Details to the Proof of 11.2

With the help of the Diamond Lemma for Ring Theory [Bg] we will construct a free basis for ${\bigwedge}_{R,q}(n)$. Our order $\prec$ on $\underline{n}$ induces a lexcographical order on multi-indices ${\bf i} \in I(n,r)$ and on the correspondinmg monomials $v_{{\bf i}} \in V^{\otimes r}$, which will be denoted by the same symbol $\prec$. On the monomials of ${\cal T}(V)$ we get an induces partial order if we declare monomials of different degree to be uncomparable. It is clear that $\prec$ is compatible with the semigroup structure of the set of monomials of ${\cal T}(V)$ as required in [Bg].

We now introduce a system of reductions of degree two in ${\cal T}(V)$ extracted from the relations (13), (14) and (16) of the exterior algebra and divide them accordingly into three types. As in [Bg] we write them as pairs consisting of a monomial and a substitution expression:

$$\begin{array}{cll} (R1) & (v_{i}v_{j},-q^{{\rm sign}(j-i)}v_{j}v... ... i < m \\ (R3) & (v_{i}v_{i},0) & \mbox{ if } 1 \leq i \leq n. \end{array}$$

Since all monomials of the reduction system are greater than the monomials in the corresponding substitution expressions our partial order $\prec$ on ${\cal T}(V)$ is compatible with the reduction system.

The set of monomials in $V^{\otimes r}$ which don't contain any monomial of the reduction system as a subexpression clearly is

$$F_r:=\{v_{{\bf i}}\vert {\bf i}=(i_1, \ldots , i_r) \in I(n,r) ,\; i_1 \prec \ldots \prec i_r \}.$$

Obviously $F_r$ generates the $r$-th homogeneous summand ${\bigwedge}_{R,q}(n,$) of the exterior algebra as an $R$-module. To see that these sets are linear independent we must show that all ambiguities of the reduction system are solvable. Since all monomials are of degree two only overlapping ambiguities occur and we can reduce to the case of degree three. Ambiguities between reductions of type (R3) are trivially solvable and such ones where both reductions are of type (R2) do not occur. Thus we have to handle the following remaining cases:

1. Both reductions are of type (R1):
   $v_{i}v_{j}v_{k}$ where $k \prec j \prec i$ and $i \neq j' \neq k$.
2. Ambiguities between (R1) und (R3):
   (a) $v_{i}v_{i}v_{j}$ where $j \prec i$ and $i\neq j'$
   (b) $v_{j}v_{i}v_{i}$ where $i \prec j$ and $i\neq j'$
3. Ambiguities between (R2) und (R3):
   (a) $v_{i'}v_{i}v_{i}$ where $1\leq i\leq m$
   (b) $v_{i'}v_{i'}v_{i}$ where $1\leq i\leq m$
4. Ambiguities between (R1) und (R2):
   (a) $v_{i}v_{j'}v_{j}$ where $j' \prec i$ and $1 \leq j \leq m$
   (b) $v_{j'}v_{j}v_{i}$ where $i \prec j$ and $1 \leq j \leq m$

In order to prove solvability of these ambiguities we will write an application of a reduction as: monomial $\mapsto$ substitution expression. The first case can be solved in the following way beginning with reduction of the left hand side pair

$$v_{i}v_{j}v_{k} \mapsto -q^{{\rm sign}(j-i)}v_{j}v_{i}v_{k} \ma... ...\mapsto -q^{{\rm sign}(j-i)+{\rm sign}(k-i)+{\rm sign}(k-j)}v_{k}v_{j}v_{i},$$

and then begining with the right hand side pair

$$v_{i}v_{j}v_{k} \mapsto -q^{{\rm sign}(k-j)}v_{i}v_{k}v_{j} \map... ...\mapsto -q^{{\rm sign}(j-i)+{\rm sign}(k-i)+{\rm sign}(k-j)}v_{k}v_{j}v_{i}.$$

The treatment of case 2 is very easy and must not be written down. In order to treat case 3 (a) we have to show that starting with a reduction of type (R2) on the left hand side pair finally reduces to zero.

$$\begin{array}{lll} v_{i'}v_{i}v_{i} & \mapsto & -q^{-2}v... ...-2}-1)\sum_{j=i+1}^mq^{i-j}v_{i}v_{j}v_{j'} =0. \end{array}$$

Part (b) of case 3 is similar and we can proceed to case 4. Condition $j' \prec i$ means that $i < j$ (the case $i=j$ has been treated above) or $i > j'$ whereas $i \prec j$ means that $j < i < j'$. As above we reduce begining with the left hand side pair in (a)

$$\begin{array}{ll} v_{i}v_{j'}v_{j} \mapsto \ldots \mapst... ...rm sign}(j'-i)+{\rm sign}(j-i)}v_{k}v_{k'}v_{i}. \end{array}$$

and then beginning with the right hand side pair

$$\begin{array}{ll} v_{i}v_{j'}v_{j} \mapsto \ldots \mapst... ...rm sign}(k'-i)+{\rm sign}(k-i)}v_{k}v_{k'}v_{i}, \end{array}$$

Since $i < j$ or $i > j'$ and $j < k$ we have ${\rm sign}(j'-i)={\rm sign}(k'-i)$ and ${\rm sign}(j-i)={\rm sign}(k-i)$. Thus both reductions lead to the same expression. Turning to part (b) the calculation of both reductions lead to similar expressions but we have to divide the sum into a $i\prec k$ and a $k\prec i$ section. First we begin with the right hand side pair in (b)

$$\begin{array}{ll} v_{j'}v_{j}v_{i} \mapsto \ldots \mapst... ...m sign}(k'-i)+{\rm sign}(k-i)} v_{k}v_{k'}v_{i} \end{array}$$

and then begining with the left hand side pair

$$\begin{array}{ll} v_{j'}v_{j}v_{i} \mapsto \ldots \mapst... ...m sign}(k'-i)+{\rm sign}(k-i)} v_{k}v_{k'}v_{i} \end{array}$$

Since $j < i < j'$ the expression ${\rm sign}(i-j)+{\rm sign}(i-j')$ is always zero. Thus both reduction coincide and the proof is finished.