---- > [!definition] Definition. ([[tensor product of modules]]) > Let $M$, $N$, $P$ be [[module|modules]] over a [[commutative ring|commutative]] [[ring]] $R$. The **tensor product** of $M$ and $N$ is a new $R$-[[module]] $M \otimes_{R} N$, with an $R$-[[bilinear map]] $\otimes: M \times N \to M \otimes_{R} N,$ > satisfying the [[universal property]] that any $R$-[[bilinear map]] $\varphi: M \times N \to P$ factors uniquely through this new module $M \otimes_{R} N$ > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAFkACAHW7wFt4nAHIgAvqXSZc+QigCM5KrUYs2ABXGSQGbHgJEy85fWatEHHtwgD4AfQBKI8cphQA5vCKgAZgCcIfiQyEBwIJAAmalM1C14bLEEEagY6ACMYBnVpfTkQPyx3AAscLV8AoMQQsKRFFTM2Xno-NCKsMpB-QMjqGsQ6mPMQXhgADyw4HDhOAEIrZtb2lPTM7L1ZNgLi0rEKMSA > \begin{tikzcd} > M \times N \arrow[d, "\otimes"'] \arrow[r, "\varphi"] & P \\ > M \otimes_R N \arrow[ru, "\exists ! \overline{\varphi}"'] & > \end{tikzcd} > \end{document} > ``` > in such a way that the map $\overline{\varphi}$ is *linear*.[^2] > > This defines $M \otimes_{R} N$ [[terminal objects are unique up to a unique isomorphism|up to]] [[isomorphism]], if it exists. > > Indeed, tensor products *do* exist in $R$-$\mathsf{Mod}$, as witnessed by the following construction. With $F^{R}(M \times N)$ denoting the [[free module]] on $M \times N$ and $j$ its [[free module|its (universal) set map]] > > Let $K$ be the [[submodule]] of $F^{R}(M \times N)$ '[[submodule generated by a subset|generated by]] bilinearities', that is, the $R$-[[submodule]] of $F^{R}(M \times N)$ generated by all elements $j(m, r_{1}n_{1}+r_{2}n_{2})-r_{1}j(m,n_{1})-r_{2}j(m, n_{2})$ > and $j(r_{1}m_{1} + r_{2}m_{2}, n)- r_{1}j(m_{1},n)-r_{2}j(m_{2},n)$ > as $m,m_{1},m_{2}$ range in $M$ and $n,n_{1},n_{2}$ range in $N$ and $r_{1},r_{2}$ range in $R$. Then $M \otimes_{R} N := \frac{F^{R}(M \times N)}{K},$ > endowed with the map $\otimes:M \times N \to M \otimes_{R}N$ obtained by composing $j$ with the [[kernel iff submodule|natural projection]] $\otimes: M \times N \xrightarrow{j} F^{R}(M \times N) \to M \otimes_{R} N = \frac{F^{R}(M \times N)}{K}$ > is an explicit construction of the tensor product. The element $\otimes(m,n)$ — that is, the class of basis element $j(m,n)$ modulo $K$ — is denoted $m \otimes n$. > > Note that elements of $M \otimes_{R} N$ arise from elements of $F^{R}(M \times N)$; in particular, an arbitrary element of the tensor product looks like a finite [[linear combination]] $\sum_{i}r_{i}(m_{i} \otimes n_{i})=\sum_{i}r_{i} [j(m_{i},n_{i})]$ > with $r_{i} \in R$, $m_{i} \in M$, and $n_{i} \in N$. Note that the coefficients $r_{i}$ are not *really* necessary, since they can be absorbed thus:[^1] $\sum_{i}r_{i}(m_{i} \otimes n_{i})=\sum_{i}(r_{i}m_{i}) \otimes n_{i}.$ > Elements of the form $m \otimes n$ are called **pure tensors**. ^definition > [!basicnonexample] Warning. > Note that $i$ is understood in these expressions to index the *pair* $(m,n)_{i}$. Perhaps a more precise notation would be $\sum_{i,j}r_{ij}(m_{i} \otimes n_{j})$ (and this often used in concrete settings), but just writing the pure tensors as $m_{i} \otimes n_{i}$ is cleaner for this level of abstraction. > ^warning > [!note] Remark. > If $M$ (resp. $N$) is [[submodule generated by a subset|generated by]] a subset $S \subset M$ (resp. $T \subset N$), then $M \otimes_{R} N$ is manifestly [[submodule generated by a subset|generated by]] $\{ s \otimes t: s \in S, t \in T \}$. > ^note [^1]: This means that $M \otimes_{R}N$ is generated by pure tensors not only as an $R$-module, but as a $\mathbb{Z}$-[[module]] (i.e. an [[abelian group]]). > [!justification] > We need to check that the provided explicit construction works as claimed — that $\otimes$ is bilinear and that $M \otimes_{R} N$ satisfies the [[universal property]]. > > **Bilinearity.