2 Modules

Highlights: Nakayama lemma, snake lemma, tensor product, flatness

Define ${\mathrm{Hom}}_A(M,N)=\{A\text{mbox}-modulehomomorphismsf:M\to N\}$. It has an $A$-module structure. The construction is “functorial”, namely. If $u:M^{\prime }\to M$ is a homomorphism, then it deduces a homomorphism

$$\mathrm{Hom}(M,N)\to \mathrm{Hom}(M^{\prime },N),f\mapsto f\circ u.$$

Similarly, given $v:N\to N^{\prime }$, it induces a homomorphism $\mathrm{Hom}(M,N)\to \mathrm{Hom}(M,N^{\prime })$.

Example. Compute ${\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}/4\mathbb{Z},\mathbb{Z}/6\mathbb{Z})$. Since $f(\overline{1})$ is divisible by $3$, there is $f(\overline{1})\in \{\overline{0},\overline{3}\}$. It deduces that ${\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}/4\mathbb{Z},\mathbb{Z}/6\mathbb{Z})\simeq \mathbb{Z}/2\mathbb{Z}$.

Easy Facts.

• ${\mathrm{Hom}}_A(A,M)\simeq M,f\mapsto f(1)$
• For ideal $\alpha \subseteq A$, define $\alpha M=\{{\sum }_{i=1}^n{\alpha }_i{x}_i:{\alpha }_i\in \alpha ,x_i\in M\}\subseteq M$ is a submodule. When $\alpha =(a)$, write $aM=(a)M$.

Definition

For submodules $N,P\subseteq M$, define $(N:P)=\{a\in A:aP\subseteq N\}\subseteq A$ an ideal. Denote annihilator of $M$ as $\mathrm{Ann}(M):=(0:M)$. Note $M$ as an $A$-module can also be regarded as an $A/\mathrm{Ann}(M)$-module. Say $M$ is a faithful $A$-module if $\mathrm{Ann}(M)=0$.

Proposition

$M$ is a finitely generated $A$-module iff there exists an surjective $A^n\twoheadrightarrow M$ iff $M$ is a quotient module of $A^n$.

Nakayama Lemma

Proposition

Suppose $M$ is a finitely generated $A$-module, and $\alpha \subseteq A$ is an ideal. Let $\varphi :M\to M$ be an endomorphism such that $\varphi (M)\subseteq \alpha M$. Then $\varphi$ satisfies an equation of form

$${\varphi }^n+a_1{\varphi }^{n-1}+\cdots +a_{n-1}\varphi +a_n=0$$

on $M$ for some $a_i\in \alpha$, namely, ${\sum }_{i=0}^n{a}_i{\varphi }^{n-i}(m)=0$ for all $m\in M$.

Example. Let $M=\mathbb{Z}{e}_1\oplus \mathbb{Z}{e}_2$ and $\alpha =(2)$. Define $\varphi$ by $\varphi \left(\begin{array}{c}{e}_1\\ e_2\end{array}\right)=\left(\begin{array}{cc}2& 0\\ 4& 8\end{array}\right)\left(\begin{array}{c}{e}_1\\ e_2\end{array}\right)$. Then by Cayley-Hamilton, $(\varphi -2)(\varphi -8)=0$.

\begin{proof} Choose a set of generators $\{e_1,\cdots ,e_n\}$ of $M$. Then $\varphi \left(\begin{array}{c}{e}_1\\ ⋮\\ e_n\end{array}\right)=T\left(\begin{array}{c}{e}_1\\ ⋮\\ e_n\end{array}\right)$ for some $T\in {\mathrm{Mat}}_{n\times n}(\alpha )$. Let $P(x)$ be the characteristic polynomial of $T$, then $P(\varphi )=0$ by Cayley-Hamilton. \end{proof}

Corollary

Let $M$ be a finitely generated $A$-module, and let $\alpha$ be an ideal of $A$ with $\alpha M=M$. Then there exists $x\equiv 1(\mathrm{mod}\alpha )$ such that $xM=0$.

\begin{proof} Let $\varphi ={\mathrm{id}}_M:M\to \alpha M=M$. By ^41223f, $\varphi$ satisfies

$${\varphi }^n+\sum _{i=1}^n{a}_i{\varphi }^{n-i}=0$$

on $M$ with $a_i\in \alpha$. Let $x=1+a_1+\cdots +a_n$, then $xM=0$. \end{proof}

Nakayama lemma

Let $M$ be a finitely generated $A$-module, and let $\alpha \subseteq A$ be an ideal with $\alpha \subseteq \mathrm{Jac}(A)$. If $\alpha M=M$, then $M=0$.

\begin{proof} By ^d05ae0 there exists $x=1+a$ with $a\in \mathrm{Jac}(A)$ such that $xM=0$. By ^h2e8bp, $1+a\in A^{\times }$ and so $M=0$. \end{proof}

Remark. The most useful situation is when $A$ is a local ring with $\alpha =m_A=\mathrm{Jac}(A)$. ^c2a5d0 tells if $M$ is finitely generated and $m_{A}M=M$, then $M=0$.

Corollary

Let $M$ be a finitely generated $A$-module with submodule $N\subseteq M$. Let $\alpha \subseteq \mathrm{Jac}(A)$ be an ideal. Then $M=\alpha M+N$ yields that $M=N$.

\begin{proof} Apply ^c2a5d0 to $M/N$. Since $M=\alpha M+N$, there is $M/N=\alpha (M/N)$ and then $M/N=0$. Therefore, $M=N$. \end{proof}

Proposition

Let $(A,m_A)$ be a local ring. Let $k=A/m_A$. Let $M$ be a finitely generated $A$-module, then $M/m_{A}M$ is a vector space over $k$. Let $x_1,\cdots ,x_n\in M$. Suppose $\overline{x_1},\cdots ,\overline{x_n}$ is a basis of $M/m_{A}M$, then $x_1,\cdots ,x_n$ generate $M$.

\begin{proof} Let $N={⟨x_1,\cdots ,x_n⟩}_A\subseteq M$. Then the composite $N\hookrightarrow M\twoheadrightarrow M/m_{A}M$ is surjective. It deduces that $M=N+m_{A}M$. By ^ffbd3c, $M=N$ and so $x_1,\cdots ,x_n$ generate $M$. \end{proof}

Remark. This is quite similar to ^k9gxk1—in fact, the Jacobson radical can be seen as an analogue of the Frattini subgroup, but in the setting of rings instead of groups.

Exact Sequence

• If $0\stackrel{0}{\to }X\stackrel{f}{\to }Y\stackrel{g}{\to }Z\stackrel{0}{\to }0$ is exact, then we say it is short exact.

