Ranks, Tensors and Abelian Groups

2016-11-09 ( 2020-11-20)

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Abelian Groups

Given $a\in A$, the homomirphism $i_ a \from \ZZ \to A$ defined by $i_ a(k) = ka$ is mono (injective) iff $a$ has infinite order. Of $S$ is a set, the free Abelian group generated by $S$ is $$ \ZZ[S] = \ZZ S = \bigoplus_ {s \in S} \ZZ, $$ and it is the group of all finite linear combinations (with integer coefficients) of elements in $S$. The elements of $S$ are a basis (base) of $\ZZ[S]$. If $A$ is an Abelian group, and $S\subset A$ is a set of elements such that the homomorphism $$\bigoplus_ {s\in S} i_ s\from \bigoplus_ {s \in S} \ZZ \to A$$ is an isomorphism, then $A$ is a free group generated by the elements in $S$ (its basis). More on abelian groups in the monograph of Fuchs (2015)¹.

Universal property: If $S\subset A$ is a basis, $B$ is any Abelian group, for every function $\varphi \from S \to B$ there exists a unique group homomorphism $f\from A \to B$ such that $f(s) = \varphi(s),$ for all $s\in S.$ In other words, homomorphisms from a free group are completely and arbitrarily determined by their values on the elements of a basis.

(a)

Every subgroup of a free abelian group is free.

Proof: This theorem is credited to Richard Dedekind in Johnson - 1980 (EBSCO LINK), page 8. For a nice introduction to Abelian groups, and the proof of (a) which I took as source, see Kurosh², §19. A more general approach: Lang³, Chapter 1, §7 and page 41. We give here a proof for a countable basis. Let $a_ 1, a_ 2, \ldots, a_ n, \ldots$ be elements of a basis $S$ of a free group $F$. Hence $F$ is isomorphic to the group of all functions $S\to \ZZ$ with finite support. With an abuse of notation, assume $F$ is the group of such functions. $\newcommand{\supp}{\operatorname{supp}}$ For each $f\in F$, $\supp(f) = \{ s \in S : f(s) \neq 0 \}$ denotes the support of $f$. Being finite, for each $f\in F$, the maximum $\max \supp(f)$ exists whenever $f\neq 0$. It is the last index of $f$. The corresponding coefficient $f(s)$ is the last coefficient of $f$.

Let $A\subset F$ be a fixed subgroup. Let $b_ 1$ be the non-zero element in $A$ with smallest last index, say $s_ 1$, and smallest positive last coefficient. Then, $b_ 1 \in A$, and if $x\in A$, $x\neq 0$,then either $\max \supp(x) > \max \supp(b_ 1)=s_ 1$ or $\max \supp(x) = \max \supp(b_ 1 = s_ 1)$. Let $A_ 1 = \{ a \in A : \max \supp(a) = s_ 1 \} $. If $a\in A_ 1$, then by definition $\max \supp(a) = s_ 1$, then $|a(s_ 1)| \geq b_ 1(s_ 1)$, and hence there are integers $q$ and $r$ such that $0\leq r \lt b_ 1(s_ 1)$ and $|a(s_ 1)| = q b_ 1(s_ 1) + r $. But if $r\neq 0$ then $q b_ 1 - a$ has last coefficient equal to $r$, which is smaller than $b_ 1(s_ 1)$, so it must be $r=0$. It therefore exists $q$ such that $a(s_ 1) = q b_ 1(s_ 1)$. Now, if $q b_ 1 - a$ were not $0$ in $A$, its last index would be smaller than $s_ 1$, which would be a contradiction, and hence there exists $q\in \ZZ$ such that $a= q b_ 1$. In other words, $A_ 1 = \langle b_ 1 \rangle = \ZZ[b_ 1]$.

