Tensor Products of Z-Modules

Tensoring a Module with the Ring.

Theorem (Tensoring with the Ring). For any $\mathbb{Z}$-module $M$, there is a canonical isomorphism $M{\otimes }_{\mathbb{Z}}\mathbb{Z}\cong M$.

Proof. Define a map $\varphi :M\times \mathbb{Z}\to M$ by $\varphi (m,n)=nm$. This map is $\mathbb{Z}$-bilinear. By the universal property of the tensor product, $\varphi$ induces a unique $\mathbb{Z}$-module homomorphism $\Phi :M{\otimes }_{\mathbb{Z}}\mathbb{Z}\to M$ such that $\Phi (m\otimes n)=nm$. The map $\Psi :M\to M{\otimes }_{\mathbb{Z}}\mathbb{Z}$ defined by $\Psi (m)=m\otimes 1$ is a $\mathbb{Z}$-module homomorphism and is the inverse of $\Phi$. For any $m\in M$, $(\Phi \circ \Psi )(m)=\Phi (m\otimes 1)=1m=m$. For any simple tensor $m\otimes n$, $(\Psi \circ \Phi )(m\otimes n)=\Psi (nm)=nm\otimes 1=m\otimes n(1)=m\otimes n$. Thus, $\Phi$ is an isomorphism. $\square$

Tensoring with a Quotient Module.

Theorem (Tensor Product with a Quotient). For any $\mathbb{Z}$-module $M$ and any integer $n\ne 0$, there is a canonical isomorphism $M{\otimes }_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z})\cong M/nM$.

Proof. Consider the short exact sequence of $\mathbb{Z}$-modules:

$$0\to \mathbb{Z}\stackrel{\times n}{\to }\mathbb{Z}\to \mathbb{Z}/n\mathbb{Z}\to 0$$

Applying the right-exact functor $M{\otimes }_{\mathbb{Z}}-$ yields the exact sequence:

$$M{\otimes }_{\mathbb{Z}}\mathbb{Z}\stackrel{\mathrm{id}\otimes (\times n)}{\to }M{\otimes }_{\mathbb{Z}}\mathbb{Z}\to M{\otimes }_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z})\to 0$$

Using the isomorphism $M{\otimes }_{\mathbb{Z}}\mathbb{Z}\cong M$, this sequence becomes:

$$M\stackrel{\times n}{\to }M\to M{\otimes }_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z})\to 0$$

By exactness, the map $M\to M{\otimes }_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z})$ is surjective, and its kernel is the image of the preceding map, which is $nM$. The First Isomorphism Theorem implies $M{\otimes }_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z})\cong M/nM$. $\square$

Tensor Product of Cyclic Groups.

Theorem (Tensor Product of Cyclic Groups). For integers $m,n>0$, there is a canonical isomorphism $(\mathbb{Z}/m\mathbb{Z}){\otimes }_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z})\cong \mathbb{Z}/\gcd (m,n)\mathbb{Z}$.

Proof. Let $M=\mathbb{Z}/m\mathbb{Z}$ and apply the previous theorem. The submodule $n(\mathbb{Z}/m\mathbb{Z})$ is the ideal generated by $n$ in $\mathbb{Z}/m\mathbb{Z}$, which is $(n\mathbb{Z}+m\mathbb{Z})/m\mathbb{Z}$. By the Third Isomorphism Theorem, we have the following chain of isomorphisms:

$$\begin{array}{rl}(\mathbb{Z}/m\mathbb{Z}){\otimes }_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z})& \cong (\mathbb{Z}/m\mathbb{Z})/(n(\mathbb{Z}/m\mathbb{Z}))\\ & \cong (\mathbb{Z}/m\mathbb{Z})/((n\mathbb{Z}+m\mathbb{Z})/m\mathbb{Z})\\ & \cong \mathbb{Z}/(n\mathbb{Z}+m\mathbb{Z})\\ & \cong \mathbb{Z}/\gcd (m,n)\mathbb{Z}\end{array}$$

An immediate consequence is that if $\gcd (m,n)=1$, then $(\mathbb{Z}/m\mathbb{Z}){\otimes }_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z})\cong \{0\}$. $\square$

Torsion and Divisible Modules.

Definition (Divisible Module). A $\mathbb{Z}$-module $D$ is divisible if for any $d\in D$ and any non-zero integer $n$, there exists $d^{\prime }\in D$ such that $d=n{d}^{\prime }$.