Link to original

Remark. Any exact sequence can be split into short exact sequences. Namely, given

$$\cdots \to M_{i-2}\stackrel{f_{i-1}}{\to }{M}_{i-1}\stackrel{f_i}{\to }{M}_i\stackrel{f_{i+1}}{\to }{M}_{i+1}\stackrel{f_{i+2}}{\to }{M}_{i+2}\to \cdots ,$$

then we get $0\to \mathrm{im}{f}_i\to M_i\to M_i/\mathrm{im}{f}_i\to 0$ and $0\to \mathrm{im}{f}_{i+1}\to M_{i+1}\to M_{i+1}/\mathrm{im}{f}_{i+1}\to 0$, where $M_i/\mathrm{im}{f}_i=M_i/\ker f_{i+1}\simeq \mathrm{im}{f}_{i+1}$.

Hom Functor

Proposition

• Let $M^{\prime }\stackrel{u}{\to }M\stackrel{v}{\to }{M}^{\prime \prime }\to 0$ $(\ast )$ be a sequence of $A$-module homomorphism. Then it is exact iff for any $A$-module $N$, the induced sequence

$$0\to \mathrm{Hom}(M^{\prime \prime },N)\stackrel{\overline{v}}{\to }\mathrm{Hom}(M,N)\stackrel{\overline{u}}{\to }\mathrm{Hom}(M^{\prime },N)\text{\hspace{1em}(**)}$$

is exact, where $\overline{v}:f\mapsto f\circ v$ and $\overline{u}:g\mapsto g\circ u$.

• (dual statement) Let $0\to N^{\prime }\stackrel{u}{\to }N\stackrel{v}{\to }{N}^{\prime \prime }(☺\stackrel{◯}{R})$ be a sequence. It is exact iff for any $A$-module $M$, the induced sequence

$$0\to \mathrm{Hom}(M,N^{\prime })\stackrel{\overline{u}}{\to }\mathrm{Hom}(M,N)\stackrel{\overline{v}}{\to }\mathrm{Hom}(M,N^{\prime \prime })\text{\hspace{1em}(☺R◯☺R◯)}$$

is exact.

\begin{proof} i) "→" Assume $(\ast )$ is exact. It is enough to show $\overline{v}$ is injective and $\mathrm{im}\overline{v}=\ker \overline{u}$. If $\overline{v}(f)=0$, then $M\stackrel{v}{\to }{M}^{\prime \prime }\stackrel{f}{\to }N$ is zero. But $v$ is surjective, then $f=0$. Note that $\overline{u}\circ \overline{v}=0$ as $v\circ u=0$, and it deduces that $\mathrm{im}\overline{v}\subseteq \ker \overline{u}$. Conversely, for any $g\in \ker \overline{u}$, we aim to find $f$ such that $g=f\circ v$. For $x\in M^{\prime \prime }$, let $\hat{x}\in v^{-1}(x)$ and set $f(x)=g(\hat{x})$. We can verify that $f$ is well-defined and is a homomorphism of $A$-modules. Hence $\mathrm{im}\overline{v}=\ker \overline{u}$.

"←" It suffices to show $v$ is surjective and $\mathrm{im}u=\ker v$. If $v$ is not surjective, then let $N=M^{\prime \prime }/\mathrm{im}v\ne 0$ and apply to $(\ast \ast )$, that is,

$$0\to \mathrm{Hom}(M^{\prime \prime },M^{\prime \prime }/\mathrm{im}v)\stackrel{\overline{v}}{\to }\mathrm{Hom}(M,M^{\prime \prime }/\mathrm{im}v)\stackrel{\overline{u}}{\to }\mathrm{Hom}(M^{\prime },M^{\prime \prime }/\mathrm{im}v)$$

is exact. The surjective map $\pi :M^{\prime \prime }\to M^{\prime \prime }/\mathrm{im}v$ is a non-zero map, but $\overline{v}(\pi )$ is a zero map, which is a contradiction. Hence $v$ is surjective.

Take $N=M^{\prime \prime }$ and $\mathrm{id}\in \mathrm{Hom}(M^{\prime \prime },M^{\prime \prime })$. Then $(\ast \ast )$ is exact yields that $\mathrm{id}\circ v\circ u=0$ and so $\mathrm{im}u\subseteq \ker v$. Finally, we check $\mathrm{im}u\supseteq \ker v$. Now we take $N=M/\mathrm{im}u$ and

$$0\to \mathrm{Hom}(M^{\prime \prime },M/\mathrm{im}u)\stackrel{\overline{v}}{\to }\mathrm{Hom}(M,M/\mathrm{im}u)\stackrel{\overline{u}}{\to }\mathrm{Hom}(M^{\prime },M/\mathrm{im}u)$$

is exact. Define canonical map $\pi :M\to M/\mathrm{im}u$. Notice that $\pi \circ u=0$, then $\pi \in \ker \overline{u}=\mathrm{im}\overline{v}$. Then there exists $\varphi \in \mathrm{Hom}(M^{\prime \prime },M/\mathrm{im}u)$ such that $\varphi \circ v=\pi$. It deduces that $\varphi (v(x))=\pi (x)=0$ and $x\in \mathrm{im}u$. Now we finish the proof.

ii) "→" Assume that $(☺\stackrel{◯}{R})$ is exact. It is enough to show $\overline{u}$ is injective and $\ker \overline{v}=\mathrm{im}\overline{u}$. If $\overline{u}(f)=0$, then $u\circ f=0$. Since $u$ is injective, we have $f=0$ and so $\overline{u}$ is injective. Note that $\overline{v}\circ \overline{u}=0$ as $v\circ u=0$, so $\mathrm{im}\overline{u}\subseteq \ker \overline{v}$. For any $g\in \ker \overline{v}$, we aim to find $f$ such that $g=f\circ u$. For any $m\in M$, define $f(m)=u^{-1}(g(m))$. Since $g(m)\in \ker v=\mathrm{im}u$ and $u$ is injective, $f$ is well-defined and $g=f\circ u$. Hence $\mathrm{im}\overline{u}=\ker \overline{v}$.

"←" Assume that $(☺\stackrel{◯}{R}☺\stackrel{◯}{R})$ is exact. It suffices to show $u$ is injective and $\ker v=\mathrm{im}u$. If $u$ is not injective, then $\ker u\ne 0$. Let $M=\ker u$ and apply it to $(☺\stackrel{◯}{R}☺\stackrel{◯}{R})$. The inclusion map $\iota :\ker u\hookrightarrow N^{\prime }$ is included in $\mathrm{Hom}(M,N^{\prime })$ and $\overline{u}(\iota )=0$, which contradicts with $\overline{u}$ injective. Take $M=N^{\prime }$ and ${\mathrm{id}}_{N^{\prime }}\in \mathrm{Hom}(N^{\prime },N^{\prime })$ satisfies $\overline{v}\circ \overline{u}\circ f=v\circ u=0$, that is, $\mathrm{im}u\subseteq \ker v$. Conversely, take $M=\ker v$ and

$$0\to \mathrm{Hom}(\ker v,N^{\prime })\stackrel{\overline{u}}{\to }\mathrm{Hom}(\ker v,N)\stackrel{\overline{v}}{\to }\mathrm{Hom}(\ker v,N^{\prime \prime })$$

is exact. Since ${\iota }^{\prime }:\ker v\to N$ is an element of $\mathrm{Hom}(\ker v,N)$ and $\overline{v}({\iota }^{\prime })=0$, there is ${\iota }^{\prime }\in \ker \overline{v}=\mathrm{im}\overline{u}$ and so there exists $\varphi$ such that $u\circ \varphi ={\iota }^{\prime }$. Then for any $m\in \ker v$, we have $m={\iota }^{\prime }(m)=u(\varphi (m))$ and $m\in \mathrm{im}u$. Now we finish the proof.
\end{proof}

Remark.

