Tensor product of modules ($M{\otimes }_{R}N$)

For a right module $M_R$ and a left module ${}_{R}N$, their tensor product of modules is the abelian group generated by symbols $m\otimes n$, modulo additivity in each variable andThe universal property of the tensor product of modules identifies additive maps from it with balanced maps on $M\times N$.

When $R$ is a commutative ring, two left modules can be used by viewing the first as a right module. The tensor product is then an $R$-module with $r(m\otimes n)=(rm)\otimes n=m\otimes (rn)$. For example, $\mathbb{Z}/a\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Z}/b\mathbb{Z}\cong \mathbb{Z}/\gcd (a,b)\mathbb{Z}$: the generator $1\otimes 1$ is killed by both $a$ and $b$, and the universal property identifies exactly those relations. For a noncommutative ring, the left and right input actions must be specified.

For a short exact sequence $0\to U\stackrel{i}{\to }V\stackrel{q}{\to }W\to 0$ and any module $E$, the sequence $U{\otimes }_{A}E\to V{\otimes }_{A}E\to W{\otimes }_{A}E\to 0$ is exact. The last map is surjective since every pure tensor $w\otimes e$ can be lifted using a preimage of $w$. Put $T=(V{\otimes }_{A}E)/\mathrm{im}(i\otimes 1_E)$. There is a well-defined balanced map $W\times E\to T$, sending $(q(v),e)$ to the class of $v\otimes e$: replacing $v$ by $v+i(u)$ changes it by an element of the discarded image. This induces $W{\otimes }_{A}E\to T$, inverse to the map induced by $q\otimes 1_E$. Thus the kernel of $q\otimes 1_E$ is exactly the image of $i\otimes 1_E$. Consequently a module is flat exactly when tensoring with it also preserves injections.

For a homomorphism of commutative rings $R\to A$ and two $A$-modules, the tensor product of modules over $A$ is the quotient of the tensor product of modules over $R$ imposing all relations $am\otimes n=m\otimes an$. The quotient map is $A$-linear when $A$ acts through the first factor on the source. This expresses the extra balancing imposed by change of rings.

For commutative $R$-algebras, their module tensor product becomes a commutative algebra by $(a\otimes b)(a^{\prime }\otimes b^{\prime })=a{a}^{\prime }\otimes b{b}^{\prime }$. Its universal property is that pairs of compatible $R$-algebra maps from $A$ and $B$ into a commutative algebra $C$ correspond to maps $A{\otimes }_{R}B\to C$. This coproduct of algebras gives, contravariantly, the fibre product of schemes on affine charts.

If $L,L^{\prime }$ are field extensions of $K$, choose a $K$-basis of $L$ containing $1$. After tensoring with $L^{\prime }$, that basis expresses $L{\otimes }_K{L}^{\prime }$ as a nonzero direct sum of copies of $L^{\prime }$, and $1\otimes 1$ is nonzero. A prime ideal of this nonzero ring supplies a point used in the point-lifting property of a scheme fibre product.

A pure tensor is an element of a tensor product that can be written as $m\otimes n$. General tensors are finite sums of pure tensors and need not themselves be pure.

If $M{\otimes }_{R}N\cong R^n$ for a positive integer $n$, then both $M$ and $N$ are projective modules. Choose a finite expression for the inverse image of one basis vector. It produces a split surjection $N^r\to R$; tensoring the splitting with $M$ exhibits $M$ as a direct summand of the finite free module $(M{\otimes }_{R}N{)}^r$. Symmetry gives the result for $N$.

An $R$-module $M$ is tensor-nilpotent when $M^{\otimes k}=0$ for some positive integer $k$. A nonzero finitely generated module cannot be tensor-nilpotent: localize at a maximal ideal in its support and reduce modulo that maximal ideal, obtaining a nonzero tensor power of a nonzero vector space.

For a right module $M_R$, a left module ${}_{R}N$, and an abelian group $P$, a balanced map is a map $b:M\times N\to P$ additive in each variable and satisfyingIt is exactly the sort of map represented by the tensor product of modules. The balancing equation transfers a scalar between the two inputs without assuming that $R$ is commutative.

When $R$ is a commutative ring and $P$ is an $R$-module, an $R$-bilinear map is balanced and induces an $R$-linear map on the tensor product. Mere additive balancedness gives an additive induced map; it does not by itself require linearity in the target. For example, $(z,w)\mapsto \overline{zw}$ is additive and balanced over $\mathbb{C}$, but its induced map is conjugate-linear.

For a right module $M_R$, a left module ${}_{R}N$, and an abelian group $P$, every balanced map $b:M\times N\to P$ factors uniquely through an additive homomorphismExistence follows because additivity and the balancing equation are exactly the relations imposed on the generating symbols $m\otimes n$. Uniqueness follows because those symbols generate the tensor product as an abelian group.

For modules over a commutative ring, if $b$ is $R$-bilinear and the target is an $R$-module, the induced map is an R-module homomorphism. This is the linear version of the universal property; the additive version alone must not silently impose an action on the target.

For modules over a commutative ring, there is a natural isomorphismgiven by $f\mapsto (m\mapsto (n\mapsto f(m\otimes n)))$.

Tensor product need not commute with an infinite direct product. For example,The element $(1\mathrm{mod}n{)}_n$ has infinite order in the product, so it survives rational localization.