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Choose algebra generators $b_1,\dots ,b_n$ for $B$ over $A$. Because the extension is integral, each $b_i$ satisfies a monic equationRepeatedly reduce any exponent at least $e_i$ using this equation. Since the generators commute, every polynomial in them becomes an $A$-linear combination of the finitely many monomialsThus $B$ is a finitely generated module over $A$, with at most $D={\prod }_i{e}_i$ generators. This proves the relevant subalgebra generated by finitely many integral elements assertion rather than merely invoking it.

For a fixed prime ideal $\mathfrak{p}\subset A$, form the fiber ring $C=B{\otimes }_A\kappa (\mathfrak{p})$. Tensoring the finite generating list shows ${\dim }_{\kappa (\mathfrak{p})}C\le D$: its pure tensors with one span the fiber over the residue field $\kappa (\mathfrak{p})$. Part (b) therefore gives only finitely many primes in $C$. By the canonical bijection in part (a), these are precisely the primes of $B$ contracting to $\mathfrak{p}$. HenceNo Noetherian hypothesis on $A$ is needed. For an empty algebra-generating list, $B=A$ and the empty product $D=1$ gives the same conclusion.

Take $V=U={\mathbb{R}}^3$ and identify $V{\otimes }_{\mathbb{R}}U$ with the space of $3\times 3$ real matrices, where pure tensors correspond to matrices of rank at most one. Letand let $f$ be the quotient map. It is not injective because its kernel is the nonzero line $\mathbb{R}{I}_3$.

If $f(v_1\otimes u_1)=f(v_2\otimes u_2)$, thenfor some $\lambda \in \mathbb{R}$. The left side has matrix rank at most two. If $\lambda \ne 0$, the right side has rank three, which is impossible. Thus $\lambda =0$ and the two pure tensors were equal. Hence $f$ is injective on the set of pure tensors while failing to be injective linearly.

In a two-factor tensor product over a field, a pure tensor is also called a separable or decomposable tensor: it can be expressed as one product $v\otimes w$. Every tensor is a finite sum of pure tensors, but a sum need not itself be a pure tensor.^[ref]

For finite-dimensional vector spaces, the tensor-product coefficient matrix gives a decisive test: a nonzero tensor is a pure tensor exactly when its matrix rank is one. Indeed, every column of $v{w}^T$ is a scalar multiple of $v$. Conversely, if all columns of a nonzero matrix are multiples of one nonzero column $v$, collect their multipliers in $w$ to obtain $C=v{w}^T$. The zero tensor is also a pure tensor, for example $0\otimes w$, but its matrix rank is zero.

In ${\mathbb{R}}^2{\otimes }_{\mathbb{R}}{\mathbb{R}}^2$, the tensoris nonseparable: its matrix rank is two and its determinant is one, whereas $\det (v{w}^T)=0$. Directly, trying $t=(a{e}_1+b{e}_2)\otimes (c{f}_1+d{f}_2)$ requires $ac=bd=1$ and $ad=bc=0$. The first two equations force all four scalars nonzero, contradicting the others. In contrast, products sharing a factor do combine: $e_1\otimes f_1+e_1\otimes f_2=e_1\otimes (f_1+f_2)$. Thus being written as two summands alone does not establish nonseparability.

A tensor product turns a bilinear map into a linear map. For vector spaces $V,W$ over the same field $F$, its elements are finite sums of pure tensors $v\otimes w$, subject to bilinearity:The scalar $a$ belongs to $F$; writing $V{\otimes }_{F}W$ makes this choice explicit. These rules preserve separate linearity while allowing arbitrary linear combinations of products.^[ref]

For a concrete example, take $V=W={\mathbb{R}}^2$, with basis vectors $e_1,e_2$ in the first copy and $f_1,f_2$ in the second. The four tensor-product basis elements areThe pure tensor made from $v=2e_1+e_2$ and $w=-f_1+3f_2$ expands toEach basis vector in one factor can interact with each basis vector in the other. This is why dimensions multiply; the tensor-product coefficient matrix displays these four interactions as a two-by-two grid.

The tensor product is different from the Cartesian product $V\times W$. With componentwise vector addition and scalar multiplication, the latter contains pairs and has dimension $\dim V+\dim W$. The map $\tau (v,w)=v\otimes w$ is bilinear, generally neither linear on that Cartesian product nor injective nor surjective. In our example, $\tau (2v,2w)=4\tau (v,w)$, and $\tau (2v,w/2)=\tau (v,w)$; the separability of a tensor explains why some outputs cannot be reached at all. Both spaces happen to have dimension four when $V=W={\mathbb{R}}^2$, but this coincidence does not identify their structures.

If $(e_1,\dots ,e_m)$ and $(f_1,\dots ,f_n)$ are bases of finite-dimensional vector spaces $V,W$ over $F$, then the $mn$ products $e_i\otimes f_j$ form a basis of $V{\otimes }_{F}W$. There is one independent coordinate for every pair of indices.^[ref]

For a proof outline, bilinearity expands each pure tensor asso these products form a spanning set. For linear independence, take the coordinate functionals $e^k,f^{\ell }$ from the two dual bases. The bilinear map $(v,w)\mapsto e^k(v)f^{\ell }(w)$ induces a linear functional ${\lambda }_{k\ell }$ on the tensor product, satisfying ${\lambda }_{k\ell }(e_i\otimes f_j)={\delta }_{ki}{\delta }_{\ell j}$. Applying it to a proposed zero linear combination selects just $c_{k\ell }$, forcing every coefficient to vanish. Thus dimensions multiply, rather than add. This finite-dimensional count includes a zero factor, whose tensor product has dimension zero.

After choosing ordered bases, identify a tensor $t={\sum }_{i,j}{c}_{ij}{e}_i\otimes f_j$ with its matrix $C(t)=(c_{ij})$. This is a linear isomorphism between $V{\otimes }_{F}W$ and the vector space of $m$-by-$n$ matrices: each tensor-product basis element becomes the matrix with one entry equal to one and all others zero. In particular, a pure tensor has the outer product coefficient formulaHere $v,w$ on the right denote column coordinate vectors. The identification depends on the chosen bases, while the tensor itself does not. Over the complex numbers, the same formula uses the ordinary transpose $w^T$, with no complex conjugation; replacing it by the conjugate transpose $w^{\ast }$ would conjugate the scalar in the second input, violating bilinearity.

The two-factor tensor rank is the smallest number of pure tensors whose sum equals $t$, with zero allowed as the empty sum. For two finite-dimensional vector spaces over the same field, it equals the matrix rank of the tensor-product coefficient matrix. A sum of $r$ outer products has column space contained in the span of their $r$ first factors, so its matrix rank is at most $r$. Conversely, choose a basis $u_1,\dots ,u_r$ for the column space, express each column in that basis, and collect the coefficients into row vectors $w_k^T$. This yields $C={\sum }_{k=1}^r{u}_k{w}_k^T$, hence a sum of $r$ pure tensors.

The pure tensor $(2,1)\otimes (-1,3)$ therefore has two-factor tensor rank one, whereas $e_1\otimes f_1+e_2\otimes f_2$ has two-factor tensor rank two. Both have two factors: the number of factors and the minimum number of product summands are different counts. Over the real numbers, the singular value decomposition supplies a useful decomposition $C={\sum }_k{\sigma }_k{u}_k{v}_k^T$, with one nonzero term per matrix rank. The two-factor rank equality does not extend to three or more factors by merely choosing one matrix layout.