• "$\to 0$" in $(\ast )$ is necessary. For example, consider $\mathbb{Z}\stackrel{\times 0}{\to }\mathbb{Z}\stackrel{\times 2}{\to }\mathbb{Z}$ is exact but right exact. Then $\mathrm{Hom}(\mathbb{Z},\mathbb{Z})\stackrel{\times 2}{\to }\mathrm{Hom}(\mathbb{Z},\mathbb{Z})\stackrel{\times 0}{\to }\mathrm{Hom}(\mathbb{Z},\mathbb{Z})$.
• "$0\to$" on LHS of $(\ast )$ is useless, as it cannot add "$\to 0$" to $(\ast \ast )$. Consider $0\to \mathbb{Z}\stackrel{\times 2}{\to }\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}\to 0$. Apply $\mathrm{Hom}(\ast ,\mathbb{Z})$, then $0\to 0\to \mathbb{Z}\stackrel{\times 2}{\to }\mathbb{Z}\to 0$ is only left exact.

Snake Lemma

snake lemma

Consider a commutative diagram where both rows are short exact.

$$\begin{array}{ccccccccc}0& \to & M^{\prime }& \stackrel{G}{\to }& M& \stackrel{H}{\to }& M^{\prime \prime }& \to & 0\\ & & f^{\prime }\downarrow & & f\downarrow & & f^{\prime \prime }\downarrow & & \\ 0& \to & N^{\prime }& \stackrel{g}{\to }& N& \stackrel{h}{\to }& N^{\prime \prime }& \to & 0\end{array}$$

Then we have an exact sequence

$$0\to \ker f^{\prime }\to \ker f\to \ker f^{\prime \prime }\to \mathrm{coker}{f}^{\prime }\to \mathrm{coker}f\to \mathrm{coker}{f}^{\prime \prime }\to 0.$$

\begin{proof} Notice that $\ker f^{\prime }\to \ker f\to \ker f^{\prime \prime }$ are restriction of $M^{\prime }\to M\to M^{\prime \prime }$. We can check that $g$ and $h$ induce maps $\mathrm{coker}{f}^{\prime }\to \mathrm{coker}f$ and $\mathrm{coker}f\to \mathrm{coker}{f}^{\prime \prime }$. So it suffices to construct $d:\ker f^{\prime \prime }\to \mathrm{coker}{f}^{\prime }$. For any $x\in M^{\prime \prime }$, take $\hat{x}\in M$ and so $f(\hat{x})\in \ker h$. Then define $\check{x}\in N^{\prime }$ such that $g(\check{x})=\hat{x}$ and $\overline{\check{x}}\in \mathrm{coker}{f}^{\prime }$ is what we desire. It is easy to check this map is well-defined.

It remains to check the sequence is exact. For convenience, define

$$0\to \ker f^{\prime }\stackrel{G^{\prime }}{\to }\ker f\stackrel{H^{\prime }}{\to }\ker f^{\prime \prime }\stackrel{d}{\to }\mathrm{coker}{f}^{\prime }\stackrel{g^{\prime }}{\to }\mathrm{coker}f\stackrel{h^{\prime }}{\to }\mathrm{coker}{f}^{\prime \prime }\to 0.$$

For any $x\in \ker d⩽M^{\prime \prime }$, there exists $y\in M$ such that $H(y)=x$. By the definition of $d$, we know $f(y)=g(d(x))=0$ and so $y\in \ker f$. It deduces that $x\in \mathrm{im}(H^{\prime })$. Conversely, for any $x\in \mathrm{im}(H^{\prime })$, there exists $y\in \ker f$ such that $x=H^{\prime }(y)$. By the definition of $d$, there is $g(d(x))=f(y)=0$ and $d(x)=0$. Thus $x\in \ker d$. Now we have proved $\ker d=\mathrm{im}{H}^{\prime }$.

For any $x\in \ker g^{\prime }$, there is $g^{\prime }(x)=0$ and so $g(\hat{x})\in \mathrm{im}f$ for any given preimage $\hat{x}\in N^{\prime }$. Then there is $y\in M$ such that $f(y)=g(\hat{x})$. By the definition of $d$, we know $d(H(y))=x$. In addition, since $f^{\prime \prime }(H(y))=h(f(y))=h(g(\hat{x}))=0$, $H(y)\in \ker f^{\prime \prime }$. It yields that $x\in \mathrm{im}d$. Conversely, for any $x\in \mathrm{im}d$, there exists $y$ such that $d(y)=x$ and $z\in M$ such that $H(z)=y$ and $f(z)=g(\hat{x})$ for any given $x\in N^{\prime }$. By the definition of $d$, there is $g^{\prime }(x)=\overline{f(z)}\in \mathrm{coker}f$ and so $g^{\prime }(x)=0$, $x\in \ker g^{\prime }$. Now we have proved $\mathrm{im}d=\ker g^{\prime }$. \end{proof}

Remark. It is a general case of short five lemma.

Tensor Product

Motivations.

• linear algebra: bilinear map
• algebraic geometry: fibre product of geometry spaces
• flatness: useful for localization

Linear algebra. Let $k$ be a field. Notice that bilinear map and linear map are different.

• Consider bilinear map $f:k^m\times k^n\to k^s$. It is different from $k$-linear map $k^{m+n}\stackrel{f^{\prime }}{\to }{k}^s$, because $f(ax,ay)=a^{2}f(x,y)$ and $f^{\prime }(ax,ay)=af(x,y)$.
• Define $g:k\times k\to k,(a,b)\to a$, which is linear but not bilinear.
• A $k$-linear map $g:k\times k\to k^s$ is determined by $f(1,0)$ and $f(0,1)$; a $k$-bilinear map $g^{\prime }:k\times k\to k^s$ is determined by $f(1,1)$.
• A $k$-bilinear map $h:k^m\times k^n\to k^s$ looks like a $k$-linear map $h:k^{mn}\to k^s$, as we will see $k^m\otimes k^n\simeq k^{mn}$.

Algebraic geometry. Note a fact that $\mathrm{Spec}(A\times B)=\{p\times B,A\times q:p\in \mathrm{Spec}(A),q\in \mathrm{Spec}(B)\}$. Hence, $\mathrm{Spec}(\mathbb{C}[x]\times \mathbb{C}[y])=\mathrm{Spec}(\mathbb{C}[x])\bigsqcup \mathrm{Spec}(\mathbb{C}[y])$. By ^6rh0cw, the set of maximal ideal of $\mathbb{C}[x,y]$ is $\mathrm{Spm}(\mathbb{C}[x,y])=\{(x-a,y-b):a,b\in \mathbb{C}\}$. In fact, $\mathbb{C}[x]\otimes \mathbb{C}[y]\simeq \mathbb{C}[x,y]$.

