R)$, saying that $m=0$ in $M_p$ means that there exist $s\in R$ such that $s\notin p$ with $sm=0$. In other words, this set is $\mathrm V(\mathrm{ann}_R(m))$. This shows that the support of $M$ is the union of the closed subsets $\mathrm V(\mathrm{ann}_R(m))$; it is in particular stable under specialization. If $M$ is finitely generated, this also shows its support is $\mathrm V(\mathrm{ann}_R(M))$, hence is closed.

At this point, one can go either following Vasconcelos (1969) who shows that a projective module $M$ of the form $S^{-1}R$ is finitely generated if and only if its trace ideal is. In particular, if $R$ is noetherian and $S^{-1}R$ is a projective $R$-module, then $\Spec(S^{-1}R)$ is closed. It is thus open and closed, and we are in the situation of the basic example above.

One can also use a topological argument explained to me by Daniel Ferrand: a minimal prime ideal of $R$ that meets $\Spec(S^{-1}R)$ is disjoint from $S$, hence belongs to $\Spec(S^{-1}R)$. Consequently, $\Spec(S^{-1}R)$ is the union of the irreducible components of $\Spec(R)$ that it meets. If this set of irreducible components is finite (or locally finite), for example if $\Spec(R)$ is noetherian, for example if $R$ is a noetherian ring, then $\Spec(S^{-1}R)$ is closed.

I did not find the time to think more about this question, and it would be nice to have an example of a projective localization which does not come from this situation.

Combinatorics of the nilpotent cone

$\global\def\Card{\operatorname{Card}}\global\def\GL{\mathrm{GL}}\global\def\im{\operatorname{im}}\gdef\KVar{\mathrm{KVar}}$

Let $n$ be an integer and $F$ be a field. Nilpotent matrices $N\in \mathrm M_n(F)$ are those matrices for which there exists an integer $p$ with $N^p=0$. Their characteristic polynomial is $\chi_N(T)=T^n$, and they satisfy $N^n=0$, which shows that the set $\mathscr N_n$ of nilpotent matrices is an algebraic variety. The equation $N^n=0$ is homogeneous of degree $n$, so that $\mathscr N_n$ is a cone.

The classification of nilpotent matrices is an intermediate step in the theory of Jordan decomposition: In an adequate basis, a nilpotent matrix can be written as a diagonal block matrix of “basic” nilpotent matrices, $p \times p$ matrices of the form \[ \begin{pmatrix} 0 & 0 & \dots & 0 & 0 \\ 1 & 0 & & & \vdots \\ 0 & 1 & \ddots & & 0 \\ \vdots & \ddots & \ddots & \ddots & 0 \\ 0 & & 0 & 1 & 0\end{pmatrix} \] whose minimal polynomial is $T^p$. The sum of the sizes of these blocks is $n$ and in this way, it is associated with any $n\times n$ nilpotent matrix a partition $\pi$ of~$n$. It is known that two nilpotent matrices are conjugate if and only if they are associated with the same partition. For any partition $\pi$ of~$n$, let us denote by $J_\pi$ the corresponding matrix whose sizes of blocks are arranged in increasing order, and $\mathscr N_\pi$ the set of nilpotent matrices that are associated with the partition $\pi$.

The theorem of Fine and Herstein (1958)

Having to teach “agrégation” classes made me learn about a classic combinatorial result: counting the number of nilpotent matrices when $F$ is a finite field.

Theorem (Fine, Herstein, 1958). — Let $F$ be a finite field with $q$ elements. The cardinality of $\mathscr N_n(F)$ is $q^{n^2-n}$. Equivalently, the probability that an $n\times n$ matrix with coefficients in $F$ be nilpotent is $q^{-n}$.

The initial proof of this results relies on the action of $\GL_n(F)$ on $\mathscr N_n(F)$: we recalled that the orbits correspond with the partitions of $n$, hence a decomposition \[ \Card(\mathscr N_n(F)) = \sum_{\pi} \Card(\mathscr N_\pi(F)). \] We know that $\mathscr N_\pi(F)$ is the orbit of the matrix $J_\pi$ under the action of $\GL_n(F)$. By the classic orbit-stabilizer formula, one thus has \[ \Card(\mathscr N_\pi(F)) = \frac{\Card(\GL_n(F))}{\Card(C_\pi(F))}, \] where $C_\pi(F)$ is the set of matrices $A\in\GL_n(F)$ such that $AJ_\pi=J_\pi A$. The precise description of $C_\pi(F)$ is delicate but their arguments go as follow.

