The tensor product of $R$-algebras has multiplication $(a\otimes b)(a^{\mathrm{\prime }}\otimes b^{\mathrm{\prime }})=a{a}^{\mathrm{\prime }}\otimes b{b}^{\mathrm{\prime }}$

The map $(a,b,a^{\mathrm{\prime }},b^{\mathrm{\prime }})\mapsto a{a}^{\mathrm{\prime }}\otimes b{b}^{\mathrm{\prime }}$ is $R$-multilinear: centrality in [L1] lets a scalar move among the four variables without changing the value.

By [L2], step 1.1 induces an $R$-bilinear multiplication $(A{\otimes }_{R}B)\times (A{\otimes }_{R}B)\to A{\otimes }_{R}B$ with the displayed pure-tensor formula; [L4] ensures that the formula has descended before any ring laws are used.

Both sides of associativity are trilinear in the three tensor arguments and agree on pure tensors by associativity in $A$ and $B$; uniqueness in [L2] makes them equal everywhere. Both distributive laws hold because the multiplication from step 2.1 is bilinear.

Left and right multiplication by $1_A\otimes 1_B$ are linear maps that agree with the identity on every pure tensor, so uniqueness in the two-factor case of [L2] makes them the identity maps.

If $A$ and $B$ are commutative, then $(a\otimes b)(a^{\mathrm{\prime }}\otimes b^{\mathrm{\prime }})=a{a}^{\mathrm{\prime }}\otimes b{b}^{\mathrm{\prime }}=a^{\mathrm{\prime }}a\otimes b^{\mathrm{\prime }}b=(a^{\mathrm{\prime }}\otimes b^{\mathrm{\prime }})(a\otimes b)$. The two bilinear multiplication maps therefore induce the same four-variable multilinear map, so [L2] gives commutativity for arbitrary tensors.

The map $R\to A{\otimes }_{R}B$, $r\mapsto r(1_A\otimes 1_B)$, is a unital ring homomorphism and its image is central, checked on pure tensors using [L1] and [L3]. Thus the resulting ring is an $R$-algebra.

Uniqueness in [L2] forces the multiplication from its displayed pure-tensor formula, while the identity and structure map are then forced by the displayed elements. Steps 2.1 through 4.1 prove existence and all asserted properties.