Flatness. Consider $\mathbb{Z}$-bilinear map $f:\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\to P$, which is determined by $f(1,\overline{1})$. Note that $f(1,\overline{1})+f(1,\overline{1})=f(1,\overline{0})=0$, then $f(1,\overline{1})\in P[2]$. In summary, the above bilinear map is same as a $\mathbb{Z}$-linear map $\tilde{f}:\mathbb{Z}/2\mathbb{Z}\to P$. Later we will see $\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Z}/2\mathbb{Z}\simeq \mathbb{Z}/2\mathbb{Z}$.

Definition

Let $A$ be a ring, and let $M,N,P$ be $A$-modules. A map $f:M\times N\to P$ is $A$-bilinear if $f(ax+by,cz+dw)=acf(x,z)+adf(x,w)+bcf(y,z)+bdf(y,w)$.

Proposition

Let $M$ and $N$ be $A$-modules. Then there exists a pair $(T,g)$ where $T$ is an $A$-module and $g:M\times N\to T$ is an $A$-bilinear map, with the universal property: Given any $A$-module $P$, and any $A$-bilinear map $f:M\times N\to P$, there exists unique $A$-linear map $f^{\prime }:T\to P$ such that $f=f^{\prime }\circ g$, that is, $f$ factors through $g$.

Moreover if $(T,g)$ and $(T^{\prime },g^{\prime })$ are two such pairs with above property, then there exists unique isomorphism $j:T\to T^{\prime }$ such that $j\circ g=g^{\prime }$.

\begin{proof} Uniqueness. Suppose $(T,g)$ and $(T^{\prime },g^{\prime })$ are two distinct pairs with the property above. Apply $T$ with $P=T^{\prime }$ and $T^{\prime }$ with $P=T$, then there exist $j:T\to T^{\prime }$ and $j^{\prime }:T^{\prime }\to T$ such that $j^{\prime }\circ j=j\circ j^{\prime }=\mathrm{id}$.

Existence. Let $C$ be the free $A$-module $A^{M\times N}$, namely, a free $A$-module with basis $\{(x,y):x\in M,y\in Y\}$. So an element in $C$ has unique expression ${\sum }_{i=1}^n{a}_i(x_i,y_i)$. Let $D$ be the submodule of $C$ generated by the following elements:

• $(x+x^{\prime },y)-(x,y)-(x^{\prime },y)$,
• $(x,y+y^{\prime })-(x,y)-(x,y^{\prime })$,
• $(ax,y)-a(x,y)$,
• $(x,ay)-a(x,y)$.

Let $T=C/D$. For $(x,y)\in C$, denote $x\otimes y$ as image in $T=C/D$. Thus, $T$ is generated by $\{x\otimes y:x\in M,y\in N\}$. Now define

$$g:M\times N\to T,(m,n)\mapsto m\otimes n,$$

which is a $A$-bilinear map. We now verify the universal property. Let $f:M\times N\to P$ be a bilinear map. It suffices to construct $f^{\prime }:T\to P$ such that $f=f^{\prime }\circ g$. We define $\overline{f}:C=A^{M\times N}\to P,(m,n)\mapsto f(m,n)$, and we can easily see $D\subseteq \ker \overline{f}$. So $\overline{f}$ induces a map $f^{\prime }:C/D=T\to P$. Remark that

$$f^{\prime }\circ g(m,n)=f^{\prime }(m\otimes n)=f(m,n)$$

and so $f^{\prime }\circ g=f$. Hence, for each bilinear map $f:M\times N\to P$, we can find $f^{\prime }$ such that $f^{\prime }\circ g=f$ and so $(T,g)$ is what we desire. \end{proof}

Definition

Call the module $T$ above as the tensor product of $M$ and $N$, also denote as $T=M{\otimes }_{A}N$. Sometimes, just $M\otimes N$ if the context is clear. Also the map $g:M\times N\to M\otimes N$ is simply $g((m,n))=m\otimes n$. In particular, the $A$-module $M{\otimes }_{A}N$ is generated by $\{m\otimes n:m\in M,n\in N\}$. Indeed, if $\{m_i:i\in I\}$ generates $M$ and $\{n_i\}$ generates $N$, then $\{m_i\otimes n_j:i\in I,j\in J\}$ generates $M{\otimes }_{A}N$.

Remark. Not all elements of $M{\otimes }_{A}N$ is of the form $x\otimes y$. In other word, $g:M\times N\to M\otimes N$ is in general not surjective. For example, let $M=\mathbb{Z}e\oplus \mathbb{Z}f$ and $N=\mathbb{Z}g\oplus \mathbb{Z}h$, then $e\otimes g+f\otimes h\in M{\otimes }_{\mathbb{Z}}N$ cannot equal to some $x\otimes y$ with $x\in M$ and $y\in Y$.

Proposition

Let $M,N,P$ be $A$-modules. Then there exists unique isomorphisms:

• $M\otimes N\to N\otimes M$ such that $x\otimes y\mapsto y\otimes x$;
• $(M\otimes N)\otimes P\to M\otimes (N\otimes P)\to M\otimes N\otimes P$;
• $(M\oplus N)\otimes P\to (M\otimes P)\oplus (N\otimes P)$;
• $A\otimes M\to M$ such that $a\otimes x\mapsto ax$.

\begin{proof} i) Consider $f:M\times N\to N\otimes M,(x,y)\mapsto y\otimes x$ and $g:M\times N\to M\otimes N,(x,y)\mapsto x\otimes y$. By ^aa5afc, there exists unique $f^{\prime }:M\otimes N\to N\otimes M$ such that the diagram commutes.

Now, in a parallel way, use the universal property of $N\otimes M$, then we get $\widetilde{f^{\prime }}(n\otimes m)=m\otimes n$ such that the diagram commutes.

It is easy to check $f^{\prime }$ and $\widetilde{f^{\prime }}$ are inverse to each other.

iv) Consider

where $f:A\times M\to M,(a,m)\mapsto am$, then there exists unique $f^{\prime }:A\otimes M\to M$ such that the diagram commutes. Now define $h:M\to A\otimes M,m\mapsto 1\otimes m$. To check $f^{\prime }$ and $h$ are inverse to each other, it suffices to check $a\otimes m=1\otimes am$.

