# nLab localization of abelian groups

Contents

group theory

### Cohomology and Extensions

#### Homological algebra

homological algebra

Introduction

diagram chasing

# Contents

## Idea

The general concept of localization applied to the (derived) category of abelian groups yield the concept of localization of abelian groups.

The two main examples are

1. classical localization at/aways from primes;

2. completion at a prime

at prime numbers $p$. Here “classical $p$-localization” is localization at the morphism $0 \to \mathbb{Z}/p\mathbb{Z}$, while $p$-completion is localization at the morphism $0 \to \mathbb{Z}[p^{-1}]$.

## Definition

###### Remark

Recall that Ext-groups $\Ext^\bullet(A,B)$ between abelian groups $A, B \in$ Ab are concentrated in degrees 0 and 1 (prop.). Since

$Ext^0(A,B) \simeq Hom(A,B)$

is the plain hom-functor, this means that there is only one possibly non-vanishing Ext-group $Ext^1$, therefore often abbreviated to just “$Ext$”:

$Ext(A,B) \coloneqq Ext^1(A,B) \,.$
###### Definition

Let $K$ be an abelian group.

Then an abelian group $A$ is called $K$-local if all the Ext-groups from $K$ to $A$ vanish:

$Ext^\bullet(K,A) \simeq 0$

hence equivalently (remark ) if

$Hom(K,A) \simeq 0 \;\;\;\;\; and \;\;\;\;\; Ext(K,A) \simeq 0 \,.$

A homomorphism of abelian groups $f \colon B \longrightarrow C$ is called $K$-local if for all $K$-local groups $A$ the function

$Hom(f,A) \;\colon\; Hom(B,A) \longrightarrow Hom(A,A)$

is a bijection.

(Beware that here it would seem more natural to use $Ext^\bullet$ instead of $Hom$, but we do use $Hom$. See (Neisendorfer 08, remark 3.2).

A homomorphism of abelian groups

$\eta \;\colon\; A \longrightarrow L_K A$

is called a $K$-localization if

1. $L_K A$ is $K$-local;

2. $\eta$ is a $K$-local morphism.

We now discuss two classes of examples of localization of abelian groups

###### Classical localization at/away from primes

For $n \in \mathbb{N}$, write $\mathbb{Z}/n\mathbb{Z}$ for the cyclic group of order $n$.

###### Lemma

For $n \in \mathbb{N}$ and $A \in Ab$ any abelian group, then

1. the hom-group out of $\mathbb{Z}/n\mathbb{Z}$ into $A$ is the $n$-torsion subgroup of $A$

$Hom(\mathbb{Z}/n\mathbb{Z}, A) \simeq \{ a \in A \;\vert\; p \cdot a = 0 \}$
2. the first Ext-group out of $\mathbb{Z}/n\mathbb{Z}$ into $A$ is

$Ext^1(\mathbb{Z}/n\mathbb{Z},A) \simeq A/n A \,.$
###### Proof

Regarding the first item: Since $\mathbb{Z}/p\mathbb{Z}$ is generated by its element 1 a morphism $\mathbb{Z}/p\mathbb{Z} \to A$ is fixed by the image $a$ of this element, and the only relation on 1 in $\mathbb{Z}/p\mathbb{Z}$ is that $p \cdot 1 = 0$.

Regarding the second item:

Consider the canonical free resolution

$0 \to \mathbb{Z} \overset{p \cdot (-)}{\longrightarrow} \mathbb{Z} \longrightarrow \mathbb{Z}/p\mathbb{Z} \to 0 \,.$

By the general discusson of derived functors in homological algebra this exhibits the Ext-group in degree 1 as part of the following short exact sequence

$0 \to Hom(\mathbb{Z},A) \overset{Hom(n\cdot(-),A)}{\longrightarrow} Hom(\mathbb{Z}, A) \longrightarrow Ext^1(\mathbb{Z}/n\mathbb{Z},A) \to 0 \,,$

where the morphism on the left is equivalently $A \overset{n \cdot (-)}{\to} A$.

