Marc-André Brochu's Oueb Site

Compter des composantes irréductibles est comme compter des points

Wed, 18 Mar 2026 00:00:00 +0000

Pour des ensembles de cardinalité finie, on a le principe d’inclusion-exclusion:

@@ |A \cup B| = |A| + |B| - |A \cap B|. @@

En comparaison, si @X@ et @Y@ sont deux parties fermées d’un espace topologique ayant chacune un nombre fini de composantes irréductibles¹, et qu’on note @n(-)@ le nombre de composantes irréductibles, alors

@@ n(X \cup Y) = n(X) + n(Y) - n(X \cap Y). @@

On déduit la première équation de celle-ci en donnant aux ensembles @A@ et @B@ la topologie discrète.

Passer du clavier CSA à Ergo-L

Sat, 07 Mar 2026 00:00:00 +0000

Jusqu’à maintenant, je me suis presque¹ toujours servi de la disposition canadienne multilingue standard. Elle permet d’écrire en anglais et en français sans trop de mal. Un objectif principal de cette disposition est de minimiser les touches mortes. Par exemple, on peut entrer les caractères “à”, “ç” et “è” directement, ce qui est assez pratique. Cependant! Mes doigts partent en incartade quand je leur demande une paire de crochets ou des accolades. Ma main fait des frasques si j’ai le malheur de lui quémander un diacritique tout nu. Je fige si je dois fermer une balise HTML ou ouvrir une commande TeX.

Closing in closure in Clojure

Tue, 03 Mar 2026 00:00:00 +0000

Learning Clojure is a lot of fun. For one, the community is great.¹ On top of that, there’s many exercises available; that’s excellent for people like me, who learn by doing. For instance, rich4clojure is a great resource. It works out-of-the-box with Visual Studio Code, which is nice when you just want to start working on problems asap and you’ll just “setup the tools later”.

Tonight, I’m solving problem 84 from that set. I thought, maybe you’d like to follow along with me. Here it goes:

Direct limits, and how to compute them from cofinal subsets

Sat, 22 Feb 2025 00:00:01 +0000

Here I talked about computing a direct limit on some cofinal subset. In this post, I want to prove what my assertions.

Note that direct limits are a special case of colimits, and what I’ll talk about can easily be generalized. In particular, colimit may also be computed on cofinal subcategories of their indexing category, and yield equal (“isomorphic up to a unique canonical isomorphism”) colimits. However, in this post I’ll stay in the particular case of direct limits, and I’ll try to define things first in terms of elements, to stay as concrete as possible.

Skyscraper sheaves

Sat, 22 Feb 2025 00:00:00 +0000

Let @X@ be a topological space, let @p@ be a point of @X@, and let @S@ be any set. In the notation of Vakil, define the skyscraper sheaf supported at @p@ by the formula

@@ (i_{p,*}S)(U) = \begin{cases} S & \text{if $p \in U$;} \\ 1 & \text{otherwise.} \end{cases} @@

Here @1@ is the singleton set. We can also define such a sheaf in other categories (abelian groups, rings, etc.), replacing the singleton set by the appropriate terminal object.

Characteristic classes (Part 1, Cohomology)

Thu, 13 Feb 2025 00:00:00 +0000

TODO: I need to fix the commutative diagrams in this post because they just don’t show up right now.

I’m planning on writing a series of posts that explore the theory and applications of characteristic classes in algebraic topology, following the book from Milnor and Stasheff. Here I start with the appendix A, where homology is discussed and basic theorems are laid out. This post will also serve me well as a quick reminder, since I keep forgetting small details and ideas regarding (singular) (co)homology. Many results and explanations can be found in Hatcher’s book.

The dominance order on partitions

Sat, 08 Feb 2025 00:00:01 +0000

TODO: Fix Young diagrams as they just don’t show up right now!

Recall that a partition @\lambda@ of some fixed natural number @n \geq 0@ is a list of nonnegative integers @(\lambda_1,\lambda_2,\dots)@ such that @\lambda_1 \geq \lambda_2 \geq \dots \geq 0@ and such that @\lambda_1+\lambda_2+\dots = n@. For the purposes of this post, the actual number of parts in a partition is irrelevant, and since one can always extend a partition with zeroes we may suppose that all partitions have the same fixed number of parts, let’s say @\ell@.

A description for the ideal generated by the symmetric polynomials in two variables

Sat, 08 Feb 2025 00:00:00 +0000

The goal of this short post is to convince myself that the ideal generated by @\operatorname{Sym}^+@ inside of @\mathbb{C}[x,y]@ can be more simply described as the ideal generated by @xy@ and @x+y@, i.e.