Recall the proof of ^aa5afc, let $C$ be the free $A$-module $A^{M\times N}$ and $D$ be the $A$-submodule generated by $(x+x^{\prime },y)-(x,y)-(x^{\prime },y)$, $(x,y+y^{\prime })-(x,y)-(x,y^{\prime })$, $(ax,y)-a(x,y)$ and $(x,ay)-a(x,y)$. Let $T=C/D$, and let $g:M\times N\to T,(m,n)\mapsto m\otimes n$. So we must have

$$(a,m)-(1,am)=(a,m)-a(1,m)+a(1,m)-(1,am)\in D$$

which yields $g((a,m))=g((1,am))$ and so $a\otimes m=1\otimes am$. \end{proof}

Proposition

Let $M_1,\cdots ,M_r$ be $A$-module. Then there exists $(T,g)$ where $g:M_1\times \cdots \times M_r\to T$ is a multilinear map such that for any $A$-module $P$ and multilinear map $f:M_1\times \cdots \times M_r\to P$, there exists unique $f$ such that the diagram commutes.

\begin{proof} Similar to ^aa5afc. \end{proof}

Strange Things about Tensor Product

Consider $M^{\prime }\subseteq M$ and $N^{\prime }\subseteq N$, then $M^{\prime }\times N^{\prime }\hookrightarrow M\times N$ is submodule. However, it could happen $M^{\prime }\otimes N^{\prime }\to M\otimes N$ is not injective. Use the following diagram to define the map.

Example. Let $A=\mathbb{Z}$, $M=\mathbb{Z}\cdot e\supseteq M^{\prime }=\mathbb{Z}\cdot (2e)$, and let $N=\mathbb{Z}/2\mathbb{Z}\cdot f=N^{\prime }$. So we have

$$\mathbb{Z}\cdot (2e){\otimes }_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z})\cdot f\to \mathbb{Z}\cdot e{\otimes }_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z})\cdot f,2e\otimes f\mapsto 2e\otimes f=e\otimes 2f=0.$$

Claim that $2e\otimes f\ne 0$ in LHS. In fact, we have the following commutative diagram.

Here we use the isomorphism $(\mathbb{Z}\cdot 2e){\otimes }_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z})\cdot f\stackrel{\sim }{\to }\mathbb{Z}\cdot e\otimes (\mathbb{Z}/2\mathbb{Z}\cdot f),2e\otimes f\mapsto e\otimes f$. Recall that $(\mathbb{Z}\cdot e){\otimes }_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z}\cdot f)\simeq (\mathbb{Z}/2\mathbb{Z})\cdot f$ by ^9968ad. So the bottom row being zero map implies the top row is zero map.

Remark. A (practical) guide to tensor product:

• $x\in M{\otimes }_{A}N$ is of form $x={\sum }_{i=1}^N{m}_i\otimes n_i$
• get familiar with common operation on $\otimes$
• always keep in mind the expression $x=\sum m_i\otimes n_i$ is NOT unique
• the universal property are not really used in practice like ^9968ad (you should remember and be able to apply ^9968ad)
• common pitfall: $2x\otimes y\in M{\otimes }_{\mathbb{Z}}N$ and it is not equal to $1/2x\otimes 4y$

Bimodule

Definition

Let $A,B$ be rings. We say $M$ is an $(A,B)$-bimodule, if it is both an $A$-left module and a $B$-right module, and $(ax)b=a(xb)$ for any $a\in A$, $b\in B$ and $x\in M$. Remark that $ab$ might not exist.

Remark. It has been defined in ^559amp, but I write it again.

Text-Ex. 2.15.

Let $A,B$ be rings. Let $M$ be an $A$-module, $P$ be a $B$-module and let $N$ be a $(A,B)$-bimodule. Then $M{\otimes }_{A}N$ is a $B$-module and $N{\otimes }_{B}P$ is an $A$-module, and we have an isomorphism of $(A,B)$-bimodule

$$(M{\otimes }_{A}B){\otimes }_{B}P\simeq M{\otimes }_A(N{\otimes }_{B}P).$$

\begin{proof} It is easy to check $M{\otimes }_{A}N$ is a $B$-module and $N{\otimes }_{B}P$ is an $A$-module. The proof of isomorphism of $(A,B)$-bimodule is similar to ^9968ad, ii).

Fix $z\in P$, define $M\times N\to M{\otimes }_A(N{\otimes }_{B}P),(x,y)\to x\otimes (y\otimes z)$. It is $A$-bilinear and it induces an $A$-linear map $f_z:M{\otimes }_{A}N\to M{\otimes }_A(N{\otimes }_{B}P),x\otimes y\mapsto x\otimes (y\otimes z)$.

Define $(M{\otimes }_{A}N)\times P\to M{\otimes }_A(N{\otimes }_{B}P),(t,z)\mapsto f_z(t)$, and it is $B$-bilinear. Son it induces a $B$-linear map $(M{\otimes }_{A}N){\otimes }_{B}P\to M{\otimes }_A(N{\otimes }_{B}P)$. To prove it is isomorphism, it suffices to construct its inverse, which can be done similarly. \end{proof}

Corollary

Suppose ${\sum }_{i=1}^n{x}_i\otimes y_i=0$ in $M{\otimes }_{A}N$, with $x_i\in M,y_i\in N$. Then there exists some finitely generated submodule $M_0\subseteq M$ and $N_0\subseteq N$ such that ${\sum }_{i=1}^n{x}_i\otimes y_i=0$ in $M_0{\otimes }_A{N}_0$. Remark that $M_0\otimes N_0\to M\otimes N$ is not injective, see Strange Things about Tensor Product as example.

\begin{proof} Since ${\sum }_{i=1}^n{x}_i\otimes y_i=0$ in $M{\otimes }_{A}N=C/D$ with $C=A^{M\times N}$, there is ${\sum }_{i=1}^n{x}_i\otimes y_i\in D$. Then

$$\begin{array}{rl}\sum _{i=1}^n{x}_i\otimes y_i=& a_1((x_1+x_1^{\prime },y_1)-(x_1,y_1)-(x_1^{\prime },y_1))+a_2((x_2+x_2^{\prime },y_2)-(x_2,y_2)-(x_2^{\prime },y_2))+\cdots \\ +& b_1(({\alpha }_1,{\beta }_1+{\beta }_1^{\prime })-({\alpha }_1,{\beta }_1)-({\alpha }_1,{\beta }_1^{\prime }))+b_2(({\alpha }_2,{\beta }_2+{\beta }_2^{\prime })-({\alpha }_2,{\beta }_2)-({\alpha }_2,{\beta }_2^{\prime }))+\cdots \\ +& c_1((a\gamma ,\delta )-a(\gamma ,\delta ))+\cdots \\ +& d_1((ϵ,b\theta )-b(ϵ,\theta ))+\cdots \end{array}\text{\hspace{1em}(*)}$$

Let $M_0$ be the submodule generated by $(x_1+x_1^{\prime },x_2+x_2,\cdots ,{\alpha }_1,\cdots ,a\gamma ,\cdots ,{ϵ}_1,\cdots )$ and $x_1,\cdots ,x_n$. Similarly define $N_0$ be the submodule generated by the second coordinates on RHS and $y_1,\cdots ,y_n$. We claim that RHS of $(\ast )$ in $M_0\otimes N_0$ is also an element in $D_{M_0\times N_0}\subseteq C_{M_0\times N_0}$. \end{proof}

Restriction and Extension of Scalars

Definition

Let $f:A\to B$ be a ring homomorphism, and let $N$ be a $B$-module. Then we can also regard $N$ as an $A$-module, so that, for any $a\in A$ and $x\in N$, define $ax=f(a)x$. This $A$-module is said to be obtained from $N$ via restriction of scalars.