###### Example

An abelian group $A$ is $\mathbb{Z}/p\mathbb{Z}$-local precisely if the operation

$p \cdot (-) \;\colon\; A \longrightarrow A$

of multiplication by $p$ is an isomorphism, hence if “$p$ is invertible in $A$”.

###### Proof

By the first item of lemma we have

$Hom(\mathbb{Z}/p\mathbb{Z}, A) \simeq \{ a \in A \;\vert\; p \cdot a = 0 \}$

By the second item of lemma we have

$Ext^1(\mathbb{Z}/p\mathbb{Z},A) \simeq A/p A \,.$

Hence by def. $A$ is $\mathbb{Z}/p\mathbb{Z}$-local precisely if

$\{ a \in A \;\vert\; p \cdot a = 0 \} \simeq 0$

and if

$A / p A \simeq 0 \,.$

Both these conditions are equivalent to multiplication by $p$ being invertible.

###### Definition

For $J \subset \mathbb{N}$ a set of prime numbers, consider the direct sum $\underset{p \in J}{\oplus} \mathbb{Z}/p\mathbb{Z}$ of cyclic groups of order $p$.

The operation of $\underset{p \in J}{\oplus} \mathbb{Z}/p\mathbb{Z}$-localization of abelian groups according to def. is called inverting the primes in $J$.

Specifically

1. for $J = \{p\}$ a single prime then $\mathbb{Z}/p\mathbb{Z}$-localization is called localization away from $p$;

2. for $J$ the set of all primes except $p$ then $\underset{p \in J}{\oplus} \mathbb{Z}/p\mathbb{Z}$-localization is called localization at $p$;

3. for $J$ the set of all primes, then $\underset{p \in J}{\oplus} \mathbb{Z}/p\mathbb{Z}$-localizaton is called rationalization..

###### Definition

For $J \subset Primes \subset \mathbb{N}$ a set of prime numbers, then

$\mathbb{Z}[J^{-1}] \hookrightarrow \mathbb{Q}$

denotes the subgroup of the rational numbers on those elements which have an expression as a fraction of natural numbers with denominator a product of elements in $J$.

This is the abelian group underlying the localization of a commutative ring of the ring of integers at the set $J$ of primes.

If $J = Primes - \{p\}$ is the set of all primes except $p$ one also writes

$\mathbb{Z}_{(p)} \coloneqq \mathbb{Z}[Primes - \{p\}] \,.$

Notice the parenthesis, to distinguish from the notation $\mathbb{Z}_{p}$ for the p-adic integers (def. below).

###### Remark

The terminology in def. is motivated by the following perspective of arithmetic geometry:

Generally for $R$ a commutative ring, then an $R$-module is equivalently a quasicoherent sheaf on the spectrum of the ring $Spec(R)$.

In the present case $R = \mathbb{Z}$ is the integers and abelian groups are identified with $\mathbb{Z}$-modules. Hence we may think of an abelian group $A$ equivalently as a quasicoherent sheaf on Spec(Z).

The “ring of functions” on Spec(Z) is the integers, and a point in $Spec(\mathbb{Z})$ is labeled by a prime number $p$, thought of as generating the ideal of functions on Spec(Z) which vanish at that point. The residue field at that point is $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$.

Inverting a prime means forcing $p$ to become invertible, which, by this characterization, it is precisely away from that point which it labels. The localization of an abelian group at $\mathbb{Z}/p\mathbb{Z}$ hence corresponds to the restriction of the corresponding quasicoherent sheaf over $Spec(\mathbb{Z})$ to the complement of the point labeled by $p$.

Similarly localization at $p$ is localization away from all points except $p$.

###### Proposition

For $J \subset \mathbb{N}$ a set of prime numbers, a homomorphism of abelian groups $f \;\colon\; A \lookrightarrow B$ is local (def. ) with respect to $\underset{p \in J}{\oplus} \mathbb{Z}/p\mathbb{Z}$ (def. ) if under tensor product of abelian groups with $\mathbb{Z}[J^{-1}]$ (def. ) it becomes an isomorphism

$f \otimes \mathbb{Z}[J^{-1}] \;\colon\; X \otimes \mathbb{Z}[J^{-1}] \overset{\simeq}{\longrightarrow} Y \otimes \mathbb{Z}[J^{-1}] \,.$

Moreover, for $A$ any abelian group then its $\underset{p \in J}{\oplus} \mathbb{Z}/p\mathbb{Z}$-localization exists and is given by the canonical projection morphism

$A \longrightarrow A \otimes \mathbb{Z}[J^{-1}] \,.$
###### Formal completion at primes

We have seen above in remark that classical localization of abelian groups at a prime number is geometrically interpreted as restricting a quasicoherent sheaf over Spec(Z) to a single point, the point that is labeled by that prime number.