Definition (Torsion Module). A $\mathbb{Z}$-module $T$ is a torsion module if for every $t\in T$, there exists a non-zero integer $n$ such that $nt=0$.

Theorem (Torsion Tensored with Divisible is Trivial). If $D$ is a divisible $\mathbb{Z}$-module and $T$ is a torsion $\mathbb{Z}$-module, then $D{\otimes }_{\mathbb{Z}}T\cong \{0\}$.

Proof. Let $d\otimes t$ be an arbitrary simple tensor in $D{\otimes }_{\mathbb{Z}}T$. Since $T$ is a torsion module, there exists a non-zero integer $n$ such that $nt=0$. Since $D$ is a divisible module, for this same integer $n$, there exists an element $d^{\prime }\in D$ such that $d=n{d}^{\prime }$.

$$d\otimes t=(n{d}^{\prime })\otimes t=d^{\prime }\otimes (nt)=d^{\prime }\otimes 0=0$$

Since every simple tensor is zero and these elements generate the module, the tensor product is the zero module. $\square$

Tensor Product of Rational Numbers.

Theorem (Tensor Square of Rationals). There is a $\mathbb{Z}$-module isomorphism $\mathbb{Q}{\otimes }_{\mathbb{Z}}\mathbb{Q}\cong \mathbb{Q}$.

Proof. Define the map $\varphi :\mathbb{Q}\times \mathbb{Q}\to \mathbb{Q}$ by $\varphi (x,y)=xy$. This map is $\mathbb{Z}$-bilinear, so it induces a unique $\mathbb{Z}$-module homomorphism $\Phi :\mathbb{Q}{\otimes }_{\mathbb{Z}}\mathbb{Q}\to \mathbb{Q}$ where $\Phi (x\otimes y)=xy$. This map is surjective, since for any $q\in \mathbb{Q}$, $q=\Phi (q\otimes 1)$. To show injectivity, let $z={\sum }_i(x_i\otimes y_i)$ be an arbitrary element of $\mathbb{Q}{\otimes }_{\mathbb{Z}}\mathbb{Q}$. Let $d$ be a common denominator for all $y_i$, such that $y_i=c_i/d$ for integers $c_i$. Then $z={\sum }_i(x_i\otimes c_i/d)={\sum }_i(x_i{c}_i\otimes 1/d)=({\sum }_i{x}_i{c}_i)\otimes 1/d$. This shows any element can be written as a simple tensor $q\otimes (1/d)$. If such an element is in the kernel of $\Phi$, then $\Phi (q\otimes (1/d))=q/d=0$, which implies $q=0$. Thus the element is $0\otimes (1/d)=0$. The kernel is trivial, so $\Phi$ is injective and therefore an isomorphism. $\square$

Tensor Products and Direct Sums.

Theorem (Tensor Products Distribute Over Direct Sums). Let $M$ be a $\mathbb{Z}$-module and $\{N_i{\}}_{i\in I}$ be a collection of $\mathbb{Z}$-modules. There is a canonical isomorphism $M{\otimes }_{\mathbb{Z}}({⨁}_{i\in I}{N}_i)\cong {⨁}_{i\in I}(M{\otimes }_{\mathbb{Z}}{N}_i)$.

This property allows complex modules to be decomposed for calculation. For example, to compute $(\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}){\otimes }_{\mathbb{Z}}(\mathbb{Q}\oplus \mathbb{Z}/3\mathbb{Z})$:

$$\begin{array}{rl}(\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}){\otimes }_{\mathbb{Z}}(\mathbb{Q}\oplus \mathbb{Z}/3\mathbb{Z})& \cong [(\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}){\otimes }_{\mathbb{Z}}\mathbb{Q}]\oplus [(\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}){\otimes }_{\mathbb{Z}}\mathbb{Z}/3\mathbb{Z}]\\ & \cong [\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Q}]\oplus [\mathbb{Z}/2\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Q}]\oplus [\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Z}/3\mathbb{Z}]\oplus [\mathbb{Z}/2\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Z}/3\mathbb{Z}]\\ & \cong \mathbb{Q}\oplus \{0\}\oplus \mathbb{Z}/3\mathbb{Z}\oplus \{0\}\\ & \cong \mathbb{Q}\oplus \mathbb{Z}/3\mathbb{Z}\end{array}$$

Tensor Products and Flat Modules.