Example. Define $f:\mathbb{Z}\to \mathbb{Q}$ and $N$ is a vector space over $\mathbb{Q}$. Then $N$ can be regarded as a $\mathbb{Z}$-module.

Proposition

Let $f:A\to B$ and $N$ be a $B$-module as above. Suppose $B$ is finitely generated as $A$-module and $N$ is a finitely generated $B$-module. Then $N$ is also finitely generated as $A$-module.

\begin{proof} Let $B={⟨b_1,\cdots ,b_n⟩}_A$ and $N={⟨n_1,\cdots ,n_s⟩}_B$. Then $N={⟨b_i{n}_j:1⩽i⩽n,1⩽j⩽n⟩}_A$. \end{proof}

Definition

Let $f:A\to B$ be a ring homomorphism. Since $M$ is a $A$-module, we can define $M_B=B{\otimes }_{A}M$. This is an $A$-module, and it is indeed an $(B,A)$-bimodule via $b(b^{\prime }\otimes m)a:=b{b}^{\prime }{\otimes }_{A}am$. We say this $B$-module $M_B$ is obtained via extension of scalars.

Example. Define $f:\mathbb{Z}\to \mathbb{Q}$ and $M=\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$. Then $M_{\mathbb{Q}}=(\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}){\otimes }_{\mathbb{Z}}\mathbb{Q}=\mathbb{Q}$, because

• $\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Q}=\mathbb{Q}$ and
• $\mathbb{Q}{\otimes }_{\mathbb{Z}}\mathbb{Z}/2\mathbb{Z}=0$ by $x\otimes y=x/2\otimes 2y=0$.

Proposition

Assume $f:A\to B$ be a ring homomorphism and $M$ is a finitely generated $A$-module. Then $M_B=B{\otimes }_{A}M$ is a finitely generated $B$-module.

\begin{proof} Since $M={⟨x_1,\cdots ,x_n⟩}_A$, we have $M_B={⟨1\otimes x_1,\cdots ,1\otimes x_n⟩}_B$. \end{proof}

Exactness Property of the Tensor Product

Theorem

Let $M,N,P$ be $A$-modules, then we have natural isomorphism

$$\varphi :{\mathrm{Hom}}_A(M{\otimes }_{A}N,P)\to {\mathrm{Hom}}_A(M,{\mathrm{Hom}}_A(N,P)).$$

\begin{proof} Take $f$ in LHS, and define $\varphi (f)$ as follows. For each fixed $x\in M$, the map $f_x:N\to P,n\mapsto f(x\otimes n)$. It is $A$-linear and so $f_x\in {\mathrm{Hom}}_A(N,P)$. It is easy to check the map

$$\varphi (f):M\to {\mathrm{Hom}}_A(N,P),x\mapsto f_x$$

is $A$-linear.

Conversely, given $g$ in RHS. Define

$$M\times N\to P,(m,n)\mapsto g(m)(n),$$

which is $A$-bilinear and it induces $\psi (g):M\otimes N\to P$.

Finally, check $\psi$ and $\varphi$ are inverse each other. \end{proof}

Example. Recall $\mathbb{Z}/m{\otimes }_{\mathbb{Z}}\mathbb{Z}/n\simeq \mathbb{Z}/(m,n)$ (See here or here). We claim that ${\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}/m\mathbb{Z},\mathbb{Z}/n\mathbb{Z})\simeq \mathbb{Z}/(m,n)\mathbb{Z}$. Note that any $f$ in LHS is completely determined by $f(\overline{1})$. Since $f(\overline{m})=0$, $f(\overline{1})$ is killed by $m$. That is, for any lift $x\in \mathbb{Z}$ of $f(\overline{1})$, we have $n\mid mx$ and so $n/(m,n)\mid x$. Therefore, ${\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}/m\mathbb{Z},\mathbb{Z}/n\mathbb{Z})\simeq n/(m,n)\mathbb{Z}/n\mathbb{Z}\simeq \mathbb{Z}/(m,n)\mathbb{Z}$.

Example. Put $\mathbb{Z}/12\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/6\mathbb{Z}$ to ^60fa98, then we have

$$\mathrm{Hom}(\mathbb{Z}/12\mathbb{Z}\otimes \mathbb{Z}/2\mathbb{Z},\mathbb{Z}/6\mathbb{Z})\simeq \mathrm{Hom}(\mathbb{Z}/12\mathbb{Z},\mathrm{Hom}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/6\mathbb{Z}))$$

where LHS $\simeq \mathrm{Hom}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/6\mathbb{Z})\simeq \mathbb{Z}/2\mathbb{Z}$ and RHS $\simeq \mathrm{Hom}(\mathbb{Z}/12\mathbb{Z},\mathbb{Z}/2\mathbb{Z})\simeq \mathbb{Z}/2\mathbb{Z}$.

Proposition

Let $M^{\prime }\to M\to M^{\prime \prime }\to 0(\ast )$ be a right exact sequence of $A$-modules. Let $N$ be any $A$-module. Then the sequence

$$M^{\prime }{\otimes }_{A}N\stackrel{\tilde{f}}{\to }M{\otimes }_{A}N\stackrel{\tilde{g}}{\to }{M}^{\prime \prime }{\otimes }_{A}N\to 0\text{\hspace{1em}(**)}$$

is still right exact.

\begin{proof} Note that it is easy to see $\tilde{g}$ is surjective. It remains to check $\ker \tilde{g}\simeq \mathrm{im}\tilde{f}$. By ^jj28ws, it suffices to show for any $A$-module $P$,

$$0\to \mathrm{Hom}(M^{\prime }\otimes N,P)\to \mathrm{Hom}(M\otimes N,P)\to \mathrm{Hom}(M^{\prime \prime }\otimes N,P)$$

is left exact. By ^60fa98, it is equivalent to show

$$0\to \mathrm{Hom}(M^{\prime },\mathrm{Hom}(N,P))\to \mathrm{Hom}(M,N\otimes P)\to \mathrm{Hom}(M^{\prime \prime },N\otimes P)$$

is left exact. By ^jj28ws, we have done. \end{proof}

Remark.

• $\text{right}\to 0$ is necessary. For example, $0\to \mathbb{Z}\stackrel{\times 2}{\to }\mathbb{Z}$ is exact, but after tensor $\mathbb{Z}/2\mathbb{Z}$ we get $0\to \mathbb{Z}/2\mathbb{Z}\to 0$, which is not exact.
• a left $0\to$ does not product similar result. For example, $0\to \mathbb{Z}\stackrel{\times 2}{\to }\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}\to 0$ is exact, but after tensor $\mathbb{Z}/2\mathbb{Z}$ we get $0\to \mathbb{Z}/2\mathbb{Z}\stackrel{\times 2}{\to }\mathbb{Z}/2\mathbb{Z}\to Z/2\mathbb{Z}\to 0$ is right exact but not left exact.

