---- > [!definition] Definition. ([[tensor product of algebras]]) > Let $R$ be a [[commutative ring|commutative]] [[ring]], and let $A,B$ be $R$-[[algebra|algebras]]. The **tensor product** of $A$ and $B$ is a new $R$-[[algebra]] $A \otimes_{R} B$, together with an $R$-[[algebra homomorphism|algebra homomorphisms]] $i_{1}:A \to A \otimes_{R} B$ and $i_{2}:B \to A \otimes_{R} B$, satisfying the [[universal property]] that any pair of $R$-[[algebra homomorphism|algebra homomorphisms]] $A \xrightarrow{f_{1}} C$, $B \xrightarrow{f_{2}} C$ factors uniquely through $A \otimes_{R} B$. In other words, one has the following commutative diagram in $\mathsf{CRing}$: > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZAJgBpiBdUkANwEMAbAVxiRACUQBfU9TXfIRQBGclVqMWbAILdeIDNjwEiZYePrNWiEACE5fJYKKj11TVJ0AZAwv7KhyAAyknGydpABhbuJhQAc3giUAAzACcIAFskFxAcCCRhHjDImMQ4hKRiFJAI6KTqLMQAZnMPNiwAfWFbfPSyeMTS8q1KquIQagY6ACMYBgAFe2MdcKwAgAscOrTCpqQAFmp+sCgkEriLT1Ca2YLERuLlkB7+oZGVMYnprpBV9cQAWk3WyzyO-fSyhcQThiwYE8UDocEm-ju2zYAB1oTAAB5YOA4OAAAgAhL4uEA > \begin{tikzcd} > C & & \\ > & A \otimes_R B \arrow[lu, "\exists !", dashed] & B \arrow[l, "i_2"'] \arrow[llu, "f_2"', bend right] \\ > & A \arrow[u, "i_1"] \arrow[luu, "f_1", bend left] & R \arrow[l] \arrow[u] > \end{tikzcd} > \end{document} > ``` > > This defines $A \otimes_{R} B$ [[terminal objects are unique up to a unique isomorphism|up to]] [[isomorphism]], if it exists. In particular, it defines it to be the [[categorical coproduct|coproduct]] of $A$ and $B$ in $R\mathsf{Alg}$, equivalently the [[categorical pushout|fibered coproduct]] $A \times_{R} B$ of $A$ and $B$ over $R$ in the [[category]] $\mathsf{CRing}$, if it exists.[^1] > > Indeed, $A \otimes_{R} B$ does exist.[^2] It is constructed by taking the [[tensor product of modules|tensor product]] of $A$ and $B$ *as $R$-[[module|modules]]*, endowing it with [[ring]] structure by defining multiplication on pure tensors as $(a \otimes b) \cdot (a' \otimes b'):=(a a') \otimes (bb')$ and extending linearly, and then endowing it with $R$-[[algebra]] structure via the structure map $\begin{align} > R &\to A \otimes_{R} B \\ > r & \mapsto (r \cdot 1_{A}) \otimes 1_{B}. > \end{align}$ > [!justification] > We need to check that our definition of $A \otimes_{R} B$ indeed satisfies the [[universal property]]. If the unique map making the diagram commute (call it $h$) exists, then > > $h(a \otimes b)=h(a \otimes 1)h(1 \otimes a)=h(i_{1}(a))h(i_{2}(b))=f_{1}(a)f_{2}(b).$ > Thus, $h$ is determined on the level of pure tensors by commutativity of the diagram, and therefore determined uniquely. > > Regarding existence of $h$, first note that, as a [[linear map|homomorphism]] of $R$-[[module|modules]], $h$ is induced by the $R$-[[bilinear map|bilinear]] map $A \times B \to C$ sending $(a,b) \mapsto f_{1}(a)f_{2}(b)$: by the [[universal property]] of $A \otimes_{B} B$ as an $R$-[[module]], there is an induced $R$-[[linear map|linear map]] $h:A \otimes_{R} B \to C$ such that $h(a \otimes b)=f_{1}(a)f_{2}(b)$. Clearly $f_{1}=h \circ i_{1}$ and $f_{2}=h \circ i_{2}$. It remains to show that $h$ is an $R$-[[algebra]] [[linear map|homomorphism]]. First, $h(\underbrace{1_{A \otimes_{R} B}}_{=1_{A} \otimes 1_{B}})=1_{C}$. Second, for $a_{1},a_{2} \in A$ and $b_{1},b_{2} \in B$, we have $\begin{align} > h(a_{1}a_{2} \otimes b_{1}b_{2})&= f_{1}(a_{1}a_{2}) f_{2}(b_{1}b_{2}) \\ > &= [f_{1}(a_{1})f_{2}(b_{1})] \cdot [f_{1}(a_{2}) f_{2}(b_{2})] \\ > & = h(a_{1} \otimes b_{1}) h (a_{2} \otimes b_{2}) > \end{align}$ > thus $h$ is an $R$-algebra homomorphism by the corollary in [[upgrading a linear map to an algebra homomorphism]]. - [ ] examples 3.22 and 3.26 in notes - [ ] not sure if the properties on page 22 are in Obsidian? > [!basicexample] > We have an $R$-[[algebra]] [[isomorphism]] $\varphi: R[X_{1},\dots,X_{n}] \otimes_{R} R[T_{1},\dots,T_{r}] \xrightarrow{ \sim} R[X_{1},\dots,X_{n},T_{1},\dots,T_{r}].