Add tensor product constructions for algebras and modules

Introduced definitions and properties of tensor products for R-algebras, including universal property, homomorphisms, and symmetric monoidal structure. Expanded module notes with bimodule examples, pure tensors, and bimodule structure on tensor products.

Author: hooyuser

associative_algebra.tex

@@ -89,7 +89,120 @@ \subsection{Graded Object}
     Let $R$ be a commutative ring and $X_1,\cdots,X_n$ be indeterminates. Then $R[X_1,\cdots,X_n]$ is an $\mathbb{N}$-graded $R$-algebra with grading $R[X_1,\cdots,X_n]_i$ being the set of homogeneous polynomials of degree $i$.
 \end{example}
 
+\subsection{Tensor Product}
+\begin{definition}{Tensor Product of Algebras}{}
+    Let $R$ be a commutative ring and $A$, $B$ be $R$-algebras. The \textbf{tensor product of $R$-algebras $A$ and $B$} is defined by the following universal property: for any triple $(C, f_A, f_B)$, where $C$ is an $R$-algebra and $f_A:A\to C$, $f_B:B\to C$ are $R$-algebra homomorphisms which satisfy
+    \[
+        f_A(a)f_B(b)=f_B(b)f_A(a),\quad \forall a\in A, b\in B,
+    \]
+    the tensor product
+    \[
+    (A\otimes_R B, \iota_A: A \times B \to A\otimes_R B, \iota_B: A \times B \to A\otimes_R B)
+    \]
+    is initial among such triples, i.e. there exists a unique $R$-algebra homomorphism \begin{align*}
+        \phi: A\otimes_R B &\longrightarrow C
+    \end{align*}
+    
+    such that the following diagram commutes
+    \[
+        \begin{tikzcd}
+        A \arrow[r, "\iota_A"] \arrow[rd, "f_A"'] & A\otimes_R B \arrow[d, "\exists! \phi", dashed] & B \arrow[l, "\iota_B"'] \arrow[ld, "f_B"] \\[0.5em]
+                                                & C                                               &                                          
+        \end{tikzcd}
+    \]
+
+    Concretely, $A\otimes_R B$ can be constructed as the tensor product of $R$-modules $A\otimes_R B$ together with multiplication defined as
+    \[
+        (a_1\otimes b_1)(a_2\otimes b_2):=(a_1a_2)\otimes(b_1b_2),\quad \forall a_1,a_2\in A, b_1,b_2\in B
+    \]
+    and unity 
+    \[
+          1_{A\otimes_R B}:=1_A\otimes 1_B.
+    \]
+    And the $R$-algebra homomorphisms $\iota_A$, $\iota_B$ are defined as
+    \begin{align*}
+        \iota_A : A &\longrightarrow A\otimes_R B\\
+        a &\longmapsto a\otimes 1_B,
+    \end{align*}
+    \begin{align*}
+        \iota_B : B &\longrightarrow A\otimes_R B\\
+        b &\longmapsto 1_A\otimes b.
+    \end{align*}
+    The unique $R$-algebra homomorphism $\phi:A\otimes_R B\to C$ is defined as
+    \begin{align*}
+        \phi: A\otimes_R B &\longrightarrow C\\
+        a\otimes b &\longmapsto f_A(a)f_B(b).
+    \end{align*}
+\end{definition}
+\begin{remark}
+    It is straightforward to check that the multiplication defined above is well-defined and makes $A\otimes_R B$ an $R$-algebra. According to \Cref{th:pure_tensors_generate_tensor_product}, since $(a,b)\mapsto f_A(a)f_B(b)$ is $\mathbb{Z}$-bilinear, $\phi$ is a well-defined abelian group homomorphism. We can further check that $\phi$ is an $R$-algebra homomorphism:
+    \[
+    \phi(r(a\otimes b))=\phi((r a)\otimes b)=f_A(r a)f_B(b)=r f_A(a)f_B(b)=r \phi(a\otimes b)
+    \]
+    \[
+    \phi\left((a_1\otimes b_1)(a_2\otimes b_2)\right)=\phi\left((a_1a_2)\otimes(b_1b_2)\right)= f_A(a_1a_2)f_B(b_1b_2)=f_A(a_1)f_A(a_2)f_B(b_1)f_B(b_2)=\phi(a_1\otimes b_1)\phi(a_2\otimes b_2)
+    \]
+    \[
+\phi\left(1_{A\otimes_R B}\right)=\phi\left(1_A\otimes 1_B\right)=f_A(1_A)f_B(1_B)=1_C
+    \]
+\end{remark}
 