Alternatively one may restrict to the “infinitesimal neighbourhood” of such a point. Technically this is called the formal neighbourhood, because its ring of functions is, by definition, the ring of formal power series (regarded as an adic ring or pro-ring). The corresponding operation on abelian groups is accordingly called formal completion or adic completion or just completion, for short, of abelian groups.

It turns out that if the abelian group is finitely generated then the operation of p-completion coincides with an operation of localization in the sense of def. , namely localization at the p-primary component $\mathbb{Z}(p^\infty)$ of the group $\mathbb{Q}/\mathbb{Z}$ (def. below). On the one hand this equivalence is useful for deducing some key properties of p-completion, this we discuss below. On the other hand this situation is a shadow of the relation between localization of spectra and nilpotent completion of spectra, which is important for characterizing the convergence properties of Adams spectral sequences.

###### Definition

For $p$ a prime number, then the p-adic completion of an abelian group $A$ is the abelian group given by the limit

$A^\wedge_p \coloneqq \underset{\longleftarrow}{\lim} \left( \cdots \longrightarrow A / p^3 A \longrightarrow A / p^2 A \longrightarrow A/p A \right) \,,$

where the morphisms are the evident quotient morphisms. With these understood we often write

$A^\wedge_p \coloneqq \underset{\longleftarrow}{\lim}_n A/p^n A$

for short. Notice that here the indexing starts at $n = 1$.

###### Example

The p-adic completion (def. ) of the integers $\mathbb{Z}$ is called the p-adic integers, often written

$\mathbb{Z}_p \coloneqq \mathbb{Z}^\wedge_p \coloneqq \underset{\longleftarrow}{\lim}_n \mathbb{Z}/p^n \mathbb{Z} \,,$

which is the original example that gives the general concept its name.

With respect to the canonical ring-structure on the integers, of course $p \mathbb{Z}$ is a prime ideal.

Compare this to the ring $\mathcal{O}_{\mathbb{C}}$ of holomorphic functions on the complex plane. For $x \in \mathbb{C}$ any point, it contains the prime ideal generated by $(z-x)$ (for $z$ the canonical coordinate function on $\mathbb{z}$). The formal power series ring $\mathbb{C}[ [(z.x)] ]$ is the adic completion of $\mathcal{O}_{\mathbb{C}}$ at this ideal. It has the interpretation of functions defined on a formal neighbourhood of $X$ in $\mathbb{C}$.

Analogously, the p-adic integers $\mathbb{Z}_p$ may be thought of as the functions defined on a formal neighbourhood of the point labeled by $p$ in Spec(Z).

###### Lemma

There is a short exact sequence

$0 \to \mathbb{Z}_p \overset{p \cdot (-)}{\longrightarrow} \mathbb{Z}_p \longrightarrow \mathbb{Z}/p\mathbb{Z} \to 0 \,.$
###### Proof

Consider the following commuting diagram

$\array{ \vdots && \vdots && \vdots \\ \downarrow && \downarrow && \downarrow \\ \mathbb{Z}/p^3\mathbb{Z} &\overset{p\cdot (-)}{\longrightarrow}& \mathbb{Z}/p^4 \mathbb{Z} &\longrightarrow& \mathbb{Z}/p\mathbb{Z} \\ \downarrow && \downarrow && \downarrow \\ \mathbb{Z}/p^2\mathbb{Z} &\overset{p\cdot (-)}{\longrightarrow}& \mathbb{Z}/p^3 \mathbb{Z} &\longrightarrow& \mathbb{Z}/p\mathbb{Z} \\ \downarrow && \downarrow && \downarrow \\ \mathbb{Z}/p\mathbb{Z} &\overset{p\cdot (-)}{\longrightarrow}& \mathbb{Z}/p^2 \mathbb{Z} &\longrightarrow& \mathbb{Z}/p\mathbb{Z} \\ \downarrow && \downarrow && \downarrow \\ 0 &\longrightarrow& \mathbb{Z}/p\mathbb{Z} &\longrightarrow& \mathbb{Z}/p\mathbb{Z} } \,.$

Each horizontal sequence is exact. Taking the limit over the vertical sequences yields the sequence in question. Since limits commute over limits, the result follows.

