Noncommutativity of tensor algebra My question is simple. 

Let $M$ be an $A$-module and let $T(M)$ be its tensor algebra. I saw that it is noncommutative in general... but I can' understand this fact...

I think that by commutativity of tensor product it is commutative... Help me... 
 A: It is true that, for $A$-modules $M$ and $N$, we have an isomorphism $M\otimes N \cong N\otimes M$.  But this is unrelated to the commutativity of the tensor algebra, and it might be confusing you.
Think about the module $M\otimes M$, which is the degree $2$ piece of the tensor algebra $T(M)$.  If $a,b\in M$, is always true that $a\otimes b = b\otimes a$?
Let's look at a simple example: let $A=k$ be a field, and $M=Av\oplus Aw$ be a $k$-vector space of dimension $2$.  What is the dimension of $M\otimes M$?
Depending on your familiarity with tensor products, you may know that $M\otimes M$ has dimension $4$, with basis $\{v\otimes v,v\otimes w, w\otimes v,w\otimes w \}$.  Clearly, this cannot be a basis if $v\otimes w = w\otimes v$!
If you are not very familiar with explicit constructions of tensor products, then consider the universal property: $\operatorname{Hom}(M\otimes M,N)$ should be in bijection with the set of bilinear maps $M\times M\to N$.  Letting $N=k$, then a bilinear form $M\times M = k^2 \times k^2 \to k$ is given by a 2-by-2 matrix $C$: $(a,b) \mapsto a^T C b$.  But then $(a,b)$ and $(b,a)$ only map to the same place when $C$ is symmetric—and clearly not all matrices are symmetric!  So in general, the tensors $a\otimes b$ and $b\otimes a$ had better be different.
A: Let me change notation: $R$ rather than $A$ will denote the (commutative) base ring, and I will abbreviate the tensor product $\otimes_R$ on the category of left $R$-modules to $\otimes$. 
Keep in mind that commutativity of multiplication $m$ on an $R$-algebra $A$ is expressed by a commutative triangle which says that the composite 
$$A \otimes A \stackrel{\sigma}{\to} A \otimes A \stackrel{m}{\to} A$$ 
equals $m$, where $\sigma$ is the natural symmetry isomorphism on the tensor product. However, if you look at how the multiplication on the tensor algebra $A = \sum_{n \geq 0} M^{\otimes n}$ is defined: 
$$A \otimes A \cong (\sum_i M^{\otimes i}) \otimes (\sum_j M^{\otimes j}) \cong \sum_n \sum_{i + j = n} M^{\otimes (i+j)} \stackrel{\sum_n \nabla}{\to} \sum_n M^{\otimes n} = A,$$ 
then you see that nowhere is the symmetry isomorphism used. In other words, the only thing used in this construction is the distributivity of $\otimes$ over coproducts, and it would make perfect sense even in contexts where $\otimes$ is not symmetric monoidal (e.g., the tensor product of bimodules). 
It migt help to run through this in the case where $R = k$, a field, and where $M = V$ is an $k$-dimensional vector space over $k$ with basis elements $e_1, \ldots, e_k$. In this case the construction $T(V) = \sum_n V^{\otimes n}$ may be identified with the noncommutative polynomial algebra generated by $k$ indeterminates $e_1, \ldots, e_k$, with $V^{\otimes n}$ the homogeneous component of degree $n$ monomials in $k$ variables; this has dimension $nk$ as expected. In the commutative polynomial algebra case, the dimension of the homogeneous component would be given rather by a multinomial coefficient. 