Definition

We say an $A$-module is a flat $A$-module, if for any exact sequence

$$\cdots \to M_{i-1}\to M_i\to M_{i+1}\to \cdots ,\text{\hspace{1em}(*)}$$

the tensor product sequence $(\ast )\otimes N$ is still exact.

Examples.

• $\mathbb{Z}/2\mathbb{Z}$ is not flat by it.
• The $A$-module $A$ is a flat $A$-module.
• $A^{\oplus n}$ is flat $A$-module.

Proposition

Let $N$ be a $A$-module. TFAE:

• $N$ is flat.
• For any short exact sequence $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0(\ast )$, $(\ast ){\otimes }_{A}N$ is still short exact.
• For any injective map $f:M^{\prime }\hookrightarrow M$, $f\otimes 1$ is also injective.
• For any injective map $f:M^{\prime }\hookrightarrow M$ with $M^{\prime },M$ finitely generated, $f\otimes 1$ is still injective.

\begin{proof} i)←>ii) is trivial as each long exact sequence can be split into short exact sequence.

iii)→iv) is easy.

i)=ii)→iii) by considering $0\to M^{\prime }\to M$ exact.

iii)→ii) When iii) holds, $\otimes$ preserves right exactness. Furthermore, ${\otimes }_{A}N$ preserves left exactness by ^8m15t3 and so for short exact sequence.

It remains to show iv)→iii). Suppose $u={\sum }_{i=1}^n{x}_i^{\prime }\otimes y_i\in \ker (M^{\prime }\otimes N\to M\otimes N)$, that is, $f(u)={\sum }_{i=1}^{n}f(x_i^{\prime })\otimes y_i=0$. We aim to show $u=0$. Let $M_0^{\prime }={⟨x_1^{\prime },\cdots ,x_n^{\prime }⟩}_A\subseteq M^{\prime }$. Let $u_0={\sum }_{i=1}^n{x}_i^{\prime }\otimes y_i$ be an element in $M_0^{\prime }\otimes N$. Remark that we do not know if $u_0$ is also zero, as $M_0\otimes N$ may not a submodule of $M^{\prime }\otimes N$ by Strange Things about Tensor Product.

^6zw2e4 says that for ${\sum }_{i=1}^{n}f(x_i^{\prime })\otimes y_i=0\in M\otimes N$, there exists finitely generated submodule $M_0\subseteq M$ containing $f(M_0^{\prime })$ and finitely generated submodule $N_0\subseteq N$ such that ${\sum }_{i=1}^{n}f(x_i^{\prime })\otimes y_i=0$ in $M_0\otimes N_0$. So in particular, from injective map $f:M^{\prime }\hookrightarrow M$, we have injective map $f:M_0^{\prime }\hookrightarrow M_0$. Since $M_0^{\prime }$ and $M_0$ are finitely generated, by iv) we have $M_0^{\prime }\otimes N\to M_0\otimes N$ is injective. Note that ${\sum }_{i=1}^{n}f(x_i)\otimes y_i=0$ in $M_0\otimes N$, then we have $u_0={\sum }_{i=1}^n{x}_i^{\prime }\otimes y_i$ is $0$ in $M_0^{\prime }\otimes N$. It deduces that ${\sum }_{i=1}^n{x}_i^{\prime }\otimes y_i=0$ in $M^{\prime }\otimes N$. \end{proof}

Text-Ex. 2.20.

Let $f:A\to B$ be a ring homomorphism and let $N$ be a flat $A$-module. Then $N_B=N{\otimes }_{A}B$ is a flat $B$-module. Remark that $N_B$ in general is not flat $A$-module.

\begin{proof} Let $M^{\prime }\hookrightarrow M$ be an injective map of $B$-modules. It remains to show $M^{\prime }{\otimes }_B{N}_B\to M{\otimes }_B{N}_B$ is still injective. But

$$M^{\prime }{\otimes }_B{N}_B=M^{\prime }{\otimes }_B(B{\otimes }_{A}N)=(M^{\prime }{\otimes }_{B}B){\otimes }_{A}N=M^{\prime }{\otimes }_{A}N$$

and then the map $M^{\prime }{\otimes }_{A}N\to M{\otimes }_{A}N$ is injective by $N$ being a flat $A$-module. \end{proof}

Remark. Let $f:\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ and $N=\mathbb{Z}$. But $N_B=\mathbb{Z}/2\mathbb{Z}$ is not flat over $\mathbb{Z}$.

Algebra and Tensor Product of Algebra

Definition

Given a ring homomorphism $f:A\to B$, we say $B$ is an $A$-algebra. Note that $B$ has an $A$-module structure which is compatible with ring struction on $B$, that is, $(f(a)b)c=f(a)(bc)$.

Examples.

• Any ring is a $\mathbb{Z}$-algebra.
• Field extension $E/K$ is a $K$-algebra with $K\hookrightarrow E$.

Definition

Given $f:A\to B$ and $g:A\to C$, a ring homomorphism $h:B\to C$ is called a homomorphism of $A$-algebra if it is also a $A$-module homomorphism (i). Equivalently, $h$ is an $A$-algebra homomorphism iff the following diagram commute (ii).

\begin{proof} i)→ii). ii) means $h(f(a))=g(a)$. Note that if i) holds, then

$$h(f(a))=h(a\cdot 1_B)=a\cdot h(1_B)=a\cdot 1_C=g(a).$$

ii)→i) It suffices to show $h(ab)=ah(b)$. By

$$h(ab)=h(f(a)b)=h(f(a))h(b)=g(a)h(b)=ah(b)$$

we finish the proof. \end{proof}

Example. Recall given field extensions $E/K$ and $F/K$, $E$ and $F$ are $K$-algebras and $f:E\to F$ is a field homomorphism. $f$ is also a homomorphism of $K$-vector spaces iff $f$ is a homomorphism between $k$-algebra iff $f:E\to F$ is a homomorphism such that $f(k)=k$ for any $k\in K$ iff the diagram commutes.

Definition

• Say $f:A\to B$ is a finite morphism of rings, if $B$ is a finitely generated $A$-module. In this case, say $B$ is a finite $A$-algebra.
• Say $f:A\to B$ is of finite type, if there exists $x_1,\cdots ,x_n\in B$ such that any $b\in B$ can be written as a polynomial in $x_1,\cdots ,x_n$ with coefficient in $f(A)$. It is equivalent to there exists surjective map $A[t_1,\cdots ,t_n]\twoheadrightarrow B,t_i\to x_i$. In this case, say $B$ is a finitely generated $A$-algebra.

So finite $A$-algebra is a finitely generated $A$-algebra.

Examples.

• $\mathbb{Q}\to \mathbb{Q}(i)$ is a finite $\mathbb{Q}$-algebra
• $\mathbb{Q}\to \mathbb{Q}[x,y]$ is of finite type
• $\mathbb{Q}\to \mathbb{Q}[x,y]/(x^2+y^2)$ is of finite type
• $\mathbb{Q}\to \mathbb{Q}[x,y]/(x^2,y^2)$ is a finite $\mathbb{Q}$-algebra

Definition

For a ring $A$, we say it is finitely generated if it is finitely generated as $\mathbb{Z}$-algebra.