We now consider a concept of $p$-completion that is in general different from def. , but turns out to coincide with it in finitely generated abelian groups.

###### Definition

For $p$ a prime number, write

$\mathbb{Z}[1/p] \coloneqq \underset{\longrightarrow}{\lim} \left( \mathbb{Z} \overset{p \cdot (-)}{\longrightarrow} \mathbb{Z} \overset{p \cdot (-)}{\longrightarrow} \mathbb{Z} \overset{}{\longrightarrow} \cdots \right)$

for the colimit (in Ab) over iterative applications of multiplication by $p$ on the integers.

This is the abelian group generated by formal expressions $\frac{1}{p^k}$ for $k \in \mathbb{N}$, subject to the relations

$(p \cdot n) \frac{1}{p^{k+1}} = n \frac{1}{p^k} \,.$

Equivalently it is the abelian group underlying the ring localization $\mathbb{Z}[1/p]$.

###### Definition

For $p$ a prime number, then localization of abelian groups (def. ) at $\mathbb{Z}[1/p]$ (def. ) is called $p$-completion of abelian groups.

###### Lemma

An abelian group $A$ is $p$-complete according to def. precisely if both the limit as well as the lim^1 of the sequence

$\cdots \overset{}{\longrightarrow} A \overset{p}{\longrightarrow} A \overset{p}{\longrightarrow} A \overset{p}{\longrightarrow} A$

vanishes:

$\underset{\longleftarrow}{\lim} \left( \cdots \overset{}{\longrightarrow} A \overset{p}{\longrightarrow} A \overset{p}{\longrightarrow} A \overset{p}{\longrightarrow} A \right) \simeq 0$

and

$\underset{\longleftarrow}{\lim}^1 \left( \cdots \overset{}{\longrightarrow} A \overset{p}{\longrightarrow} A \overset{p}{\longrightarrow} A \overset{p}{\longrightarrow} A \right) \simeq 0 \,.$
###### Proof

By def. the group $A$ is $\mathbb{Z}[1/p]$-local precisely if

$Hom(\mathbb{Z}[1/p], A) \simeq 0 \;\;\;\;\;\;\; and \;\;\;\;\;\;\; Ext^1(\mathbb{Z}[1/p], A) \simeq 0 \,.$

Now use the colimit definition $\mathbb{Z}[1/p] = \underset{\longrightarrow}{\lim}_n \mathbb{Z}$ (def. ) and the fact that the hom-functor sends colimits in the first argument to limits to deduce that

\begin{aligned} Hom(\mathbb{Z}[1/p], A) & = Hom( \underset{\longrightarrow}{\lim}_n \mathbb{Z}, A ) \\ & \simeq \underset{\longleftarrow}{\lim}_n Hom(\mathbb{Z},A) \\ & \simeq \underset{\longleftarrow}{\lim}_n A \end{aligned} \,.

This yields the first statement. For the second, use that for every cotower over abelian groups $B_\bullet$ there is a short exact sequence of the form

$0 \to \underset{\longleftarrow}{\lim}^1_n Hom(B_n, A) \longrightarrow Ext^1( \underset{\longrightarrow}{\lim}_n B_n, A ) \longrightarrow \underset{\longleftarrow}{\lim}_n Ext^1( B_n, A) \to 0$

(by this lemma).

In the present case all $B_n \simeq \mathbb{Z}$, which is a free abelian group, hence a projective object, so that all the Ext-groups out of it vannish. Hence by exactness there is an isomorphism

$Ext^1( \underset{\longrightarrow}{\lim}_n \mathbb{Z}, A ) \simeq \underset{\longleftarrow}{\lim}^1_n Hom(\mathbb{Z}, A) \simeq \underset{\longleftarrow}{\lim}^1_n A \,.$

This gives the second statement.