For the next step, define as $b_ 2$, the element in $A \smallsetminus A_ 1$ with smallest last index (say, $s_ 2$) and smallest nonzero positive last coefficient. Since $b_ 2 \not\in A_ 1$, $s_ 1 \lt s_ 2$. Let $A_ 2 = \{ a \in A : \max \supp(a) \leq s_ 2 \} $. Note that $A_ 1 \cap \langle b_ 2 \rangle = 0$, and therefore it is a direct sum $A_ 1 \oplus \langle b_ 2 \rangle \subset A_ 2$. Moreover, if $a \in A_ 2\smallsetminus A_ 1$, then $\max \supp (a) = s_ 2$, and as before there exist $q \in \ZZ$ such that $a (s_ 2) = q b_ 2(s_ 2)$ (otherwise $b_ 2(s_ 2)$ is not the smallest positive last coefficient), which means that $q b_ 2 - a \in A_ 1=\langle b_ 1\rangle $, or that $a \in \langle b_ 1 \rangle \oplus \langle b_ 2 \rangle$. Thus, $A_ 2 = \ZZ[b_ 2] \oplus A_ 1$.

By induction, we can continue with a sequence of subgroups $$ A_ 1 \subset A_ 2 \subset \ldots \subset A_ n \subset \ldots \subset A $$ such that $A_ n$ is the free group generated by the elements $b_ 1, \ldots, b_ n, \ldots$ (which are a basis for $A_ n$). This sequence either stops, when $A_ n = A$, and hence $A$ is a free group of rank $n$, or it never stops, and $A$ is the free group generated by the set of all $b_ j$ (each element $a \in A$ has a last index; there must be an $s_ j$ such that $s_ j$ is greater than this last index, and therefore $a\in A_ j$). /qed/

(b)

Every subgroup of a finitely generated abelian group is finitely generate.

Proof: See Kurosh, §20, or Fuchs Theorem 2.5 pag. 82, or Theorem 2.6 pag. 83 (Stacked Basis Theorem). /qed/

From the previous propositions and the Smith Normal Form of a $\ZZ$-matrix, one can (constructively) prove that

(c)

A finitely generated Abelian group is the finite direct sum of cyclic subgroups $C_ j\subset A$, $$ A \cong \bigoplus_ {j=1}^l C_ j $$ with $C_ j$ cyclic of order $\nu_ j$ (or order infinity if $\nu_ j= 0$, i.e. $C_ j = \ZZ / \nu_ j \ZZ$, with $\nu_ j \geq 0$.

The terms with $\nu_ j\geq 1$ contribute to the torsion subgroup of $A$, while the terms with $\nu_ j = 0$ contribute to the torsion-free rank of $A$. In other words, $A= T \oplus \ZZ^r$, where $r$ is the rank of $A$. We will show some properties and alternative definitions of the rank (for example, if $A$ is not finitely generated).

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Tensor product of $\ZZ$-modules (AKA Abelian Groups)

Let $A$, $B$ be two Abelian groups. The tensor product $A\otimes_ \ZZ B = A \otimes B $ is defined as follows: let $\ZZ[A\times B]$ denote the free abelian group generated by elements in the cartesian product $A\times B$ (i.e., finite $\ZZ$-linear combinations of symbols in $A\times B$). Elements of $A\times B$ will be written as $a\times b$ instead of $(a,b)$. Then, consider in $\ZZ[A\times B]$ the subgroup $R$ generated by $$ \begin{aligned} (a_ 1 + a_ 2) \otimes b - (a_ 1 \otimes b + a_ 2 \otimes b) & \text{ with } a_ 1, a_ 2\in A, b \in B \\\ a \otimes (b_ 1 + b_ 2) - (a \otimes b_ 1 + a \otimes b_ 2) & \text{ with } a\in A, b_ 1, b_ 2 \in B \\\ a k \otimes b - a \otimes k b & \text{ with } a\in A, b\in B, k \in \ZZ
\end{aligned} $$ (note: the third line is not independent, actually, for $\ZZ$-modules) The tensor product is the quotient $$ A \otimes B = \dfrac{ \ZZ[A\times B] }{R}~. $$ The coset containing the element $(a,b) \sim a \otimes b$ will be denoted again as $a \otimes b$. It is an Abelian Group (a $\ZZ$-module), as quotient of a free abelian group and $R$.

