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For two unital associative algebras over a field $k$, their tensor product carries the multiplication in the title, extended by bilinearity, with identity $1\otimes 1$. Choosing a basis $(b_j)$ of the second factor gives the basis of a module $(1\otimes b_j)$ for the resulting left $A$-module; dual basis functionals recover the coefficients uniquely.

The tensor product over $k$ of two central simple $k$-algebras is central simple. After extending scalars to an algebraic closure, both factors and their tensor product become full matrix algebras; faithful flatness then descends simplicity and the center.

The tensor product of homologically graded chain complexes has differential $D(x_p\otimes y)=d{x}_p\otimes y+(-1{)}^p{x}_p\otimes dy$. The sign makes the two mixed terms in $D^2$ cancel. Total degree is the sum of the two degrees.

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.

The tensor product crystal uses the $i$-string data ${\epsilon }_i$ and ${\phi }_i$ to decide on which tensor factor a Kashiwara operator acts. Its connected components are highest-weight crystals and encode the decomposition of the tensor product representation.

This tensor-product operation combines one crystal basis with another, with factor selection governed by their string parameters.

The tensor product of two globally generated line bundles is globally generated. At a point choose generator germs arising from a global section of each factor. Their tensor is the germ of a global tensor-product section and generates the rank-one tensor-product stalk. Iteration gives the same property for every positive tensor power, on an arbitrary scheme.

For representations $V$ and $W$ of a group $G$, the tensor product representation is the tensor product $V\otimes W$ with diagonal action

The tensor product of Lie algebra representations has action $x(v\otimes w)=xv\otimes w+v\otimes xw$. The Lie algebra representation identity follows because operators on the two different tensor factors commute. This action descends to exterior powers and symmetric powers.

The tensor product of linear maps is determined by $(A\otimes B)(x\otimes y)=Ax\otimes By$ and linear extension.

A tensor-product operator factors as $A\otimes B$ on a tensor product of two spaces. If both factors are positive semidefinite operators, their tensor-product operator is positive: a tensor-product basis of eigenvectors has eigenvalues $a_i{b}_j\ge 0$. This positivity underlies the necessary direction of the positive partial transpose criterion.

For $n\ge 2$, the defining Lie algebra representation $V$ of ${\mathfrak{sp}}_{2n}$ has the displayed tensor product decomposition. The crystal of the defining symplectic representation and the crystal tensor-product rule give exactly these three highest vertices, each occurring once. The symmetric square is $L(2{\omega }_1)$ of dimension $n(2n+1)$. The exterior square splits into its invariant inverse-form line and the primitive exterior square $L({\omega }_2)$ of dimension $n(2n-1)-1$. Their dimensions sum to $4n^2$. At $n=1$, the exterior square is only the invariant line, giving $L(2{\omega }_1)\oplus L(0)$.

For three distinguishable spin one-half systems, the Clebsch-Gordan decomposition first combines two spins into $0\oplus 1$, then couples the third: $0\otimes 1/2=1/2$ and $1\otimes 1/2=1/2\oplus 3/2$. Thus the tensor product has one quartet and two doublets. A Hamiltonian operator depending only on the total $S^2$ and $S_z$ acts identically on the two doublet copies. In particular $H={ϵ}_t{S}^2/{\hslash }^2+{ϵ}_z{S}_z/\hslash$ has energy eigenvalues ${ϵ}_{t}S(S+1)+{ϵ}_{z}M$, generically once for each quartet projection and twice for each doublet projection. Equal numerical energies have their multiplicities added.

Over a principal ideal domain, a module is flat exactly when it is torsion-free. Hence the tensor product of two torsion-free modules over a principal ideal domain is torsion-free: both tensor functors are exact, so their composite is exact.

For two collected spatial modes $a,b$, each with polarizations $H,V$, put $K^{†}=a_H^{†}{b}_V^{†}+a_V^{†}{b}_H^{†}$. The one-pair state is a Bell state. The normalized two-pair component isThis follows by expanding $(K^{†}{)}^2$ and using the bosonic creation operator identity $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$. The three orthogonal normalized Fock states have equal probability $1/3$. A tensor product of independently labelled pairs is a different multimode description; identifying its labels without renormalization changes its norm.

A quantum state on the tensor product of two two-dimensional complex state spaces. A normalized pure state has squared coefficient moduli summing to one. It is a product state exactly when $\alpha \delta =\beta \gamma$, by the two-qubit product-state determinant criterion. The weaker equality of moduli only characterizes independence of the computational-basis outcomes; phases can preserve quantum entanglement despite those independent statistics.

For the two-spin Hamiltonian operator $H=({\omega }_{1}ZI+{\omega }_{2}IZ+J(XX+YY+ZZ))/2$, juxtaposition of two Pauli matrices denotes their tensor product. Put $\Delta ={\omega }_1-{\omega }_2$. The exchange matrix element between $|01⟩$ and $|10⟩$ is $J$, whereas their diagonal separation is $\Delta$. Nondegenerate perturbation theory therefore gives admixtures $\pm J/\Delta$ and second-order frequency shifts $\pm J^2/\Delta$. Neglecting the exchange terms is justified to leading order when $|J/\Delta |\ll 1$. A large common Zeeman energy is insufficient. To approximate accumulated phases by this leading-order Hamiltonian operator, one also needs $J^{2}t/|\Delta |\ll 1$ over the observation time, or must include the second-order shifts.

For fixed nonidentical chemical shifts, increasing the static magnetic field increases $|\Delta |$ in angular-frequency units while the NMR scalar coupling changes little. If $\Delta =0$ remains exact, the approximation fails at every nonzero $J$.

For an abelian variety and an integer $N$ invertible in the base field, a torsion line bundle $M\in \hat{A}[N]$ has trivial pullback by $[N]$. Choose a trivialization. The descent action of the kernel $A[N]$ on that trivial line bundle acts through a character ${\chi }_M:A[N]\to {\mu }_N$. Define $e_N(P,M)={\chi }_M(P)$. This is independent of rescaling the trivialization and bilinear under translation and tensor product. A trivial character descends the trivial bundle, so $M$ is trivial. Since both groups have order $N^{2\dim A}$, the character map is an isomorphism and the Weil pairing for an abelian variety is perfect. The theta autoduality of a Jacobian turns it into a pairing on two copies of the Jacobian variety's torsion group.

For a two-body operator $R(u)$ acting on a tensor product of particle species spaces, the spectral equation is $R_{12}(u)R_{13}(u+v)R_{23}(v)=R_{23}(v)R_{13}(u+v)R_{12}(u)$. It guarantees agreement between the two ways to reorder three particles in factorized scattering. A Yang–Baxter operator is a related braid-form operator; multiplying by the permutation operator converts between the braided and unbraided conventions.