Definition (Flat Module). A $\mathbb{Z}$-module $F$ is flat if the functor $-{\otimes }_{\mathbb{Z}}F$ is exact. This means that for any short exact sequence of $\mathbb{Z}$-modules $0\to A\stackrel{f}{\to }B\stackrel{g}{\to }C\to 0$, the resulting sequence $0\to A{\otimes }_{\mathbb{Z}}F\stackrel{f\otimes \mathrm{id}}{\to }B{\otimes }_{\mathbb{Z}}F\stackrel{g\otimes \mathrm{id}}{\to }C{\otimes }_{\mathbb{Z}}F\to 0$ is also exact. The critical property is that $f\otimes \mathrm{id}$ remains injective.

The module $\mathbb{Q}$ is a flat $\mathbb{Z}$-module. Tensoring the sequence $0\to \mathbb{Z}\stackrel{\times n}{\to }\mathbb{Z}\to \mathbb{Z}/n\mathbb{Z}\to 0$ with $\mathbb{Q}$ yields:

$$0\to \mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Q}\stackrel{(\times n)\otimes \mathrm{id}}{\to }\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Q}\to (\mathbb{Z}/n\mathbb{Z}){\otimes }_{\mathbb{Z}}\mathbb{Q}\to 0$$

which is isomorphic to the sequence $0\to \mathbb{Q}\stackrel{\times n}{\to }\mathbb{Q}\to \{0\}\to 0$. This sequence is exact, as multiplication by a non-zero integer $n$ is an automorphism of $\mathbb{Q}$, hence it is injective.

Conversely, $\mathbb{Z}/n\mathbb{Z}$ is not a flat $\mathbb{Z}$-module. Tensoring $0\to \mathbb{Z}\stackrel{\times n}{\to }\mathbb{Z}$ with $\mathbb{Z}/n\mathbb{Z}$ yields the map $(\times n)\otimes \mathrm{id}:\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Z}/n\mathbb{Z}$, which is isomorphic to the map $\mathbb{Z}/n\mathbb{Z}\stackrel{\times n}{\to }\mathbb{Z}/n\mathbb{Z}$. This map is the zero map, which is not injective, whereas the original map $\mathbb{Z}\stackrel{\times n}{\to }\mathbb{Z}$ was injective.

Localization as a Tensor Product.

Theorem (Localization as Tensor Product). Let $M$ be a $\mathbb{Z}$-module and $p$ be a prime number. The localization of $M$ at the prime ideal $(p)$, denoted $M_{(p)}$, is canonically isomorphic to the tensor product of $M$ with the localized ring ${\mathbb{Z}}_{(p)}$, where ${\mathbb{Z}}_{(p)}=\{a/b\in \mathbb{Q}\mid p\nmid b\}$. That is, $M_{(p)}\cong M{\otimes }_{\mathbb{Z}}{\mathbb{Z}}_{(p)}$.

This recasts the construction of the module of fractions as a tensor product. For example, the localization of $\mathbb{Z}/n\mathbb{Z}$ at $(p)$ is $(\mathbb{Z}/n\mathbb{Z}{)}_{(p)}\cong (\mathbb{Z}/n\mathbb{Z}){\otimes }_{\mathbb{Z}}{\mathbb{Z}}_{(p)}\cong {\mathbb{Z}}_{(p)}/n{\mathbb{Z}}_{(p)}$. There are two cases for the structure of this quotient module:

1. If $p\nmid n$, then $n$ is a unit in the ring ${\mathbb{Z}}_{(p)}$. The ideal $n{\mathbb{Z}}_{(p)}$ is the entire ring ${\mathbb{Z}}_{(p)}$, so the quotient is $\{0\}$.
2. If $p\mid n$, let $n=p^{k}m$ where $\gcd (p,m)=1$. Then $m$ is a unit in ${\mathbb{Z}}_{(p)}$, so the ideal $n{\mathbb{Z}}_{(p)}$ equals the ideal $p^k{\mathbb{Z}}_{(p)}$. The quotient is ${\mathbb{Z}}_{(p)}/p^k{\mathbb{Z}}_{(p)}\cong \mathbb{Z}/p^k\mathbb{Z}$. This result recovers the $p$-primary component of the cyclic group, matching the classical definition of localization for such modules.