@@\langle \operatorname{Sym}^+\rangle = \langle xy,x+y\rangle.@@

As a shorthand, set @A = \mathbb{C}[x,y]@. The ring @A@ is graded:

@@A = \bigoplus_{d = 0}^\infty A^{(d)},@@

where @A^{(d)}@ is the @\mathbb{C}@-module consisting of the homogeneous polynomials of degree @d@:

@@A^{(d)} = \mathbb{C}\{x^ay^b \mid a + b = d\}.@@

In general, a polynomial is said to be symmetric when it is invariant under any permutation of the variables. In our case, a polynomial @p(x,y) \in A@ is symmetric when @p(x,y) = p(y,x)@. For instance, @x^3 + y^3 + 2xy@ is symmetric while @x+y^2@ is not. The product and difference of two symmetric polynomials is also a symmetric polynomial. Also, @1@ is a trivial example of a symmetric polynomial. Hence the set of all symmetric polynomials is a subring of @A@, which we denote by @\operatorname{Sym}@. This subring is naturally graded:

Galois connections

Thu, 06 Feb 2025 00:00:00 +0000

Let @X@ and @Y@ be two partially ordered sets (posets). Two functions @f : X \to Y@ and @g : Y \to X@ are said to form a Galois connection when the following is true for all @a \in X@ and all @b \in Y@: \begin{equation*}f(a) \leq b \iff a \leq g(b).\tag{C}\end{equation*} We say that @f@ is the lower (or left) adjoint, and @g@ is the upper (or right) adjoint. Oftentimes the lower adjoint is marked with a lower-star @f_*@ and the upper adjoint with an upper-star @f^*@. We will see for instance that the direct and inverse image form such a connection, and this hopefully explains the usual notation for these concepts. The shorthand I’m going to use for saying two functions are in Galois connection is @f_* \dashv f^*@.

The determinant of a matrix is just the product of its columns

Wed, 05 Feb 2025 00:00:00 +0000

A fact I’ve known for a very long time, but never bothered to prove for myself, is the unicity of the determinant for square matrices. More precisely, let @A@ be any commutative ring with identity; then the determinant is the only @A@-valued function on matrices that: (i) is linear on each column, (ii) is such that switching two columns changes the sign, and (iii) sends the identity matrix to 1.

Filters are generalized subsets

Mon, 03 Feb 2025 00:00:00 +0000

TODO Add small proofs of known facts about limits using the language of filters.

Given any set @X@, a filter on @X@ is a collection of subsets @\mathcal{F} \subseteq 2^X@ such that:

@X@ itself is in @\mathcal{F}@;
if @A \in \mathcal{F}@ and @B \in \mathcal{F}@, then @A \cap B \in \mathcal{F}@;
if @A \in \mathcal{F}@ and @C@ is any subset with @A \subseteq C@, then @C \in \mathcal{F}@ (this is refered to by the expression upward closure).

There are always at least two (possibly identical) filters on any set @X@: one is @\top = \{X\}@ and the other is @\bot = 2^X@. Note that @\bot@ is the only filter which contains the empty set, because of the upward closure property. We can put a partial order on @\operatorname{Filter}(X)@, the set of filters on @X@, by declaring @\mathcal{F} \leq \mathcal{G}@ if and only if @\mathcal{G} \subseteq \mathcal{F}@. At first glance, it may seem weird to “reverse” the order, but it makes sense if you think of a filter as a way of approximating something, or as some kind of locating scheme. The usual intersection and union of sets makes @\text{Filter}(X)@ into a complete lattice, i.e. any collection of filters has a least upper bound.

Le radical est l'intersection des idéaux maximaux

Tue, 11 Apr 2023 00:00:00 +0000

L’exercice 1.1 d’Ottem et Ellingsrud demande:

In any ring the radical of an ideal @\mathfrak{a}@ equals the intersection of the prime ideals containing the ideal. Show, using the Nullstellensatz, that in the polynomial ring @k[x_1,\dots,x_n]@ the radical @\sqrt{\mathfrak{a}}@ equals the intersection of all the maximal ideals containing @\mathfrak{a}@.