** For linearity in the first argument, we have > $\begin{align} > \otimes(r_{1}m_{1} + r_{2}m_{2}, n) = & [ j(r_{1}m_{1} + r_{2}m_{2}, n)] \\ > = & [r_{1}j(m_{1},n) + r_{2}j(m_{2},n)] \\ > = & r_{1}[j(m_{1},n)] + r_{2}[j(m_{2}, n)] \\ > = & r_{1} \otimes(m_{1},n) + r_{2} \otimes(m_{2}, n), > \end{align}$ > where in the penultimate step we used that $\pi$ is a [[linear map]] and in the step before we used that, by construction, $j(r_{1}m_{1} + r_{2}m_{2}, n)$ and $r_{1}j(m_{1}, n)+r_{2}j(m_{2},n)$ represent the same [[coset]] (their difference lives in the [[kernel of a module homomorphism|kernel]] $K$). A symmetric strategy proves linearity in the second argument. > > **Universal property.** Let $\varphi:M \times N \to P$ be any [[bilinear map]]. The [[free module|universal property of free modules]] yields a unique factorization of $\varphi$ through $F^{R}(M \times N)$ via a map $\tilde{\varphi}$: > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAFkACAHW7wFt4nAHIgAvqXSZc+QigCM5KrUYs2ABXGSQGbHgJEy85fWatEIAGIA9AEoAKLrwFDhASnHKYUAObwioABmAE4Q-EhkIDgQSIoqZmy89MFoABZYWkGh4YiR0UgATNSmahYAVpkgIWGF1PmIcSXmILwwAB5YcDhwnACEPHxYDLDASXQp6WIg1Ax0AEYwDOrS+nIgwVg+qTieYkA > \begin{tikzcd} > M \times N \arrow[r, "\varphi"] \arrow[d, "j"] & P \\ > F^R(M \times N) \arrow[ru, "\exists ! \tilde{\varphi}"'] & > \end{tikzcd} > \end{document} > ``` > The claim is that $\tilde{\varphi}(K)= (0$). If we can show this, then $K \subset \ker \tilde{\varphi}$, and hence the [[characterization of quotienting a group|universal property of quotient groups]] will give us in turn a unique factorization of $\tilde{\varphi}$: > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAFkACAHW7wFt4nAHIgAvqXSZc+QigCM5KrUYs2ABXGSQGbHgJEy85fWatEIAGIA9AEoAKLrwFDhASi1S9sw6QBMJqrmHDzcEC5wAPq2IpwAvJw2Dk58WIJwIm6cAPScANLiyjBQAObwRKAAZgBOEPxIZCA4EEiKKmZsvPTVaAAWWJ4gNXUN1M1IftSmahYAViDUDHQARjAM6tL6ciDVWCW9OIPD9YiTTS2IbdPBvDAAHlhwOBkAhKF4DLDAXXQ9-WILEBLVbrTY+CxYMDYWBHWonM7jRAAZimQU63DQA0WKzWG28Bgsu32hwkVTho3OSBRIFWYCgSAAtAA2RrXdHhNLlbEgvEyAk7PYHWEjZFjC5tBiQ4JQOhwXrFQFsiy3B5POBvXgQGgwaqSsAwb7cbp9LAA7m4sH8olCsQUMRAA > \begin{tikzcd} > M \times N \arrow[r, "\varphi"] \arrow[d, "j"'] \arrow[dd, "\otimes"', bend right=60] & P \\ > F^R(M \times N) \arrow[ru, "\tilde{\varphi}" description] \arrow[d, "\pi"'] & \\ > M \otimes_R N = F^R(M \times N) / K \arrow[ruu, "\exists! \overline{\varphi}"', dashed] & > \end{tikzcd} > \end{document} > ``` > > and we will be done. > > To verify that $\tilde{\varphi}|_{K}=0$, it suffices to show that $\tilde{\varphi}$ kills every generator of $K$. This follows straight from the fact that $\tilde{\varphi}$ is $R$-[[bilinear map|bilinear]]. > ^justification > [!definition] Definition. (Tensor product of finitely many modules) > Now let $M_{1},\dots,M_{\ell},P$ be $R$-[[module|modules]]. Their **tensor product** is a new $R$-[[module]] $M_{1} \otimes_{R} \cdots \otimes_{R} M_{\ell}$ satisfying the [[universal property]] that any $R$-[[multilinear map|multilinear map]] $\varphi:M_{1} \times \dots \times M_{\ell} \to P$ factors uniquely through it as $\overline{\varphi} \circ \otimes$ in such a way that $\overline{\varphi}$ is [[linear map|linear]]. > Concretely, $M_{1} \otimes_{R} \dots \otimes_{R}M_{\ell}$ may