+\begin{definition}{Tensor Product of $R$-algebra Homomorphisms}{}
+    Let $R$ be a commutative ring and $A_1$, $A_2$, $B_1$, $B_2$ be $R$-algebras. Given two $R$-algebra homomorphisms $f:A_1\to A_2$ and $g:B_1\to B_2$, the \textbf{tensor product of $R$-algebra homomorphisms} is defined as the $R$-algebra homomorphism 
+    \begin{align*}
+        f\otimes_R g: A_1\otimes_R B_1 &\longrightarrow A_2\otimes_R B_2\\
+        a\otimes b &\longmapsto f(a)\otimes g(b).
+    \end{align*}
+    which is induced by the universal property of tensor product $A_1\otimes_R B_1$ through the following commutative diagram:
+    \[
+    \begin{tikzcd}
+        A_1 \arrow[r, "\iota_{A_1}"] \arrow[d, "f"']                              & A_1\otimes_R B_1 \arrow[d, "f\otimes_R g", dashed] & B_2 \arrow[l, "\iota_{B_1}"'] \arrow[d, "g"] \\[0.7em]
+        A_2  \arrow[r, "\iota_{A_2}"'] & A_2\otimes_R B_2                                   & B_2 \arrow[l, "\iota_{B_2}"]                
+        \end{tikzcd}
+    \] 
+\end{definition}
+
+
+\begin{proposition}{Symmetric Monoidal Structure on $R$-$\mathsf{Alg}$}{}
+    Let $R$ be a commutative ring. The tensor product $\otimes_R$ defines a symmetric monoidal structure on the category $R$-$\mathsf{Alg}$, with unit object $R$.
+    \begin{enumerate}[(i)]
+        \item Tensor product: the tensor product functor is 
+        \[
+        \begin{tikzcd}[ampersand replacement=\&]
+            R\text{-}\mathsf{Alg}\times R\text{-}\mathsf{Alg}\&[-25pt]\&[+10pt]\&[-30pt] R\text{-}\mathsf{Alg}\&[-30pt]\&[-30pt] \\ [-15pt] 
+            (A_1, B_1)  \arrow[dd, "f\times g"{name=L, left}] 
+            \&[-25pt] \& [+10pt] 
+            \& [-30pt] A_1\otimes_R B_1\arrow[dd, "f\otimes_R g"{name=R}] \&[-30pt]\\ [-10pt] 
+            \&  \phantom{.}\arrow[r, "\otimes_R", squigarrow]\&\phantom{.}  \&   \\[-10pt] 
+            (A_2, B_2) \& \& \&  A_2\otimes_R B_2\&
+        \end{tikzcd}
+        \]  
+        \item Associator: for any $R$-algebras $A$, $B$, $C$, there is a natural isomorphism
+        \begin{align*}
+            \alpha_{A,B,C}:(A\otimes_R B)\otimes_R C &\xlongrightarrow{\sim}A\otimes_R (B\otimes_R C)\\
+            (a\otimes b)\otimes c &\longmapsto a\otimes (b\otimes c)
+        \end{align*}
+        \item Unit object: $R$.
+        \item An isomorphism in $R$-$\mathsf{Alg}$:
+        \begin{align*}
+            \iota: R \otimes_R R &\xlongrightarrow{\sim} R\\
+            r\otimes r' &\longmapsto rr'
+        \end{align*}
+        \item Symmetry: for any $R$-algebras $A$, $B$, there is a natural isomorphism
+        \begin{align*}
+            \gamma_{A,B}: A\otimes_R B &\xlongrightarrow{\sim} B\otimes_R A\\
+            a\otimes b &\longmapsto b\otimes a
+        \end{align*}
+    \end{enumerate}
+\end{proposition}
+
+\begin{proposition}{}{}
+    Let $R$ be a commutative ring and $A_1$, $A_2$ be $R$-algebras. Let $I_1\subseteq A_1$, $I_2\subseteq A_2$ be two-sided ideals of $A_1$, $A_2$ respectively. Then we have an $R$-algebra isomorphism
+    \[
+        (A_1\otimes_R A_2)/(I_1\otimes_R A_2 + A_1\otimes_R I_2) \cong (A_1/I_1)\otimes_R (A_2/I_2).
+    \]
+    
+\end{proposition}
 
