Bundle a family of objects into a single 'universal' object

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Bundle a family of objects into a single 'universal' object

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[note article incomplete]More examples needed.[/note]

[QUICK DESCRIPTION]

There are many situations in mathematics, especially in algebraic contexts, where a question about some infinite family of objects is equivalent to a question about a single 'universal' object of the same type.  Often, it is easier to consider the question just for the universal object.

[PREREQUISITES]

[EXAMPLE]

[note section contributions wanted]The Hasse principle seems a good example here, but I don't know enough number theory to do it justice.  This example needs to be finished off.[/note]

Let $f$ be a polynomial, and suppose we wish to determine if there are any integers $x$ for which $f(x)=0$.

Certainly, if $f(x) = 0$ then $f(x) \equiv 0 \mod n$, for any positive integer $n$.  In other words, the equation $f(x)=0$ has solutions in the ring $\mathbb{Z}/n \mathbb{Z}$.  Consider the other direction: if we have solutions in $\mathbb{Z}/n \mathbb{Z}$, when do we have solutions in $\mathbb{Z}$?  Ideally, we want to incorporate all the 'modulo $n$' cases into a single case.  The way to do this is by forming an inverse limit, as explained below.

We have a family of rings $\mathbb{Z}/n \mathbb{Z}$, but this family has some structure to it: given positive integers $m$ and $n$, there is a natural map from $\mathbb{Z}/mn \mathbb{Z}$ to $\mathbb{Z}/n \mathbb{Z}$ given by 'reduction modulo $n$'.  Now consider sequences $(x_n)_{n \in \mathbb{N}}$, where $x_n$ is an element of $\mathbb{Z}/n \mathbb{Z}$.  Define addition and multiplication of sequences pointwise, and note that this forms a commutative ring, with identity given by the 'all 1' sequence.  We say such a sequence is compatible if, for any positive integers $m$ and $n$, we have $x_{mn} \equiv x_n$ modulo $n$.  Observe that the compatible sequences form a subring.  This is the inverse limit of the rings $\mathbb{Z}/n \mathbb{Z}$, and is written $\hat{\mathbb{Z}}$.

It is clear that listing all the reductions modulo $n$ of a given integer will define a compatible sequence, so $\hat{\mathbb{Z}}$ contains $\mathbb{Z}$ as a subring.  Hence if want to rule out solutions in $\mathbb{Z}$, it suffices to rule out solutions in $\hat{\mathbb{Z}}$.

We could instead have only looked at solutions in the rings $\mathbb{Z}/p^n \mathbb{Z}$, where $p$ is some prime.  This time, the inverse limit is $\mathbb{Z}_p$, known as the $p$-adic integers, which is the ring of integers of the field $\mathbb{Q}_p$ of $p$-adic numbers.  Despite the appearance of the prime $p$ here, this field actually has characteristic zero, and contains a copy of the rational numbers.  It turns out that $\hat{\mathbb{Z}}$ is isomorphic as a ring to the (unrestricted) Cartesian product of the rings $\mathbb{Z}_p$, as $p$ ranges over all primes.  So finding solutions in $\hat{\mathbb{Z}}$ is equivalent to finding them separately in $\mathbb{Z}_p$ for each prime $p$.

[EXAMPLE]

(This is an example of a popular trick in group theory, 'try it for the free group'.  This could perhaps be spun off to its own article.)

Let $G$ be a group generated by $d$ elements, and let $H$ be a subgroup of index $n$.  How many elements could it take to generate $H$?

At first sight, this seems an unreasonable question, as for $d>1$ the family of $d$-generated groups is rather diverse, and apparently has little structure in common.  For instance, every finite simple group is generated by $2$ elements.  Moreover, $G$ could be infinite; in this case, how do we even know that $H$ is finitely generated?  We don't want to be considering such a wide range of different examples.

However, we have a group that is in some sense universal for $d$-generator groups: this is the free group on $d$ generators.  Let's call our set of generators $X = \{ x_1, \dots, x_d \}$; we also need the symbols $X^{-1} = \{ x^{-1}_1, \dots, x^{-1}_d \}$.  Then the free group $F(X)$ on $X$ consists of all equivalence classes of strings of symbols in $X \cup X^{-1}$, with the rule that we can 'cancel out' $x_i$ and $x^{-1}_i$ if we see them next to each other.  Multiplication is given by concatenating the strings.

Now $F(X)$ is universal in the following sense: given any function from the set $X$ to a group $G$, this function extends to a homomorphism from $F(X)$ to $G$.  In particular, if the function takes $X$ to a generating set for $G$, we obtain a surjective homomorphism $\phi$ from $F(X)$ to $G$.  With a little more thought, it is clear how to reduce our original question to the special case of the free group: if $H$ is a subgroup of $G$ of index $n$, then its preimage $\phi^{-1}(H)$ must have index $n$ in $F(X)$.  Now if $Y$ is a generating set for $\phi^{-1}(H)$, then $\phi(Y)$ generates $H$, and evidently $|\phi(Y)| \le |Y|$.  So we can rephrase our question as follows:

Let $G$ be a *free* group generated by $d$ elements, and let $H$ be a subgroup of index $n$.  How many elements could it take to generate $H$?

The answer turns out to be surprisingly simple (although the proof is not obvious, and involves tricks beyond the scope of this article; one possible approach is [[Use topology to study your group | via topology]]).  The number of generators needed for $H$ in this case is determined entirely by $n$ and $d$, and is always precisely $n(d-1) +1$.  This is known as the Schreier index formula.  As an immediate consequence, we get the following answer to our original question:

Let $G$ be a group generated by $d$ elements, and let $H$ be a subgroup of index $n$.  Then the number of elements needed to generate $H$ is at most $n(d-1) +1$.  The free groups show that this bound cannot be improved.

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