$ Moreover, if $I$ and $J$ are [[ideal|ideals]] of $R[X_{1},\dots,X_{n}]$ and $R[T_{1},\dots,T_{r}]$ respectively, then we have an $R$-[[linear map|linear]] isomorphism $\frac{R[X_{1},\dots,X_{n}]}{I} \otimes_{R} \frac{R[T_{1},\dots,T_{r}]}{J} \xrightarrow{\sim} \frac{R[X_{1},\dots,X_{n}, T_{1},\dots,T_{r}]}{I^{e}+J^{e}}$ where $\cdot^{e}$ is from [[extension of an ideal]]. ^basic-example > [!proof] > We show $\varphi$ exists by showing it satisfies the [[universal property]]. If we have a [[diagram]] of the form > ![[CleanShot 2025-06-01 at [email protected]|400]] > > with the unlabeled arrows the obvious inclusions, then if a ring map $\psi$ exists, it is determined by the values $\psi(X_{i})$ and $\psi(T_{j})$ for $i \in [n]$, $j \in [r]$. These are in turn determined by commutativity: $\psi(X_{i})=\iota_{X}(X_{i})$ and $\psi(T_{j})=\iota_{T}(T_{j})$. Nothing to worry about with this map wrt well-definition etc., so done. In particular, unraveling how universal objects are unique [[terminal objects are unique up to a unique isomorphism|up to unique iso.]], we must have $\varphi(p \otimes q)=pq$ (i.e., plug in $R[X_{1},\dots,X_{n},T_{1},\dots,T_{r}]$ where $S$ is in the picture, and then fill in the void with $R[X_{1},\dots,X_{n}] \otimes_{R} R[T_{1},\dots,T_{r}]$. Then there is only one possibility for the induced morphism $\varphi$ ([[terminal objects are unique up to a unique isomorphism|because here]]), which has to be an isomorphism, and is necessarily $p \otimes q \mapsto pq$ e.g. by commutativity. (fast but good) [^1]: Equivalently, the [[categorical coproduct|coproduct]] of $A$ and $B$ in $R$-$\mathsf{Alg}$. > [!note] Note. > The data of $R$-algebra homomorphisms $i_{1}$ and $i_{2}$ includes the fact that we have commutative diagrams in $\mathsf{CRing}$ > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBoAmAXVJADcBDAGwFcYkQBBEAX1PU1z5CKAIwVqdJq3YAlHnxAZseAkTIiJDFm0ScABAB0DEPAFt4AfRl6AQvP7KhRMcU1Sd+oyazm4V2-aKAirCyOSkrjRa0rp2vA6CqijhGlHusjwSMFAA5vBEoABmAE4QpkhkIDgQSOGS2uxYFiKBJWVIYlU1iMTxIG3liJ3VtX0DSAAsNCOIAMxpDbpN5CA0jPQARjCMAArBTrrFWDkAFjitpYMArNPdE2OXSDddSLPclNxAA > \begin{tikzcd} > & A \otimes_R B & B \arrow[l, "i_2"'] \\ > A \otimes_R B & & R \arrow[u] \arrow[lu] \\ > A \arrow[u, "i_1"] & R \arrow[l] \arrow[lu] & > \end{tikzcd} > \end{document} > ``` > and, putting them together, we have the following commutative square in $\mathsf{CRing}$: > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAEEACAHW4jwC28APoAlTgCEQAX1LpMufIRQBGclVqMWbKbPnY8BImRUb6zVog4y5IDAaVE1p6ue1XRMjTCgBzeESgAGYAThACSADM1DgQSABMeiCh4VExcYgqSSkRiPHpSGSaFmxYwio2wWG5aiCxha5aliBl8SDUDHQARjAMAAoKhsogIVi+ABY4XtJAA > \begin{tikzcd} > A \otimes_R B & B \arrow[l, "i_2"'] \\ > A \arrow[u, "i_1"] & R \arrow[l] \arrow[u] > \end{tikzcd} > \end{document} > ``` > In other words, specifying $i_{1}$ and $i_{2}$ is specifying a commutative square of [[ring homomorphism|ring homomorphisms]], and hence why we are discussing pushouts etc. in $\mathsf{CRing}$. > ---- #### [^2]: At first glance, it may seem strange to call this object a tensor product — after all, the [[universal property]] does not much look like that in [[tensor product of modules]]. However, the object *satisfying* the universal property is precisely tensor product of $A$ and $B$ viewed as $R$-[[module|modules]], plus some extra data, justifying the name. [^3]: It must be checked that this is [[well-defined]]. ---- #### 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 > ```