 \subsection{Tensor Algebra}
 \begin{definition}{Tensor Algebra $T^{\bullet}(M)$}{}

module.tex

@@ -199,7 +199,7 @@ \section{Basic Concepts}
 \end{proposition}
 
 \begin{prf}
-    We prove $(i)\Rightarrow(ii)\Rightarrow(iii)\Rightarrow(i)$.
+    We prove $\text{(i)}\Rightarrow(ii)\Rightarrow(iii)\Rightarrow(i)$.
 
 \paragraph{$(i)\;\Rightarrow\;(ii)$}
 Assume $M$ is simple and let $0\neq m\in M$.  
@@ -385,13 +385,26 @@ \subsection{Free Object}
 \subsection{Tensor Product}
 
 \begin{definition}{Bimodule}{bimodule}
-    Let $R$ and $S$ be rings. An ($R$,$S$)-\textbf{bimodule} is an abelian group $M$ together with a structure of left $R$-module and a structure of right $S$-module such that
+    Let $R$ and $S$ be rings. An $(R, S)$-\textbf{bimodule} is an abelian group $M$ together with a structure of left $R$-module and a structure of right $S$-module such that
     \[
         r(ms)=(rm)s     
     \]
     for all $r\in R$, $s\in S$, and $m\in M$.
 \end{definition}
 
+\begin{example}{Examples of Bimodule}{}
+    Let $R$ and $S$ be rings.
+    \begin{itemize}
+        \item If $M$ is a left $R$-module, then $M$ is an $(R,\mathbb{Z})$-bimodule. 
+        \item If $M$ is a right $S$-module, then $M$ is a $(\mathbb{Z},S)$-bimodule.
+        \item If $R$ is a commutative ring, then every left $R$-module is an $(R,R)$-bimodule, and every right $R$-module is also an $(R,R)$-bimodule. In particular, every abelian group is a $\mathbb{Z}$-module, hence a $(\mathbb{Z},\mathbb{Z})$-bimodule.
+        \item Any two-sided ideal of a ring $R$ is an $(R,R)$-bimodule, with the ring multiplication both as the left and as the right multiplication. In particular, $R$ itself is an $(R,R)$-bimodule.
+        \item If $M$ is a right $R$-module, then $M$ is an $(\mathrm{End}_{\mathsf{Mod}\text{-}R}(M),R)$-bimodule, where the left multiplication is given by function application.
+        \item If $M$ is a left $R$-module, then $M$ is an $(R,\mathrm{End}_{R\text{-}\mathsf{Mod}}(M)^{\mathrm{op}})$-bimodule, where the right multiplication is given by function application.
+    \end{itemize}
+    
+\end{example}
+
 \begin{definition}{Balanced Product}{}
     Let $R$ be a ring, $M$ be a right $R$-module, $N$ be a left $R$-module and $G$ be an abelian group. A map $b:M\times N\to G$ is called a \textbf{$R$-balanced product} if 
     \begin{enumerate}[(i)]
@@ -485,6 +498,144 @@ \subsection{Tensor Product}
     To show the uniqueness, assume there exists another group homomorphism $\tilde{b}':F/K\to G$ such that $\tilde{b}'\circ\pi\circ\iota=b$. Since $\hat{b}\circ \iota=b$, by the uniqueness of $\tilde{b}$, we have $\tilde{b}'\circ\pi=\hat{b}$. Then by the uniqueness of $\tilde{b}$, we have $\tilde{b}=\tilde{b}'$. Thus the tensor product $F/K$ is the initial object in the category $\mathsf{Bal}_R(M, N)$.
 \end{prf}
 