They first replace the group $C_\pi(F)$ by the algebra $A_\pi(F)$ of all matrices $A\in\mathrm M_n(F)$ such that $AJ_\pi=J_\pi A$. For any integer, let $m_i$ be the multiplicity of an integer $i$ in the partition $\pi$, so that $n=\sum i m_i$. The block decomposition of $J_\pi$ corresponds with a decomposition of $F^n$ as a direct sum of invariant subspaces $V_i$, where $V_i$ has dimension $i m_i$. In fact, $V_1=\ker(J_\pi)$, $V_1\oplus V_2=\ker(J_\pi^2)$, etc. This shows that $A_\pi(F)$ is an algebra of block-triangular matrices. Moreover, the possible diagonal blocks can be shown to be isomorphic to $\mathrm M_{m_i}(F)$. In other words, we have a surjective morphism of algebras \[ A_\pi(F) \to \prod_i \mathrm M_{m_i}(F), \] whose kernel consists of nilpotent matrices. In particular, the proportion of invertible elements in $A_\pi(F)$ is equal to the proportion of invertible elements in the product $\prod_i \mathrm M_{m_i}(F)$.

Ultimately, Fine and Herstein obtain an explicit sum over the set of partitions of $n$ which they prove equals $q^{n^2-n}$, after an additional combinatorial argument.

Soon after, the theorem of Fine and Herstein was given easier proofs, starting from Gerstenhaber (1961) to Kaplansky (1990) and Leinster (2021).

A proof

The following proof is borrowed from Caldero and Peronnier (2022), Carnet de voyage en Algébrie. It can be seen as a simplification of the proofs of Gerstenhaber (1961) and Leinster (2021).

Let us start with the Fitting decomposition of an endomorphism $u\in \mathrm N_n(F)$: the least integer $p$ such that $\ker(u^p)=\ker(u^{p+1})$ coincides with the least integer $p$ such that $\im(u^p)=\im(u^{p+1})$, and one has $F^n=\ker(u^p)\oplus \im(u^p)$. The subspaces $N(u)=\ker(u^p)$ and $I(u)=\im(u^p)$ are invariant under $u$, and $u$ acts nilpotently on $\ker(u^p)$ and bijectively on $\im(u^p)$. In other words, we have associated with $u$ complementary subspaces $N(u)$ and $I(u)$, a nilpotent operator of $N(u)$ and an invertible operator on $I(u)$. This map is bijective.

For any integer $d$, let $\nu_d$ be the cardinality of nilpotent matrices in $\mathrm M_d(F)$, and $\gamma_d$ be the cardinality of invertible matrices in $\mathrm M_d(F)$. Let also $\mathscr D_d$ be the set of all pairs $(K,I)$, where $K$ and $I$ are complementary subspaces of dimensions $d$, $n-d$ of $F^n$. We thus obtain \[ n^2 = \sum_{(K,I)\in\mathscr D_d} \nu_d \cdot \gamma_{n-d}. \] We need to compute the cardinality of $\mathscr D_d$. In fact, given one pair $(K,I)\in\mathscr D_d$, all other are of the form $(g\cdot K,g\cdot I)$, for some $g\in\GL_n(F)$: the group $\GL_n(F)$ acts transitively on $\mathscr D_d$. The stabilizer of $(K,I)$ can be identified with $\GL_d(F)\times \GL_{n-d}(F)$. Consequently, \[ \Card(\mathscr D_d) = \frac{\Card(\GL_n(F))}{\Card(\GL_d(F)\Card(\GL_{n-d}(F))} = \frac{\gamma_n}{\gamma_d \gamma_{n-d}}. \] We thus obtain \[ q^{n^2} = \sum_{d=0}^n \frac{\gamma_n}{\gamma_d \gamma_{n-d}} \nu_d \gamma_{n-d} = \gamma_n \sum_{d=0}^n \frac{\nu_d}{\gamma_d}. \] By subtraction, we get \[ \frac{\nu_n}{\gamma_n} = \frac {q^{n^2}}{\gamma_n} - \frac{q^{(n-1)^2}}{\gamma_{n-1}},\] or \[ \nu_n = q^{n^2} - q^{(n-1)^2} \frac{\gamma_n}{\gamma_{n-1}}. \] It remains to compute $\gamma_n$: since an invertible matrix consists of a nonzero vector, a vector which does not belong to the line generated by the first one, etc., we have \[ \gamma_n = (q^n-1) (q^n-q)\dots (q^n-q^{n-1}). \] Then, \[ \gamma_n = (q^n-1) q^{n-1} (q^{n-1}-1)\dots (q^{n-1}-q^{n-2}) = (q^n-1)q^{n-1} \gamma_{n-1}. \] We thus obtain \[ \nu_n = q^{n^2} - q^{(n-1)^2} (q^n-1) q^{n-1} = q^{n^2} - q^{(n-1)^2} q^{2n-1} + q^{(n-1)^2} q^{n-1} = q^{n^2-n}, \] as claimed.