###### Example

For $p$ a prime number, the p-primary cyclic groups of the form $\mathbb{Z}/p^n \mathbb{Z}$ are $p$-complete (def. ).

###### Proof

By lemma we need to check that

$\underset{\longleftarrow}{\lim} \left( \cdots \overset{p}{\longrightarrow} \mathbb{Z}/p^n \mathbb{Z} \overset{p}{\longrightarrow} \mathbb{Z}/p^n \mathbb{Z} \overset{p}{\longrightarrow} \mathbb{Z}/p^n \mathbb{Z} \right) \simeq 0$

and

$\underset{\longleftarrow}{\lim}^1 \left( \cdots \overset{p}{\longrightarrow} \mathbb{Z}/p^n \mathbb{Z} \overset{p}{\longrightarrow} \mathbb{Z}/p^n \mathbb{Z} \overset{p}{\longrightarrow} \mathbb{Z}/p^n \mathbb{Z} \right) \simeq 0 \,.$

For the first statement observe that $n$ consecutive stages of the tower compose to the zero morphism. First of all this directly implies that the limit vanishes, secondly it means that the tower satisfies the Mittag-Leffler condition (def.) and this implies that the $\lim^1$ also vanishes (prop.).

###### Definition

For $p$ a prime number, write

$\mathbb{Z}(p^\infty) \coloneqq \mathbb{Z}[1/p]/\mathbb{Z}$

(the p-primary part of $\mathbb{Q}/\mathbb{Z}$), where $\mathbb{Z}[1/p] = \underset{\longrightarrow}{\lim}(\mathbb{Z}\overset{p}{\to} \mathbb{Z} \overset{p}{\to} \mathbb{Z} \to \cdots )$ from def. .

Since colimits commute over each other, this is equivalently

$\mathbb{Z}(p^\infty) \simeq \underset{\longrightarrow}{\lim} ( 0 \hookrightarrow \mathbb{Z}/p\mathbb{Z} \hookrightarrow \mathbb{Z}/p^2 \mathbb{Z} \hookrightarrow \cdots ) \,.$
###### Theorem

For $p$ a prime number, the $\mathbb{Z}[1/p]$-localization

$A \longrightarrow L_{\mathbb{Z}[1/p]} A$

of an abelian group $A$ (def. , def. ), hence the $p$-completion of $A$ according to def. , is given by the morphism

$A \longrightarrow Ext^1( \mathbb{Z}(p^\infty), A )$

into the first Ext-group into $A$ out of $\mathbb{Z}(p^\infty)$ (def. ), which appears as the first connecting homomorphism $\delta$ in the long exact sequence of Ext-groups

$0 \to Hom(\mathbb{Z}(p^\infty),A) \longrightarrow Hom(\mathbb{Z}[1/p],A) \longrightarrow Hom(\mathbb{Z},A) \overset{\delta)}{\longrightarrow} Ext^1(\mathbb{Z}(p^\infty), A) \to \cdots \,.$

that is induced (via this prop.) from the defining short exact sequence

$0 \to \mathbb{Z} \longrightarrow \mathbb{Z}[1/p] \longrightarrow \mathbb{Z}(p^\infty) \to 0$

(def. ).

e.g. (Neisendorfer 08, p. 16)

###### Proposition

If $A$ is a finitely generated abelian group, then its $p$-completion (def. , for any prime number $p$) is equivalently its p-adic completion (def. )

$\mathbb{Z}[1/p] A \simeq A^\wedge_p \,.$
###### Proof

By theorem the $p$-completion is $Ext^1(\mathbb{Z}(p^\infty),A)$. By def. there is a colimit

$\mathbb{Z}(p^\infty) = \underset{\longrightarrow}{\lim} \left( \mathbb{Z}/p\mathbb{Z} \to \mathbb{Z}/p^2 \mathbb{Z} \to \mathbb{Z}/p^3 \mathbb{Z} \to \cdots \right) \,.$