The tensor product has the followin Universal Property: a function $f \from A \times B \to G$ is bilinear if $$ \begin{aligned} f (a_ 1 + a_ 2, b ) =f (a_ 1, b) + f(a_ 2, b) & \text{ with } a_ 1, a_ 2\in A, b \in B \\\ f( a , b_ 1 + b_ 2) = f( a , b_ 1) + f ( a , b_ 2) & \text{ with } a\in A, b_ 1, b_ 2 \in B \\\ f( a k , b) = f( a , k b) & \text{ with } a\in A, b\in B, k \in \ZZ
\end{aligned} $$ (note: the third line is not independent, actually, for $\ZZ$-modules) For evary abelian group $G$ and every bilinear map $f\from A\times B \to G$, there exists a unique homomorphism of abelian groups $F\from A\otimes B \to G$ such that $F(a\times b) = f(a,b)$.

(1)

$A\otimes B \cong B \otimes A$.

(2)

$\ZZ \otimes A \cong A \otimes \ZZ \cong A $.

Proof: The isomorphism is given by $k\otimes a \in \ZZ \otimes A \mapsto ka \in A$, with inverse $a \mapsto 1\otimes a$. /qed/

(3)

If $\alpha \from A \to A'$ is a homomorphism, then $\alpha_ *\from A \otimes B \to A' \otimes B$ defined by $\alpha_ *(a\otimes b) = (\alpha a) \otimes b$, for $a\in A$, $b\in B$, is a homomorphism. The same for $\beta\from B \to B'$, $\beta_ *\from A\otimes B \to A \otimes B'$ is the homomorphism defined as $\beta_ *(a\otimes b) = a \otimes (\beta b)$. Both $-\otimes B$ and $A \otimes -$ are covariant functors from the category of Abelian groups to the category of Abelian groups.

Given two morphisms $\alpha \from A \to A'$ and $\beta \from B \to B'$, the notation $\alpha \otimes \beta \from A\otimes B \to A' \otimes B'$ denotes the composition $\alpha_ * \beta_ *$.

(4)

The functor $A \otimes -$ is additive: there is a natural isomorphism $$ A \otimes
\left( \bigoplus_ {j\in J} B_ j \right) \cong A \otimes \bigoplus_ {j\in J} B_ j $$

Proof: Cfr. Proposition 7.3, page 110 of Hilton-Stammbach (1997)⁴ /qed/

(5)

The functor $A\otimes -$ is right-exact: if $B'\stackrel{\beta'}{\to} B \stackrel{\beta''}{\to} B'' \to 0$ is an exact sequence of abelian groups, then for any abelian group $A$ the sequence $$ \begin{CD} A \otimes B' @>{\beta'_ *}>> A \otimes B @>{\beta''_ *}>> A \otimes B'' @>{}>> 0 \end{CD} $$ is exact.

Proof: Cfr. Proposition 7.3, page 110 of Hilton-Stammbach (1997) /qed/

An abelian group (a $\ZZ$-module, and more generally an $R$-module) $A$ is called flat if the corresponding tensor functor $A \otimes -$ is not only right-exact, but it is exact: more precisely, for every short exact sequence $0 \to B'\stackrel{\beta'}{\to} B \stackrel{\beta''}{\to} B'' \to 0$ the sequence \begin{CD} 0 @>{}>> A \otimes B' @>{\beta'_ *}>> A \otimes B @>{\beta''_ *}>> A \otimes B'' @>{}>> 0 \end{CD}
is exact.

(6)

An abelian group $A$ is flat if and only if it is torsion-free. Hence $\QQ \otimes -$ is an exact functor.

