Soergel generators and Bott–Samelson products are finite free on both sides

The decomposition $R=R^{s_i}\oplus {\alpha }_i{R}^{s_i}$ is a direct sum of graded $(R^{s_i},R^{s_i})$-submodules, because the averaging idempotent $e=\frac{1}{2}(1+s_i)$ satisfies $e^2=e$, $s_{i}e=e{s}_i=e$ and $\ker e=(1-e)R=R(1-e)$ with $1-e=\frac{1}{2}(1-s_i)$; applied to any $f$ this gives $f=e(f)+(1-e)(f)$ with $e(f)\in R^{s_i}$ and $(1-e)(f)=\frac{1}{2}(f-s_i(f))=\frac{1}{2}{\alpha }_i{\mathrm{\partial }}_i(f)={\delta }_i{\mathrm{\partial }}_i(f)\in {\alpha }_i{R}^{s_i}$, by the formula ${\mathrm{\partial }}_i(f)={\alpha }_i^{-1}(f-s_i(f))$ and ${\delta }_i={\alpha }_i\mathrm{/}2$.

Left side: tensoring the decomposition of step 1.1 over $R^{s_i}$ in the right tensor slot gives the decomposition $R{\otimes }_{R^{s_i}}R=(R{\otimes }_{R^{s_i}}{R}^{s_i})\oplus (R{\otimes }_{R^{s_i}}{\alpha }_i{R}^{s_i})$ of graded left $R$-modules (the left action multiplies the first slot), with summands $\{r\otimes 1:r\in R\}$ and $\{r\otimes {\alpha }_i{r}^{\mathrm{\prime }}:r\in R,r^{\mathrm{\prime }}\in R^{s_i}\}$: the first is isomorphic to $R$ as a graded left $R$-module through $r\otimes 1\mapsto r$ with inverse $r\mapsto r\otimes 1$, and the second is free of rank one as a left $R$-module with generator $1\otimes {\alpha }_i$, hence isomorphic to $R$ through $r\otimes {\alpha }_i{r}^{\mathrm{\prime }}\mapsto r{r}^{\mathrm{\prime }}$ (well defined by the balanced relation $r\otimes {\alpha }_i{r}^{\mathrm{\prime }}=(r{r}^{\mathrm{\prime }})\otimes {\alpha }_i$ for $r^{\mathrm{\prime }}\in R^{s_i}$, and left $R$-linear because $r^{\mathrm{\prime }\mathrm{\prime }}(r\otimes {\alpha }_i{r}^{\mathrm{\prime }})=r^{\mathrm{\prime }\mathrm{\prime }}r\otimes {\alpha }_i{r}^{\mathrm{\prime }}$), with inverse $t\mapsto t\otimes {\alpha }_i$. Hence the elements $1\otimes 1$ and $1\otimes {\alpha }_i$ form a graded left $R$-basis, of degrees $0$ and $2$ before the shift.

Right side: since ${\alpha }_i{R}^{s_i}\subseteq R$ is also an $(R^{s_i},R^{s_i})$-sub-bimodule, tensoring the decomposition of step 1.1 in the left tensor slot gives the decomposition $R{\otimes }_{R^{s_i}}R=(R^{s_i}{\otimes }_{R^{s_i}}R)\oplus ({\alpha }_i{R}^{s_i}{\otimes }_{R^{s_i}}R)$ of graded right $R$-modules (the right action multiplies the second slot); the first summand is the set of elements $r^{\mathrm{\prime }}\otimes r$ with $r^{\mathrm{\prime }}\in R^{s_i}$ and is isomorphic to $R$ through $1\otimes r\mapsto r$ with inverse $r\mapsto 1\otimes r$, and the second summand is the set of elements ${\alpha }_i{r}^{\mathrm{\prime }}\otimes r={\alpha }_i\otimes r^{\mathrm{\prime }}r$ with $r^{\mathrm{\prime }}\in R^{s_i}$, free of rank one as a right $R$-module with generator ${\alpha }_i\otimes 1$ and isomorphic to $R$ through ${\alpha }_i\otimes r\mapsto r$. Hence $1\otimes 1$ and ${\alpha }_i\otimes 1$ form a graded right $R$-basis, again of degrees $0$ and $2$; note that $1\otimes {\alpha }_i=(1\otimes 1)\cdot {\alpha }_i$ is ${\alpha }_i$ times the first basis element and is not a right-basis element.

Applying the external shift $(1)$, which lowers degrees by one and preserves freeness with the same ranks, gives $B_i\cong R(-1)\oplus R(1)$ as a graded left $R$-module and as a graded right $R$-module, with left basis $1\otimes 1$ of degree $-1$ and $1\otimes {\alpha }_i$ of degree $1$ and right basis $1\otimes 1$ of degree $-1$ and ${\alpha }_i\otimes 1$ of degree $1$; in particular $B_i$ is finite free of rank two on each side.

Tensor products: if $M$ is finite free as a left $R$-module with basis $m_1,\dots ,m_a$ and $N$ is finite free as a left $R$-module with basis $n_1,\dots ,n_b$, expand the second factor in its left basis first: $N\cong {⨁}_{q}R{n}_q$ as a left $R$-module, hence $M{\otimes }_{R}N\cong {⨁}_{q}M$ as a left $R$-module via $m\otimes {\sum }_q{r}_q{n}_q\mapsto (m{r}_q{)}_q$. Expanding each copy of $M$ in its left basis gives the left basis $m_p\otimes n_q$; the inverse sends the $(p,q)$-th basis vector to that tensor. For right freeness, expand the first factor in its right basis and then the second factor in its right basis. These side-correct identifications respect homogeneous degrees, so an iterated tensor product of the $B_i$ is finite free of rank $2^r$ on each side.

Therefore $B_i$ is free of rank two on both sides and every Bott–Samelson product $B_{i_1}{\otimes }_R\cdots {\otimes }_R{B}_{i_r}$ is finite free on both sides, with left and right bases obtained by tensoring the two-element bases of the factors and with degrees the sums of the factor degrees.