Background
The tensor space of type $(r,s)$ associated with $V$ is the vector space $$\underbrace{V\otimes \ldots \otimes V}_{\text{r copies}} \otimes \underbrace{V^* \otimes \ldots \otimes V^*}_{\text{s copies}}.$$
The tensor algebra of $V$ is the direct sum $T(V)=\sum_{r,s} V_{r,s}$ for $r,s \geq 0.$
Let $C(V)$ be the subalgebra $\sum_{k=0}^{\infty} V_{k,0}$ of $T(V)$. Let $I(V)$ denote the two sided ideal in $C(V)$ generated by the set of elements of the form $v \otimes v,$ where $v \in V$ and set $$I_k(V)=I(V) \cap V_{k,0}.$$
The exterior algebra $\Lambda(V)$ is the graded algebra $C(V)/I(V)$. If we set $$\Lambda_k(V)=V_{k,0}/I_k(V)$$ for $k \geq 2$ and $\Lambda_0(V)=\mathbb{R}, \ \Lambda_1(V)=V$ then $$\Lambda(V)=\sum_{k=0}^{\infty} \Lambda_k(V).$$
Denote multiplication in the algebra $\Lambda(V)$ by $\wedge$
Questions
I know very little abstract algebra so a lot of this construction is strange to me. I have the following questions:
What is the point of doing the intersection $I(V) \cap V_{k,0}$ when defining $I_k(V)$?
Why is the exterior algebra graded? Here is what I attempted to do:
Let $u \in \Lambda_k(V)$ and $v \in \Lambda_l(V)$ We should show that $u \wedge v \in \Lambda_{k+l}.$ I think $u \in \Lambda_k$ means that $u$ is the equivalence class $u+I_{k}(V)$ and similarly $v$ is the equivalence class $v+I_{l}(V).$ I also think that $u \wedge v$ corresponds to the equivalence class of $u \otimes v$ in (I don't know what space).
As you can see, my understanding of this construction is very weak. I think if I see $2$ clearly I will understand how to work with the exterior algebra. To keep this question unambiguous, please consider my attempt with question $2$ above and try to explain, explicitly, what an element in $\Lambda_n(V)$ looks like (and how I can see this from the construction above) and also to explain what $u \wedge v$ means in terms of the quotient construction.