Proof: Exercise 8.7 of Hilton-Stammbach, page 115. Note that torsion-free $\neq$ free (e.g. $\QQ$). /qed/

Rank of an abelian group

Let $A$ be an abelian group. Its rank is the maximal number of $\ZZ$-linearly independent elements in $A$. Recall that $n$ elements $a_ 1, a_ 2, \ldots, a_ n$ in $A$ are dependent if there are $n$ integers $k_ 1, k_ 2, \ldots, k_ n$, not all zero, such that $k_ 1 a_ 1 + k_ 2 a_ 2 + \ldots + k_ n a_ n = 0$. And they are ($\ZZ$-linearly) independent if they are not dependent. Each element of a set of independent elements in $A$ has infinite order. A group $A$ is free if it is generated by a set of independent elements (i.e., a basis).

Remark: This definition is actually the definition of $0$-rank, or torsion-free rank, for some authors (Robinson⁵, page 96).

Example: $\QQ$ has rank $1$, and it is not finitely generated.

Example: $\ZZ^r$ has rank $r$.

Example: $\QQ/\ZZ$ is equal to its torsion subgroup (see below), and has rank $0$ (and it is not finitely generated).

(7)

The rank of an Abelian group $A$ is the $\QQ$-dimension of the $\QQ$-vector space $\QQ \otimes A$: $$ \operatorname{rk} A = \dim_\QQ \QQ \otimes A $$

Proof: See the proof of 4.2.1, page 96, of Robinson. /qed/

Let $A,B,C$ be abelian groups, and $f\from A \to B$ a monomorphism (injective homomorphism) $g\from B \to C$ an epimorphism (surjective homomophism) such that $\operatorname{Im} f = \ker g$. In other words, the following sequence is exact. \begin{CD} 0 @>{}>> A @>{f}>> B @>{g}>> C @>{}>> 0 \ \end{CD}

(8)

If the previous sequence is exact, then $\operatorname{rk} B = \operatorname{rk}{A} + \operatorname{rk}{C}$

Proof: Since the functor $\QQ \otimes -$ is exact, as $\QQ$ is flat (torsion-free), the sequence \begin{CD} 0 @>{}>> \QQ \otimes A @>{f_ *}>> \QQ \otimes B @>{g_ *}>> \QQ \otimes C @>{}>> 0 \ \end{CD} is exact (and it is a sequence of $\QQ$-vector spaces). Therefore $\dim \QQ \otimes B = \dim \QQ \otimes A + \QQ \otimes C$, $\implies$ $\operatorname{rk} B = \operatorname{rk}{A} + \operatorname{rk}{C}$ /qed/

(9)

Let $T\subset A$ the torsion subgroup of $A$, consisting of all elements with finite order in $A$. The quotient $A/T$ is torsion-free, and $\operatorname{rk}{A} = \operatorname{rk}{A/T}$.

Proof: It is easy to see that $\operatorname{rk}{T} = 0$, and since the sequence $0 \to T \to A \to A/T \to 0 $ is exact, $\operatorname{rk} A = 0 + \operatorname{A/T}$. /qed/

(A)

The rank of a free Abelian group is the number of elements in any basis. The rank of an arbitrary Abelian group $A$ is the supremum of the rank of $F$, where $F$ ranges over all the free subgroups $F\subset A$.

Proof: A free abelian group is isomorphic to $\ZZ^r$ (if finitely generated) or $\ZZ^\infty$ (if not finitely generated), and the first part follows easily. Note that if $F\subset A$, then $\operatorname{rk} A = \operatorname{rk} F + \operatorname{rk} A/F$, and hence $\operatorname{rk}{F} \leq \operatorname{rk} A$.

Now, consider the torsion subgroup $T\subset A$. The quotient $T/A$ is torsion-free, and the ranks of the free subgroups of $A$ are the same as the ranks of the free subgroups of $T/A$ (in fact, the families of free subgroups are almost the same : why?). Now the result follows by the same fact on the dimension of $\QQ$-vector spaces. /qed/

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