Tensor Products of Modules — Examples

Apply [L1] to $M=\mathbb{Z}\mathrm{/}m$ and $I=n\mathbb{Z}$ to obtain $\mathbb{Z}\mathrm{/}m{\otimes }_{\mathbb{Z}}\mathbb{Z}\mathrm{/}n\cong (\mathbb{Z}\mathrm{/}m)\mathrm{/}n(\mathbb{Z}\mathrm{/}m)$.

Define $\varphi :\mathbb{Z}\mathrm{/}d\to (\mathbb{Z}\mathrm{/}m)\mathrm{/}n(\mathbb{Z}\mathrm{/}m)$ by $\varphi ([a{]}_d)=[a{]}_m+n(\mathbb{Z}\mathrm{/}m)$. If $a-b\in d\mathbb{Z}=m\mathbb{Z}+n\mathbb{Z}$ by [L2], then $[a-b{]}_m$ lies in $n(\mathbb{Z}\mathrm{/}m)$, so $\varphi$ is well-defined.

The map $\varphi$ is surjective because every class in the target is represented by some $[a{]}_m$. If $\varphi ([a{]}_d)=0$, then $[a{]}_m=n[b{]}_m$ for some integer $b$, so $a-nb\in m\mathbb{Z}$ and hence $a\in m\mathbb{Z}+n\mathbb{Z}=d\mathbb{Z}$ by [L2]; therefore $[a{]}_d=0$, and $\varphi$ is injective.

Steps 1.1 and 2.1 give the displayed isomorphism. For $(m,n)=(2,3)$ one has $d=1$, so the tensor product is $\mathbb{Z}\mathrm{/}1=0$ although both $\mathbb{Z}\mathrm{/}2$ and $\mathbb{Z}\mathrm{/}3$ are nonzero. This refutes the claim.

By [L1], $\mathbb{Q}{\otimes }_{\mathbb{Z}}\mathbb{Z}\mathrm{/}n\cong \mathbb{Q}\mathrm{/}n\mathbb{Q}$.

For every $q\in \mathbb{Q}$, the element $q\mathrm{/}n$ belongs to $\mathbb{Q}$ by [L2] and $q=n(q\mathrm{/}n)$, so $n\mathbb{Q}=\mathbb{Q}$.

Hence $\mathbb{Q}\mathrm{/}n\mathbb{Q}=0$, and step 1.1 proves the claim.

Apply [L1] with $M=R\mathrm{/}I$ to obtain $(R\mathrm{/}I){\otimes }_R(R\mathrm{/}J)\cong (R\mathrm{/}I)\mathrm{/}J(R\mathrm{/}I)$.

The map $q:R\mathrm{/}I\to R\mathrm{/}(I+J)$ given by $q(r+I)=r+(I+J)$ is well-defined and surjective. Its kernel consists of the classes $i+j+I=j+I$ with $i\in I$ and $j\in J$, which is exactly $J(R\mathrm{/}I)$.

By [L3], step 1.2 induces $(R\mathrm{/}I)\mathrm{/}J(R\mathrm{/}I)\cong R\mathrm{/}(I+J)$; composing with step 1.1 proves the displayed isomorphism.

If $I=0$ or $J=0$, the formula reduces to the appropriate tensor-unit isomorphism. If either ideal is $R$, then both sides are zero. Thus all stated boundary cases are included.

The displayed coefficient formula is additive in both variables and satisfies $F(sf(r)\otimes p)=F(s\otimes rp)$, so it is $R$-balanced. It therefore induces an additive map $F:S{\otimes }_{R}R[x]\to S[x]$ by [L3].

Define $G:S[x]\to S{\otimes }_{R}R[x]$ by $G({\sum }_i{s}_i{x}^i)={\sum }_i{s}_i\otimes x^i$. The sum is finite by [L2]. Coefficientwise addition and convolution multiplication show that $G$ is an $S$-algebra homomorphism, using $(s\otimes x^i)(t\otimes x^j)=st\otimes x^{i+j}$ from [L4].

The map $F$ is $S$-linear, sends $1\otimes 1$ to $1$, and, using [L2] and [L4], satisfies $F((s\otimes p)(t\otimes q))=F(st\otimes pq)=F(s\otimes p)F(t\otimes q)$. Hence it is an $S$-algebra homomorphism.