+\begin{proposition}{Pure Tensors Generate Tensor Product}{pure_tensors_generate_tensor_product}
+    Let $R$ be a ring, $M$ be a right $R$-module, and $N$ be a left $R$-module. The elements of the image of the tensor product map $\otimes :M\times N\to M\otimes_R N$
+    \[
+       \mathrm{im}(\otimes) = \{m\otimes n \mid m\in M, n\in N\}.
+    \]
+    are called \textbf{pure tensors}.
+    \begin{enumerate}[(i)]
+        \item $\mathrm{im}(\otimes)$ generates $M\otimes_R N$ as an abelian group:
+        \[
+            M\otimes_R N = \langle \mathrm{im}(\otimes) \rangle_{\mathsf{Ab}}:=\left\{
+                \sum_{i=1}^k m_i\otimes n_i \mid k\in \mathbb{Z}_{\ge 1}, m_i\in M, n_i\in N
+            \right\}.
+        \]
+        \item Let $F:=\mathrm{Free}_{\mathsf{Ab}}(\mathrm{im}(\otimes))$. Then we have
+        \[
+        M\otimes_R N \cong F/\langle X_M\cup X_N\rangle,
+        \]
+        where
+        \begin{align*}
+            X_M&=\left\{[(m+m')\otimes n]-[m\otimes n]-[m'\otimes n]\in F\midv m,m'\in M,n\in N\right\},\\
+            X_N&=\left\{[m\otimes(n+n')]-[m\otimes n]-[m\otimes n']\in F\midv m\in M, n,n'\in N\right\}.
+        \end{align*}
+        \item Let $f:\mathrm{im}(\otimes)\to G$ be a map. If and only if the map
+        \begin{align*}
+            M \times N &\longrightarrow G \\
+            (m,n)&\longmapsto f(m\otimes n)
+        \end{align*}
+        is $\mathbb{Z}$-bilinear, i.e.
+        \begin{align*}
+            f\left((m_1+m_2)\otimes n\right)&=f\left(m_1\otimes n\right)+f\left(m_2\otimes n\right),\\
+            f\left(m\otimes (n_1+n_2)\right)&=f\left(m\otimes n_1\right)+f\left(m\otimes n_2\right),
+        \end{align*}
+        there exists a group homomorphism $\tilde{f}:M\otimes_R N\to G$ such that the following diagram commutes
+        \[
+        \begin{tikzcd}[ampersand replacement=\&]
+            \mathrm{im}(\otimes) \arrow[r, "i", hook] \arrow[rd, "f"']  \& M\otimes_R N \arrow[d, dashed, "\tilde{f}"] \\[0.3cm]
+            \&  G
+        \end{tikzcd}
+        \]
+        If such $\tilde{f}$ exists, then it is unique and is totally determined by $f$ as follows:
+        \begin{align*}
+            \tilde{f}:M\otimes_R N&\longrightarrow G\\
+            \sum_{i=1}^k m_i\otimes n_i &\longmapsto \sum_{i=1}^k f(m_i\otimes n_i).
+        \end{align*}
+    \end{enumerate}
+\end{proposition}
+\begin{prf}
+    \begin{enumerate}[(i)]
+        \item 
+    Let $\mathrm{pr}:M\otimes_R N\to M\otimes_R N/\langle \mathrm{im}(\otimes) \rangle_{\mathsf{Ab}}$ be the canonical projection. Then we have the following commutative diagram
+    \[
+     \begin{tikzcd}[ampersand replacement=\&]
+            M\times N \arrow[r, "\otimes"] \arrow[rd, "0"']  \& M\otimes_R N \arrow[d, "\mathrm{pr}", two heads] \\[2em]
+            \&  M\otimes_R N/\langle \mathrm{im}(\otimes) \rangle_{\mathsf{Ab}}
+        \end{tikzcd}
+    \]
+    Note $0= 0\circ \otimes$. By the uniqueness in the universal property of tensor product, we have $\mathrm{pr}=0$. Therefore, $M\otimes_R N=\langle \mathrm{im}(\otimes) \rangle_{\mathsf{Ab}}$.
+
+    \item Define
+    \begin{align*}
+        j:\mathrm{im}(\otimes)&\longrightarrow  \mathrm{Free}_{\mathsf{Ab}}(\mathrm{im}(\otimes))\\
