Dimension of a tensor algebra quotient For example, for arbitrary $V$ we can find a dimension of $ \bigwedge V = T(V) / (u \otimes u)$ by writing basis explicitly -- the dimension of $\bigwedge V$ is $2^{\dim{V}}$.
Let's consider a tensor algebra $T(V)$ of a vector space $V =  \mathbb{ C}^{2}$ with a standard basis $(e_{1}, e_{2})$. 
More generally: how to find the dimension tensor algebra quotient by arbitrary ideal? For instance, let $I = (e_{1} \otimes e_{1}, e_{2} \otimes e_{2}, e_{1} \otimes e_{2} + e_{2} \otimes e_{1} - 1)$ be an two-sided ideal, $A = T(V) / (I)$. How to find the dimension of $A$? The quotient may be described as a free algebra, generated by $e_{1}, e_{2}$ with the given relations: $e_{1} \otimes e_{1} = 0,  e_{2} \otimes e_{2}=0, e_{1} \otimes e_{2} + e_{2} \otimes e_{1}=1$ but  deriving the desired result from the given data seems to be tricky enough. Are there any hints that might help?
Any help would be much appreciated.
 A: There are better answers than mine since I have little experience with this. I'll just share how I approach these dimension questions.
I think that finding a basis really is the way to go. The tensor algebra is graded by $T(V)=\bigoplus_{n\in\mathbb{N}} T_n(V)$, where $T_n(V)=V^{\otimes n}$ has a basis given by $n$-fold tensors of basis elements $e_1,\ldots,e_n$ of $V$. I usually try to get better and better upper bounds on the dimension by considering how the relations allow me to rewrite these $n$-fold tensors of basis elements. Let me try on your example.
In your example, the third relation suggests that any $n$-fold tensor of $e_1$ and $e_2$ could be reordered so that all $e_1$'s appear before all $e_2$'s, at the expense of adding some lower grade terms. That is,
$$
e_{i_1}\otimes \cdots \otimes e_{i_n} = e_1\otimes \cdots \otimes e_1 \otimes e_2\otimes \cdots \otimes e_2+[\text{lower terms}]
$$
But then the first two relations kill off any pair of adjacent $e_1$'s or $e_2$'s. Thus the first term above is zero whenever $n>2$. So any $n$-fold tensor of $e_1,e_2$ could be written as a sum of lower grade terms if $n>2$. Thus the subspace $T_2(V)/I$ of $T(V)/I$ is in fact everything. Better yet, $1,e_1,e_2,e_1\otimes e_2$ span the algebra (i.e. you don't need $e_2\otimes e_1$).
It remains to find a basis. Are $1,e_1,e_2,e_1\otimes e_2$ linearly independent in $A$? [EDIT: The comments mention some decent ways to approach this.]