The proof of Leinster (2021)

Leinster defines a bijection from $\mathscr N_n(F)\times F^n$ to $\mathrm M_n(F)$. The definition is however not very canonical, because he assumes given, for any subspace $V$ of $F^n$, a basis of $V$.

Take a pair $(u,x)$, where $u\in\mathscr N_n(F)$ and $x\in F^n$ and consider the subspace $V_x=\langle x,u(x),\dots\rangle$, the smallest $u$-invariant subspace of $F^n$ which contains $x$. To describe $u$, we observe that we know its restriction to $V_x$, and we need to describe it on the chosen complementary subspace $V'_x$.

To that aim, we have to give ourselves an endomorphism $u'_x$ of $V'_x$ and a linear map $\phi_x\colon V'_x\to V_x$. Since we want $u$ to be nilpotent, it is necessary and sufficient to take $u'_x$ nilpotent.

Instead of considering $\phi_x\colon V'_x\to V_x$, we can consider the map $y\mapsto y+\phi_x(y)$. Its image is a complement $W_x$ of $V_x$ in $F^n$, and any complement can be obtained in this way. The nilpotent endomorphism $u'_x$ of $V'_x$ transfers to a nilpotent endomorphism $w_x$ of $W_x$.

All in all, what the given pair $(u,x)$ furnishes is a subspace $V_x$ with a basis $(x_1=x,x_2=u(x),\dots)$, a complement $W_x$, and a nilpotent endomorphism $w_x$ of $W_x$. This is more or less what the Fitting decomposition of an endomorphism gives us! Recall that $V_x$ was assumed to have been given a basis $(e_1,\dots,e_p)$. There exists a unique automorphism of $V_x$ which maps $e_i$ to $u^{i-1}(x)$ for all $i$. In other words, we have a pair of complementary subspaces $(V_x,W_x)$, a linear automorphism of $V_x$, and a nilpotent automorphism of $W_x$. By the Fitting decomposition, these data furnish in a bijective way an endomorphism of $F^n$, and that concludes the proof.

A remark about motivic integration

The framework of motivic integration suggests to upgrade these combinatorial results into equalities valid for all field $F$, which hold in the Grothendieck ring of varieties $\KVar_F$. As an abelian group, it is generated by symbols $[X]$, for all algebraic varieties $X$ over $F$, with relations $[X]=[Y]+[X\setminus Y]$, whenever $Y$ is a closed subvariety of $X$. The ring structure is defined so that the formula $[X]\cdot[Y]=[X\times Y]$ for all algebraic varieties $X$ and $Y$ over $F$.

By construction of this ring, equalities $[X]=[Y]$ in $\KVar_F$ imply that many invariants of $X$ and $Y$ coincide. In particular, when $F$ is a finite field, they will have the same number of points.

The question is thus to compute the class $[\mathscr N_n]$ of the variety $\mathscr N_n$, for any field $F$. The proofs that I described above can be more or less transferred to this context and imply the following theorem. We denote by $\mathbf L\in \KVar_F$ the class of the affine line $\mathbf A^1$.

Theorem. — One has an equality $[\mathscr N_n] \mathbf L^n = \mathbf L^{n^2}$ in the localization of the Grothendieck ring $\KVar_F$ by the element $(\mathbf L-1)\dots(\mathbf L^{n-1}-1)$.

The following question is then natural. (I have not thought about it at all.)

Question. — Does one have $[\mathscr N_n]=\mathbf L^{n^2-n}$ in $\KVar_F$?