+         m\otimes n &\longmapsto [m\otimes n]
+    \end{align*}
+    By the universal property of free abelian group, the inclusion $i:\mathrm{im}(\otimes)\hookrightarrow M\otimes_R N $ induces a group homomorphism $\widehat{i}:\mathrm{Free}_{\mathsf{Ab}}(\mathrm{im}(\otimes))\to M\otimes_R N$.
+    \[
+    \begin{tikzcd}
+                &[-1em] \mathrm{im}(\otimes) \arrow[d, "j", hook] \arrow[rd, "f"] \arrow[ld, "i"', hook]                                                                                   &   \\[3em]
+    M\otimes_R N & \mathrm{Free}_{\mathsf{Ab}}(\mathrm{im}(\otimes)) \arrow[d, "p", two heads] \arrow[r, "\widehat{f}", dashed] \arrow[l, "\widehat{i}"', dashed]                         & G \\[3em]
+                & \mathrm{Free}_{\mathsf{Ab}}(\mathrm{im}(\otimes))/\langle X_M\cup X_N\rangle_{\mathsf{Ab}}. \arrow[ru, "\overline{f}"', dashed] \arrow[lu, "\overline{i}", dashed] &  
+    \end{tikzcd}
+    \]
+    Since 
+    \[
+    \widehat{i}\left(X_M\right)=\widehat{i}\left(X_N\right)=\{0\}\implies   X_M\cup X_N \subseteq \ker \widehat{i}\implies \langle X_M\cup X_N\rangle_{\mathsf{Ab}}\subseteq \ker \widehat{i}
+    \]
+    there exists a unique group homomorphism $\overline{i}:\mathrm{Free}_{\mathsf{Ab}}(\mathrm{im}(\otimes))/\langle X_M\cup X_N\rangle_{\mathsf{Ab}}\to M\otimes_R N $ such that $\overline{i}\circ p= \widehat{i}$. Since $\widehat{i}$ is surjective, $\overline{i}$ is also surjective. Define
+    \begin{align*}
+        \beta:M\times N&\longrightarrow \mathrm{Free}_{\mathsf{Ab}}\left(\mathrm{im}(\otimes)\right)/\langle X_M\cup X_N\rangle_{\mathsf{Ab}}\\
+         (m,n)&\longmapsto \overline{\left[m \otimes n\right]}
+    \end{align*}
+    We can check that $\beta$ is a balanced product
+    \begin{align*}
+       \beta\left(m_1+m_2,n\right)&=\overline{\left[(m_1+m_2) \otimes n\right]}=\overline{\left[m_1 \otimes n\right]}+\overline{\left[m_2 \otimes n\right]} = \beta\left(m_1,n\right)+\beta\left(m_2,n\right),\\
+       \beta\left(m, n_1+n_2\right)&= \overline{\left[m \otimes \left(n_1+n_2\right)\right]}= \overline{\left[m \otimes n_1\right]}+\overline{\left[m \otimes n_2\right]}=\beta\left(m, n_1\right)+\beta\left(m, n_2\right),
+       \\
+       \beta\left(mr, n\right)&= \overline{\left[(mr) \otimes n\right]}= \overline{\left[m \otimes (rn)\right]}=\beta\left(m, rn\right).
+    \end{align*}
+    By the universal property of tensor product $M \otimes_R N$, there exists a unique group homomorphism
+    \begin{align*}
+        \varphi:M\otimes_R N&\longrightarrow \mathrm{Free}_{\mathsf{Ab}}(\mathrm{im}(\otimes))/\langle X_M\cup X_N\rangle,\\
+         m\otimes n &\longmapsto \overline{[m\otimes n]}.
+    \end{align*}
+    such that $\varphi\circ \otimes=\beta$. Note 
+    \begin{align*}
+    \varphi\circ \overline{i}\left(\overline{[m_1\otimes n_1]+\cdots [m_k\otimes n_k]}\right)&=\varphi\circ \overline{i}\circ p\left([m_1\otimes n_1]+\cdots [m_k\otimes n_k]\right)\\
+    &=\varphi\circ \widehat{i}\left([m_1\otimes n_1]+\cdots [m_k\otimes n_k]\right)\\
+    &=\varphi\left(m_1\otimes n_1\right)+\cdots +\varphi\left(m_k\otimes n_k\right)\\