be constructed by inducting upon the two-module definition: every $R$-[[multilinear map|multilinear map]] $M_{1} \times \dots \times M_{\ell} \to P$ factors uniquely through $( ( \cdots (M_{1} \otimes_{R} M_{2}) \otimes_{R} \cdots)\otimes_{R} M_{\ell-1}) \otimes_{R} M_{\ell},$ where we note that any other grouping of parentheses would also solve the universal problem (in this sense, the tensor product is associative and using the notation $M_{1} \otimes_{R} \dots \otimes_{R} M_{\ell}$ is authorized). > Elements of this module are finite linear combinations $\sum_{i}m_{1,i} \otimes \cdots \otimes m_{\ell,i}$of pure tensors. The structure map $M_{1} \times \dots \times M_{\ell} \to M_{1} \otimes_{R} \cdots \otimes_{R} M_{\ell}$ acts as $(m_{1},\dots,m_{\ell}) \mapsto m_{1} \otimes \cdots \otimes m_{\ell}$. ^definition > [!justification] > Let's perform the mentioned induction. If $\varphi:M_{1} \times \dots \times M_{\ell} \to P$ is multilinear, then (invoking the induction hypothesis) for every $m_{\ell} \in M_{\ell}$, the function $(m_{1},\dots,m_{\ell-1}) \mapsto \varphi(m_{1},\dots,m_{\ell-1}, m_{\ell })$ is multilinear and hence factors through $(\cdots (M_{1} \otimes_{R} M_{2}) \otimes_{R} \cdots)\otimes_{R}M_{\ell-1}$ via a unique linear map. That means that $\varphi$ has induced a unique [[bilinear map]] $( ( \cdots (M_{1} \otimes_{R} M_{2}) \otimes_{R} \cdots) \otimes_{R} M_{\ell-1}) \times M_{\ell} \to P$ which induced in turn a unique [[linear map]] $( (M_{1} \otimes_{R} M_{2}) \otimes_{R} \cdots) \otimes_{R} M_{\ell-1} \otimes_{R} M_{\ell} \to P$ by the [[universal property]] of $\otimes_{R}$. ^justification > [!definition] Definition. (Finite tensor power) > The $\ell$-fold tensor product of a [[module]] $M$ by itself is called the **$\ell$th tensor power of $M$** and denoted $\mathbb{T}_{R}^{\ell}(M):=M ^{\otimes \ell} = \underbrace{M \otimes_{R} \cdots \otimes_{R} M}_{\ell \text{ times}}$ > > For every $R$-[[module]] $P$, every $R$-multilinear map $M^{\ell} \to P$ factors uniquely through $M^{\otimes \ell}$: > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAFkA9AHW5gYYgAvqXSZc+QigCM5KrUYs2ABWGiQGbHgJEy0+fWatEHTsF4Q8AW3gACXvwZDh8mFADm8IqABmAJwgrJDIQHAgkWQUjNl56PzQACyw1XwCgxBCwpAAmEVTAnOosxEjDJRMHAA8sOBw4WwBCe244xOTqBjoAI35lcW0pED8sdwScFyEgA > \begin{tikzcd} > M^\ell \arrow[r, "\varphi"] \arrow[d] & P \\ > M^{\otimes \ell} \arrow[ru, "\exists ! \varphi"'] & > \end{tikzcd} > \end{document} > ``` > We set $\mathbb{T}_{R}^{1}(M)=M$, and by convention $\mathbb{T}_{R}^{0}(M)=R$. > [!basicexample] > - For all $R$-[[module|modules]] $N$, $R \otimes_{R} N \cong N$. Indeed, any [[bilinear map]] $R \times N \to P$ factors through $N$ where $\otimes(r,n):=rn$ (set $\overline{\varphi}(n):=\varphi(1,n)$, then $\overline{\varphi}(\otimes(r,n))=\varphi(1,rn)=\varphi(r,n)$). > - Consider $\mathbb{Z}/2$ and $\mathbb{Z} / 3$ as $\mathbb{Z}$-[[module|modules]]. In $\mathbb{Z} / 2 \otimes_{\mathbb{Z}}\mathbb{Z} / 3$ we have $a \otimes b=(3a) \otimes b=a \otimes(3b)=0.$ So this module is zero. ^basic-example - [ ] what about all the other usual tensor product properties (many following from left adjointness)? ---- #### [^2]: While $\varphi$ is, in general, decidedly not. Thus, $M\otimes_R N$ is the 'best approximation' to $M \times N$ available in $R$-$\mathsf{Mod}$ if we want to view $R$-bilinear maps from $M \times N$ as $R$-*linear*. [^2]. Recall that $j$ is the 'indicators/basis map'; see [[free module]]. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```