Tout d’abord, je souhaite me rafraîchir la mémoire et redémontrer la première assertion qu’ils font, c’est-à-dire que le radical d’un idéal est en fait l’intersection de tous les idéaux premiers qui le contiennent. Je trouve que c’est une caractérisation intéressante du radical. Peut-être même que je vais créer un autre article spécifiquement pour ça ! La preuve n’est pas triviale, en tout cas pour l’une des deux inclusions. La première inclusion est facile à voir: soit @\{\mathfrak{p}_i\}_{i\in I}@ la collection de tous les idéaux premiers qui contiennent @\mathfrak{a}@. Vu que pour tout idéal premier @x^n \in \mathfrak{p_i}@ implique que @x\in\mathfrak{p_i}@ (pour un certain entier @n\geq 1@), il est clair que @\sqrt{\mathfrak{a}}\subseteq \bigcap_{i\in I}\mathfrak{p_i}@. C’est la seconde inclusion qui pose problème. Je vais la montrer par contrapositive: soit @x\notin \sqrt{\mathfrak{a}}@. Je considère la collection @\mathcal{I}@ des idéaux @\mathfrak{b}\supseteq \mathfrak{a}@ tels que @x^n\notin\mathfrak{b}@ pour tout entier @n\geq 1@. Si j’ordonne partiellement cette collection par l’inclusion, le lemme de Zorn me donne l’existence d’un élément maximal @\mathfrak{p}\in\mathcal{I}@. Je veux montrer que @\mathfrak{p}@ est premier. Soit @a@ et @b@ deux éléments de l’anneau tels que @a,b\notin \mathfrak{p}@. Par maximalité, ni @\mathfrak{p}+(a)@ ni @\mathfrak{p}+(b)@ ne font partie de la collection @\mathcal{I}@. Par exemple, comme @a\notin \mathfrak{p}@, l’idéal @\mathfrak{p}+(a)@ (i.e. le plus petit idéal qui contient à la fois @\mathfrak{p}@ et @(a)@) contient strictement @\mathfrak{p}@. Donc il existe deux entiers @n@ et @m@ tels que @x^n\in \mathfrak{p}+(a)@ et @x^m\in\mathfrak{p}+(b)@. Puisque l’idéal @\mathfrak{p}+(ab)@ contient @x^{n+m}@, il n’est lui non plus pas contenu dans @\mathcal{I}@. En particulier @ab\notin\mathfrak{p}@ car sinon ajouter @(ab)@ à @\mathfrak{p}@ n’aurait rien changé, ce qui n’est pas le cas. Cela prouve que @\mathfrak{p}@ est premier. Mais alors @\bigcap_{i\in I}\mathfrak{p}_i\subseteq \mathfrak{p}@ et puisque @x\notin \mathfrak{p}@ par construction, je retrouve bien que @x\notin\bigcap_{i\in I}\mathfrak{p_i}@, ce qu’il fallait démontrer.

L'existence et l'unicité de l'expansion en série de Laurent

Fri, 17 Feb 2023 00:00:00 +0000

TODO: really rough article, clean it up!

Soit @\Omega@ un domaine et @f:\Omega\to\mathbb{C}@ une fonction holomorphe.

Je prend un @z\in A@

@\gamma_1(t) = r_1e^{-it}@ et @\gamma_2(t) = r_2e^{it}@.

(TODO: expliquer) Par la formule intégrale de Cauchy on a que \begin{equation} f(z) = \frac{1}{2\pi i}\left(\oint_{\gamma_1} \frac{f(w)}{w-z}dz + \oint_{\gamma_2} \frac{f(w)}{w-z}dz\right) \end{equation}

Existence de l’expansion

Développement de la partie principale

Lorsque @w@ est sur le chemin @\gamma_1@, on a que

@@ \begin{align} \frac{1}{w-z} &= \frac{1}{(w-a)-(z-a)} = \frac{1}{w-a}\cdot\frac{1}{1-\frac{z-a}{w-a}}\\ &= \frac{1}{w-a}\frac{-(w-a)}{z-a}\sum_{n=0}^{\infty}\left(\frac{w-a}{z-a}\right)^n\\ &= \frac{-1}{w-a}\sum_{n=1}^\infty \left(\frac{w-a}{z-a}\right)^n \end{align} @@

puisque @|\frac{z-a}{w-a}|>1@ il faut utiliser l’autre formule pour une série géométrique.

A perfect number theory proof

Fri, 08 Oct 2021 00:00:00 +0000

For my first article on this blog (or any blog for that matter), I want to present one of my favorite proofs of elementary number theory. It’s based around the concept of a perfect number. You know that a concept named like that ought to be interesting. A perfect number is any positive integer that is equal to the sum of its positive divisors, except the fact that you don’t count the number itself in that sum. You only sum its proper divisors. The whole is equal to the sum of its parts; that’s why such a number is perfect. For instance, six is a perfect number because its proper divisors are 1, 2 and 3; if you sum these up you get back six. In fact six is the smallest perfect number. The next one is 28 as its proper divisors are 1, 2, 4, 7 and 14.