Example. $A=\mathbb{Z}[x,y]/(x^2+y^2)$

a Construction on Page 31. Let $f:A\to B$ and $g:A\to C$ be two ring homomorphisms. Let $D=B{\otimes }_{A}C$. Claim that we can define ring structure on $D$ and make it an $A$-algebra.

\begin{proof} Consider $B\times C\times B\times C\to D,(b,c,b^{\prime },c^{\prime })\mapsto b{b}^{\prime }\otimes c{c}^{\prime }$. By ^ml3hwi above, it induces an $A$-linear map

By ^aa5afc, this is the same as an $A$-bilinear map $\mu :D\times D\to D$ with $\mu (b\otimes c,b^{\prime }\otimes c^{\prime })=b{b}^{\prime }\otimes c{c}^{\prime }$. One can check this $\mu$ (used as multiplication) is compatible with $+$ on $D$, giving $D$ a ring structure.

Define $A\to D$ by $a\mapsto f(a)\otimes 1$. Note that $f(a)\otimes 1=a\cdot 1_V\otimes 1_C=1_B\otimes a\cdot 1_C=1\otimes g(a)$. It is easy to see it is a homomorphism of rings.

Remark.

• page 31 of Atiyah is not correct, as $\mathbb{Z}\to \mathbb{Z}\otimes \mathbb{Z}$ is not a ring homomorphism.

• In fact there is a commutative diagram of ring homomorphism

  

Example (strange $\otimes$). Define $D=\mathbb{Q}(i){\otimes }_{\mathbb{Q}}\mathbb{Q}(i)$ is not a domain, because

$$(1\otimes i+i\otimes 1)(1\otimes i-i\otimes 1)=1\otimes (-1)-(-1)\otimes 1=0.$$

To see the above two elements are nonzero, notice that $D=(\mathbb{Q}1\oplus \mathbb{Q}i)\otimes (\mathbb{Q}1\oplus \mathbb{Q}i)$ is a vector space over $\mathbb{Q}$ and

$$\begin{array}{rl}D& =(\mathbb{Q}1\oplus \mathbb{Q}i)\otimes (\mathbb{Q}1\oplus \mathbb{Q}i)\\ & =\mathbb{Q}1\otimes \mathbb{Q}1\oplus \mathbb{Q}i\otimes \mathbb{Q}1\oplus \mathbb{Q}1\otimes \mathbb{Q}i\oplus \mathbb{Q}i\otimes \mathbb{Q}i\\ & \simeq \mathbb{Q}\oplus \mathbb{Q}\oplus \mathbb{Q}\oplus \mathbb{Q}\end{array}$$

where $\mathbb{Q}1{\otimes }_{\mathbb{Q}}\mathbb{Q}1\simeq \mathbb{Q}{\otimes }_{\mathbb{Q}}\mathbb{Q}=\mathbb{Q}$. Hence $D$ is a $4$-dimensional $\mathbb{Q}$-v.s. with a basis $1\otimes 1,1\otimes i,i\otimes 1,i\otimes i$. So the above two elements are nonzero.

Examples.

• $\mathbb{C}[x]\otimes \mathbb{C}[y]\simeq \mathbb{C}[x,y]$. Each element on LHS has a unique expression as $\mathbb{C}[x],\mathbb{C}[y]$ are linear vector spaces, then $\phi :\sum a_{ij}{x}^i\otimes y^j\mapsto \sum a_{ij}{x}^i{y}^j$ is well-defined.
• $\mathbb{C}[[x]]{\otimes }_{\mathbb{C}}\mathbb{C}[[y]]\ne \mathbb{C}[[x,y]]$. The problem is, $\{x^n:n⩾0\}$ is NOT a basis of $\mathbb{C}[[x]]$. “For example”, consider $f={\sum }_{i=0}^n(xy{)}^i$ and ${\sum }_{i=0}^n{x}^i\otimes y^i\notin \text{LHS}$.

Proposition

Let $\alpha \subseteq A$ be an ideal, and let $M$ be an $A$-module. Then $M/\alpha M\simeq (A/\alpha ){\otimes }_{A}M$.

\begin{proof} Consider short exact sequence $0\to \alpha \to A\to A/\alpha \to 0$. Tensor with $M$ and we get a right exact sequence

$$\alpha {\otimes }_{A}M\stackrel{f}{\to }M\to A/\alpha \otimes M\to 0.$$

Recall the notation $\alpha M=\{{\sum }_{i=1}^N{a}_i{m}_i:a_i\in \alpha ,m_i\in M\}\subseteq M$. It is easy to see $\mathrm{im}f=\alpha M$ (remark that in general $\alpha \otimes M\to \alpha M$ is not injective). So we get a short exact sequence $0\to \alpha M\hookrightarrow M\to A/\alpha {\otimes }_{A}M\to 0$. So $M/\alpha M\simeq (A/\alpha ){\otimes }_{A}M$. \end{proof}

Remark. This is an example where $\alpha \otimes M\to \alpha M$ is not injective. Let $\alpha =(2)\subseteq \mathbb{Z}$, and let $M=\mathbb{Z}/2\mathbb{Z}$. Then $0\ne (2\mathbb{Z}){\otimes }_{\mathbb{Z}}\mathbb{Z}/2\mathbb{Z}$ and $(2)\cdot \mathbb{Z}/2\mathbb{Z}=0$.

a useful fact

Let $\alpha \subseteq A$ be an ideal. Let $M,N$ be two $A/\alpha$-modules, then one have an $A$-module isomorphism

$$M{\otimes }_{A/\alpha }N\simeq M{\otimes }_{A}N.$$

Remark. For general $A\to B$ and two $B$-modules $X$ and $Y$, $X{\otimes }_{B}Y\not\simeq X{\otimes }_{A}Y$.

\begin{proof} Consider

Define $g_A:M\times N\to M{\otimes }_{A}N,(m,n)\mapsto m{\otimes }_{A}n$, then $g_A$ is $A/\alpha$-bilinear and there exists $f^{\prime \prime }:M{\otimes }_{A/\alpha }N\to M{\otimes }_{A}N$ such that $f^{\prime \prime }\circ g_A=g_{A/\alpha }$. Similarly, define $g_{A/\alpha }:M\times N\to M{\otimes }_{A/\alpha }N,(m,n)\mapsto m{\otimes }_{A/\alpha }n$, then $g_{A/\alpha }$ is $A$-bilinear and there exists $f^{\prime }:M{\otimes }_{A/\alpha }N\to M{\otimes }_{A}N$ such that $f^{\prime }\circ g_{A/\alpha }=g_A$. So $f^{\prime }=f^{\prime \prime -1}$ and so $M{\otimes }_{A/\alpha }N\simeq M{\otimes }_{A}N$. \end{proof}