Together this implies, by this lemma, that there is a short exact sequence of the form

$0 \to \underset{\longleftarrow}{\lim}^1 Hom(\mathbb{Z}/p^n \mathbb{Z},A) \longrightarrow X^\wedge_p \longrightarrow \underset{\longleftarrow}{\lim}_n Ext^1(\mathbb{Z}/p^n \mathbb{Z}, A) \to 0 \,.$

By lemma the lim^1 on the left is over the $p^n$-torsion subgroups of $A$, as $n$ ranges. By the assumption that $A$ is finitely generated, there is a maximum $n$ such that all torsion elements in $A$ are annihilated by $p^n$. This means that the Mittag-Leffler condition (def.) is satisfied by the tower of $p$-torsion subgroups, and hence the lim^1-term vanishes (prop.).

Therefore by exactness of the above sequence there is an isomorphism

\begin{aligned} L_{\mathbb{Z}[1/p]}X & \simeq \underset{\longleftarrow}{\lim}_n Ext^1(\mathbb{Z}/p^n \mathbb{Z}, A) \\ & \simeq \underset{\longleftarrow}{\lim}_n A/p^n A \end{aligned} \,,

where the second isomorphism is by lemma .

###### Proposition

If $A$ is a $p$-divisible group in that $A \overset{p \cdot (-)}{\longrightarrow} A$ is an isomorphism, then its $p$-completion (def. ) vanishes.

###### Proof

By theorem the localization morphism $\delta$ sits in a long exact sequence of the form

$0 \to Hom(\mathbb{Z}(p^\infty),A) \longrightarrow Hom(\mathbb{Z}[1/p],A) \overset{\phi}{\longrightarrow} Hom(\mathbb{Z},A) \overset{\delta}{\longrightarrow} Ext^1(\mathbb{Z}(p^\infty), A) \to \cdots \,.$

Here by def. and since the hom-functor takes colimits in the first argument to limits, the second term is equivalently the limit

$Hom(\mathbb{Z}[1/p],A) \simeq \underset{\longleftarrow}{\lim} \left( \cdots \to A \overset{p \cdot (-)}{\longrightarrow} A \overset{p \cdot (-)}{\longrightarrow} A \right) \,.$

But by assumption all these morphisms $p \cdot (-)$ that the limit is over are isomorphisms, so that the limit collapses to its first term, which means that the morphism $\phi$ in the above sequence is an isomorphism. But by exactness of the sequence this means that $\delta = 0$.

###### Corollary

Let $p$ be a prime number. If $A$ is a finite abelian group, then its $p$-completion (def. ) is equivalently its p-primary part.

###### Proof

By the fundamental theorem of finite abelian groups, $A$ is a finite direct sum

$A \simeq \underset{i}{\oplus} \mathbb{Z}/p_i^{k_i}\mathbb{Z}$

of cyclic groups of ordr $p_i^{k_1}$ for $p_i$ prime numbers and $k_i \in \mathbb{N}$ (thm.).

Since finite direct sums are equivalently products (biproducts: Ab is an additive category) this means that

$Ext^1( \mathbb{Z}(p^\infty), A ) \simeq \underset{i}{\prod} Ext^1( \mathbb{Z}(p^\infty), \mathbb{Z}/p_i^{k_1}\mathbb{Z} ) \,.$

By theorem the $i$th factor here is the $p$-completion of $\mathbb{Z}/p_i^{k_i}\mathbb{Z}$, and since $p \cdot(-)$ is an isomorphism on $\mathbb{Z}/p_i^{k_i}\mathbb{Z}$ if $p_i \neq p$ (because its kernel evidently vanishes), prop. says that in this case the factor vanishes, so that only the factors with $p_i = p$ remain. On these however $Ext^1(\mathbb{Z}(p^\infty),-)$ is the identity by example .

## References

• Joseph NeisendorferA Quick Trip through Localization, in Alpine perspectives on algebraic topology, Third Ariolla Conference 2008 (pdf)

• Peter May, Kate Ponto, section 10.1 of More concise algebraic topology: Localization, completion, and model categories (pdf)

Last revised on March 28, 2017 at 04:40:18. See the history of this page for a list of all contributions to it.