+    &=\overline{[m_1\otimes n_1]}+\cdots +\overline{[m_k\otimes n_k]}\\
+    &=\overline{[m_1\otimes n_1]+\cdots +[m_k\otimes n_k]}.
+    \end{align*}
+    We have $\varphi\circ \overline{i}=\mathrm{id}$, which implies that $\overline{i}$ is injective. Therefore, $\overline{i}$ is an isomorphism.
+    \item If there exists a group homomorphism $\tilde{f}:M\otimes_R N\to G$ such that $f=\tilde{f}\circ i$, then we can check that the map $(m,n)\mapsto f(m\otimes n)$ is $\mathbb{Z}$-bilinear:
+    \begin{align*}
+        f\left((m_1+m_2)\otimes n\right)&=\tilde{f}\left(i\left((m_1+m_2)\otimes n\right)\right)=\tilde{f}\left(m_1\otimes n+m_2\otimes n\right)=f\left(m_1\otimes n\right)+f\left(m_2\otimes n\right),\\
+        f\left(m\otimes (n_1+n_2)\right)&=\tilde{f}\left(i\left(m\otimes (n_1+n_2)\right)\right)=\tilde{f}\left(m\otimes n_1+m\otimes n_2\right)=f\left(m\otimes n_1\right)+f\left(m\otimes n_2\right).
+    \end{align*}
+    Conversely, if the map $(m,n)\mapsto f(m\otimes n)$ is $\mathbb{Z}$-bilinear, then we have
+    \begin{align*}
+        \widehat{f}\left([(m+m')\otimes n]-[m\otimes n]-[m'\otimes n]\right)&=\widehat{f}\left([(m+m')\otimes n]\right)-\widehat{f}\left([m\otimes n]\right)-\widehat{f}\left([m'\otimes n]\right)\\
+        &=f\left((m+m')\otimes n\right)-f\left(m\otimes n\right)-f\left(m'\otimes n\right)=0,\\
+        \widehat{f}\left([m\otimes(n+n')]-[m\otimes n]-[m\otimes n']\right)&=\widehat{f}\left([m\otimes(n+n')]\right)-\widehat{f}\left([m\otimes n]\right)-\widehat{f}\left([m\otimes n']\right)\\
+        &=f\left(m\otimes(n+n')\right)-f\left(m\otimes n\right)-f\left(m\otimes n'\right)=0,
+    \end{align*}
+    which implies
+    \[
+    \widehat{f}\left(X_M\right)=\widehat{f}\left(X_N\right)=\{0\}\implies X_M\cup X_N \subseteq \ker \widehat{f}\implies \langle X_M\cup X_N\rangle_{\mathsf{Ab}}\subseteq \ker \widehat{f}.
+    \]
+    Thus by the universal property of quotient group, there exists a unique group homomorphism 
+    \begin{align*}
+        \overline{f}: \mathrm{Free}_{\mathsf{Ab}}(\mathrm{im}(\otimes))/\langle X_M\cup X_N\rangle_{\mathsf{Ab}}&\longrightarrow G\\
+        \overline{[m\otimes n]}&\longmapsto f(m\otimes n)
+    \end{align*}
+    such that $\overline{f}\circ p=\widehat{f}$. Composing $\overline{f}$ with the isomorphism $\varphi:M\otimes_R N\to \mathrm{Free}_{\mathsf{Ab}}(\mathrm{im}(\otimes))/\langle X_M\cup X_N\rangle$ in (ii), we obtain a group homomorphism
+    \begin{align*}
+        \tilde{f}:M\otimes_R N&\longrightarrow G\\
+        m\otimes n &\longmapsto f(m\otimes n).
+    \end{align*}
+    \end{enumerate}
+\end{prf}
+
+\begin{proposition}{}{}
+    Let $Q$, $R$, and $S$ be commutative rings, $M$ be a $(Q,R)$-bimodule, and $N$ be a $(R,S)$-bimodule. Then $M\otimes_R N$ has a natural $(Q,S)$-bimodule structure given by
+    \[
+        q\cdot (m\otimes n):=(q\cdot m)\otimes n,\quad (m\otimes n)\cdot s:=m\otimes (n\cdot s)
+    \]
+    for all $q\in Q$, $s\in S$, $m\in M$, and $n\in N$.
+\end{proposition}
+
 \begin{example}{Base Change Functor}{}
     Let $\varphi:R\to S$ be a ring homomorphism. The \textbf{base change functor} $-\otimes_R S:R\text{-}\mathsf{Mod}\to S\text{-}\mathsf{Mod}$ is defined as follows:
     \[