For every polynomial ${\sum }_i{s}_i{x}^i$, one has $F(G({\sum }_i{s}_i{x}^i))={\sum }_i{s}_i{x}^i$. For an elementary tensor, balance gives $G(F(s\otimes {\sum }_i{r}_i{x}^i))={\sum }_{i}sf(r_i)\otimes x^i={\sum }_{i}s\otimes r_i{x}^i=s\otimes {\sum }_i{r}_i{x}^i$.

The two maps $GF$ and the identity induce the same balanced pairing by step 2.2, so uniqueness in [L3] makes them equal; step 2.2 already gives $FG=1$ coefficientwise. Thus $F$ and $G$ are inverse $S$-algebra homomorphisms.

For $i,j<n$, let $E_{ij}$ have entry $1$ at $(i,j)$ and $0$ elsewhere. Entrywise decomposition writes every $A\in M_n(F)$ uniquely as $A={\sum }_{i,j<n}{a}_{ij}{E}_{ij}$, so the $E_{ij}$ form an $F$-basis; when $n=0$, this is the empty basis of the zero space.

The pairing $(k,A)\mapsto (k{a}_{ij})$ is $F$-balanced, so [L4] gives a unique additive $T:K{\otimes }_F{M}_n(F)\to M_n(K)$ with $T(k\otimes A)=(k{a}_{ij})$, and $T$ is $K$-linear by [L6]. Define $S:M_n(K)\to K{\otimes }_F{M}_n(F)$ by $S(B)={\sum }_{i,j<n}{b}_{ij}\otimes E_{ij}$, which is additive and $K$-linear by [L6]. Then $T(S(B))={\sum }_{i,j<n}{b}_{ij}{E}_{ij}=B$ by step 1.1, and on a generator $S(T(k\otimes A))={\sum }_{i,j<n}k{a}_{ij}\otimes E_{ij}={\sum }_{i,j<n}k\otimes a_{ij}{E}_{ij}=k\otimes A$, moving each $F$-scalar $a_{ij}$ across the balanced tensor by [L6] and using step 1.1. Both composites are additive and agree on generators, so $T$ and $S$ are mutually inverse and the displayed map is a $K$-linear isomorphism.

Matrix multiplication gives $E_{ij}{E}_{\mathrm{\ell }r}=0$ if $j\mathrm{\ne }\mathrm{\ell }$ and $E_{ij}{E}_{jr}=E_{ir}$. The displayed map preserves these products by [L5], and it sends $1\otimes I_n$ to $I_n$; by bilinearity it is a unital algebra homomorphism.

Combining steps 2.1 and 2.2 proves the algebra isomorphism. For $n=0$ it is the unique map between one-element zero algebras, while for $n=1$ it is the tensor-unit identification $K{\otimes }_{F}F\cong K$.

The pairing $(k,(a_j))\mapsto (k{a}_j)$ is $F$-balanced, so [L3] gives a unique additive $T:K{\otimes }_F{F}^n\to K^n$ with $T(k\otimes (a_j))=(k{a}_j)$, and $T$ is $K$-linear by [L1]. It sends $1\otimes e_j$ to the $j$th standard coordinate vector.

Define $S:K^n\to K{\otimes }_F{F}^n$ by $S((k_j{)}_{j<n})={\sum }_{j<n}{k}_j\otimes e_j$, which is additive and $K$-linear by [L1]. Then $T(S((k_j)))=(k_j)$ because $T(k_j\otimes e_j)$ is $k_j$ in coordinate $j$ and zero elsewhere, and on a generator $S(T(k\otimes a))={\sum }_{j<n}k{a}_j\otimes e_j={\sum }_{j<n}k\otimes a_j{e}_j=k\otimes a$, moving each $F$-scalar $a_j$ across the balanced tensor by [L1] and expanding $a$ in the basis of [L2]. Both composites are additive and agree on generators, so $T$ is an isomorphism.

At $n=0$ both modules are zero — the empty sum defining $S$ is $0$ — so the same argument gives the unique isomorphism.

