The type-A Iwahori-Hecke presentation of the finite Hecke algebra

For every $w\in S_n$ choose a reduced word $w=s_{i_1}\cdots s_{i_{\mathrm{\ell }}}$; step by step the partial products have length $1,2,\dots ,\mathrm{\ell }$, so repeated application of [F4] gives $T_w=T_{s_{i_1}}\cdots T_{s_{i_{\mathrm{\ell }}}}$. Since the $T_w$ form a basis of $H$ by [F3], every element of $H$ is a finite linear combination of products of the generators $T_{s_1},\dots ,T_{s_{n-1}}$: clause (1).

The quadratic relation of clause (2) is [F5]. For $\mathrm{\mid }i-j\mathrm{\mid }>1$ the simple transpositions commute, and both products in $T_{s_i}{T}_{s_j}=T_{s_i{s}_j}=T_{s_j{s}_i}=T_{s_j}{T}_{s_i}$ are length-additive because $s_i{s}_j$ has exactly two inversions and $\mathrm{\ell }(s_i)=\mathrm{\ell }(s_j)=1$ by [F6]; [F4] applies. If $\mathrm{\mid }i-j\mathrm{\mid }=1$, the two permutations $s_i{s}_{i+1}{s}_i$ and $s_{i+1}{s}_i{s}_{i+1}$ are equal, as is checked by their action on $i,i+1,i+2$ and on the remaining points, and each product of the three generators is length-additive since these permutations have exactly three inversions by [F6]; applying [F4] to both sides gives $T_{s_i}{T}_{s_{i+1}}{T}_{s_i}=T_{s_i{s}_{i+1}{s}_i}=T_{s_{i+1}{s}_i{s}_{i+1}}=T_{s_{i+1}}{T}_{s_i}{T}_{s_{i+1}}$.

Let $F$ be the free unital associative $A$-algebra on $T_1,\dots ,T_{n-1}$ and $I\subseteq F$ the two-sided ideal generated by the relators of [F1], so that $H_v(n)=F\mathrm{/}I$, and put $B:=\mathbb{C}{\otimes }_A{H}_v(n)$. By right exactness of tensoring [F7], $B\cong (\mathbb{C}{\otimes }_{A}F)\mathrm{/}\mathrm{im}(\mathbb{C}{\otimes }_{A}I\to \mathbb{C}{\otimes }_{A}F)$ as $\mathbb{C}$-algebras; the base change of the free algebra is the free unital $\mathbb{C}$-algebra on the images ${\tau }_i:=1\otimes T_i$, and the image ideal is generated by the specialized relators: every element of $I$ is a finite sum of products $arb$ with $a,b\in F$ and $r$ a defining relator, so its image lies in that ideal, while every specialized relator and its products lie in the image. These relators are ${\tau }_i^2-(q-1){\tau }_i-q\cdot 1_F$, the braid relators and the commutation relators, with $1_F$ the unit. Hence $B$ is presented by the three families of relations of clause (2), that is, $B\cong H(n)$; the unit is the algebra unit in both cases, so the relation is ${\tau }_i^2=(q-1){\tau }_i+q\cdot 1$.

By [F2] the algebra $H_v(n)$ is free over $A$ with basis $\{T_w\}$, so its base change $B\cong H(n)$ has $\mathbb{C}$-basis $\{1\otimes T_w\}$ and dimension $n!$. The assignment ${\tau }_i\mapsto T_{s_i}$ defines a unital $\mathbb{C}$-algebra homomorphism $\phi :H(n)\to H$ because the three families of relations hold in $H$ by step 1.2; it is surjective by step 1.1. Since ${\dim }_{\mathbb{C}}H(n)=n!={\dim }_{\mathbb{C}}H$ by [F3], a surjection between vector spaces of equal finite dimension is an isomorphism, so $H\cong H(n)\cong \mathbb{C}{\otimes }_A{H}_v(n)$ under $v\mapsto q$: clause (3). Under $\phi$ the basis element $1\otimes T_w$ of $H_v(n)$ maps to the product $T_{s_{i_1}}\cdots T_{s_{i_{\mathrm{\ell }}}}$ along a reduced word, which is $T_w$ by [F4]; hence the images of the standard basis $T_w$ form a basis of $H$, in agreement with [F3].

Step 1.1 proves clause (1), step 1.2 proves clause (2) with the unit displayed explicitly, and steps 1.3 and 2.1 prove clause (3) and the specialization statement; all algebras are finite-dimensional over $\mathbb{C}$ or free of finite rank over $A$, the generators and permutation matrices are explicit, and no choice principle is used.