Bourbaki Algebra Exercises

Author: Eiko

Tags: algebra, bourbaki, exercises

Time: 2025-01-10 06:45:40 - 2025-01-13 07:00:39 (UTC)

This is a documentation of some interesting exercises in Bourbaki.

Chapter 1

Section 6, Exercise 30

Let $p$ be a prime and $r, t$ non-negative integers, $C$ is a cyclic group of order $p^{r + t}$ and $T \leq C$ a subgroup of order $p^{t}$ . So $p^{r} = [C : T] = | C / T |$ .

If $C$ acts on a finite set $E$ , we have

$| E | \equiv | E^{T} | \mod p^{r + 1} = p [C : T] .$

think of $E^{T}$ as something that interpolate between $E$ and $E^{C}$ , where $| E | = | E |$ and $| E | \equiv | E^{C} | \mod p$ .
As an application, we have

$(\binom{p^{r + t} m}{p^{t} n}) \equiv (\binom{p^{r} m}{n}) \mod p^{r + 1} .$

Proof

This is an application of orbit formula

$| E | = \sum [C : Stab (e)]$

we have that $e \in E^{T}$ iff $T \subset Stab (e)$ iff $[C : Stab (e)]$ is a factor of $[C : T]$ (Note this iff requires cyclic!). So that otherwise, $e \notin E^{T}$ iff $[C : Stab (e)]$ is divisible by $p [C : T]$ , thus the result.
Let $[n]$ denote a finite set of size $n$ . Consider the action of $C$ on $C \times [m]$ given by acting on the first coordinate. This induces an action of $C$ on $E = (\binom{C \times [m]}{| T | n})$ , the set of subsets of $C \times [m]$ of size $p^{t} n = | T | n$ .

The projection map $φ : C \times [m] \to C / T \times [m]$ is $C$ -equivariant and also $T$ -invariant (though these facts are not used). We claim that this induces a bijection

$φ^{'} : E^{T} = {(\binom{C \times [m]}{| T | n})}^{T} \to (\binom{C / T \times [m]}{n}) .$

If a set $F \in E$ is fixed by $T$ , it is given a $T$ -action on $F$ . This action is free (all stablizers are trivial) so every orbit is of size $| T |$ , mapping each orbit (there are $n$ of them) $T \cdot (s, x)$ to $T s \in C / T$ gives the desired bijection. This shows

$| E^{T} | = | (\binom{C / T \times [m]}{n}) | = (\binom{p^{r} m}{n}) .$

Applying the previous result, and use $| E | = (\binom{p^{r + t} m}{p^{t} n})$ , we get

$(\binom{p^{r + t} m}{p^{t} n}) \equiv (\binom{p^{r} m}{n}) \mod p^{r + 1} .$

Inspired: A New Proof of Binomial Mod $p$ Expansion Using Group Actions

The above proof inspires me to come up with a new proof of the binomial expansion mod $p$ , i.e.

$(\binom{n}{m}) \equiv (\binom{n_{k}}{m_{k}}) \dots (\binom{n_{0}}{m_{0}}) \mod p$

where $n = n_{k} p^{k} + n_{k - 1} p^{k - 1} + \dots + n_{0}$ and $m = m_{k} p^{k} + m_{k - 1} p^{k - 1} + \dots + m_{0}$ with $0 \leq m_{i}, n_{i} < p$ .

Let $C_{i}$ be the cyclic group of order $p^{i}$ . Consider the set

$X = ⨆_{i = 0}^{k} C_{i} \times [n_{i}] .$

This set is endowed with a natural $P = C_{k} \times \dots \times C_{0}$ action given by multiplying on the first coordinate. The action naturally induces an action on the set $E = (\binom{X}{m})$ of size $m$ subsets of $X$ .

Now let’s investigate what is $E^{P}$ . We have that a set $F \in E^{P}$ would imply a $P$ action on $F$ , for which the orbit sizes are computed as

$| P \cdot (c_{i}, x) | = [P : Stab ((c_{i}, x))] = | C_{i} | = p^{i} .$

This together with the uniqueness of $p$ -adic expansions tells us that the size of $F \cap (C_{i} \times [n_{i}])$ must be $p^{i} m_{i}$ , consisting of exactly $m_{i}$ orbits of size $p^{i}$ . Collapsing the orbits and mapping to the set $\prod_{i = 0}^{k} (\binom{[n_{i}]}{m_{i}})$ gives the bijection

$E^{P} = {(\binom{X}{m})}^{P} \overset{1 : 1}{\to} \prod_{i = 0}^{k} (\binom{[n_{i}]}{m_{i}}) .$

Finally, utilizing the orbit formula, for any $p$ -group $| E | \equiv | E^{P} | \mod p$ , we get the desired result

$(\binom{n}{m}) \equiv (\binom{n_{k}}{m_{k}}) \dots (\binom{n_{0}}{m_{0}}) \mod p .$

Chapter 1

Section 6, Exercise 30

Proof

Inspired: A New Proof of Binomial Mod p Expansion Using Group Actions

Inspired: A New Proof of Binomial Mod $p$ Expansion Using Group Actions