Conjugation fixes real scalars and is additive and multiplicative by the coordinate formulas in [L1]. Hence $(z,w)\mapsto (zw,\bar{z}w)$ is $\mathbb{R}$-bilinear and induces an $\mathbb{R}$-linear map $\mathrm{\Phi }$ from the tensor product.

By [L2] and [L3], $1\otimes 1,i\otimes 1,1\otimes i,i\otimes i$ form an $\mathbb{R}$-basis of the source. Their images are $(1,1),(i,-i),(i,i),(-1,1)$.

By [L4] and [L5], $\mathrm{\Phi }((z\otimes w)(z^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }}))=(z{z}^{\mathrm{\prime }}w{w}^{\mathrm{\prime }},\overline{z{z}^{\mathrm{\prime }}}w{w}^{\mathrm{\prime }})=\mathrm{\Phi }(z\otimes w)\mathrm{\Phi }(z^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }})$, and $\mathrm{\Phi }(1\otimes 1)=(1,1)$; thus $\mathrm{\Phi }$ is an $\mathbb{R}$-algebra homomorphism.

Given $(u+vi,x+yi)\in \mathbb{C}\times \mathbb{C}$, its unique coordinates in the four images of step 1.2 are $a=(u+x)\mathrm{/}2$, $b=(v-y)\mathrm{/}2$, $c=(v+y)\mathrm{/}2$, and $d=(x-u)\mathrm{/}2$. Therefore those images form a real basis and $\mathrm{\Phi }$ is bijective.

Since $\mathrm{\Phi }(i\otimes i)=(-1,1)$, the two displayed tensors map respectively to $(1,0)$ and $(0,1)$, the standard product idempotents.

Steps 2.1 and 2.2 prove the claimed algebra isomorphism, and step 2.3 identifies its idempotents.

A field is an integral domain, so [L2] makes $A$ an integral domain. Since $x\mathrm{\ne }0$ by [L1], $xa=xb$ implies $x(a-b)=0$ and hence $a=b$; therefore ${\mu }_x$ is injective.

Under [L3], the tensor map sends the class $\bar{a}$ to $\overline{xa}=0$ because $x\in (x)$. Thus ${\mu }_x\otimes 1$ is the zero map.

The module $A\mathrm{/}(x)$ is nonzero because $1\mathrm{\notin }(x)$ by [L1]. Hence step 1.2 is a zero map on a nonzero module and is not injective, despite step 1.1.

Balance in the tensor product and [L2] give $2\otimes \bar{1}=1\otimes 2\bar{1}=1\otimes \bar{0}=0$.

The integer $2$ is nonzero, and $\bar{1}$ is nonzero by [L2]. Thus neither factor in step 1.1 is zero.

Moreover, [L3] sends $1\otimes \bar{1}$ to the nonzero class $\bar{1}$, so the ambient tensor-product group is itself nonzero. Steps 1.1 and 2.1 therefore refute the claim.

Suppose $e_1\otimes f_1+e_2\otimes f_2$ were elementary. By [L1], write its factors as $a{e}_1+b{e}_2$ and $c{f}_1+d{f}_2$.

Expanding the elementary tensor gives coefficients $ac,ad,bc,bd$ on the ordered basis $e_1\otimes f_1,e_1\otimes f_2,e_2\otimes f_1,e_2\otimes f_2$. Uniqueness of coefficients in [L2] therefore yields $ac=bd=1$ and $ad=bc=0$.

From $ac=1$, both $a$ and $c$ are nonzero. Then $ad=0$ forces $d=0$, contradicting $bd=1$. Hence the displayed tensor is not elementary and the claim is false.

The map $u:\mathbb{Z}\to \mathbb{Z}$, $u(a)=2a$, is injective: if $2a=2b$, cancellation in [L1] gives $a=b$.

Under the unit identifications [L3], the map $u\otimes 1_{\mathbb{Z}\mathrm{/}2}$ is multiplication by $2$ on $\mathbb{Z}\mathrm{/}2$, hence is zero by [L2].

The zero map on $\mathbb{Z}\mathrm{/}2$ is not injective because $\bar{1}\mathrm{\ne }\bar{0}$ by [L2]. Thus step 1.1 is an injection whose tensor map is not injective, refuting the claim. This is consistent with [L4], which guarantees right exactness only.