Zariski cotangent space, as defined in Arapura's "Algebraic Geometry over the Complex Numbers" In "Algebraic Geometry over the Complex Numbers", Arapura gives the following definition:

Definition 2.5.8. When $(R, m, k)$ is a local ring satisfying the tangent space conditions, we define its cotangent space as $T_R^* = m/m^2 = m \otimes_R k$, [...]

To give an example, he restricts to $\mathbb{R}^n$. Then $R$ is the ring of germs of $C^\infty$ functions at $x=0$ and the maximal ideal $m$ consists of those germs that vanish at $0$; it is generated by the coordinate functions $\{ x_i \}$. $k = R/m = \mathbb{R}$. Given a function $f$ with Taylor expansion $f(x) = f(0) + \sum \frac{\partial f}{\partial x_i}\vert_0 x_i + \mathcal{O}(x^2)$, the second term corresponds to the elements of $m/m^2$. What I do not understand is how that space is supposed to be isomorphic to the tensor product $m \otimes_R k$.Naively, $m$ includes the "second order behavior" that was discarded in the quotient, I don't see how tensoring with $k$ improves that.
 A: One way to view this is that any element of $\mathfrak m \otimes_R k$ can be put in the form $m \otimes 1$ with $m \in \mathfrak m$. This is because elements of $k$ can be seen as equivalence classes of elements of $R$, thus can be pulled "to the left" as in 
$$
\sum_{i=1}^n m_i \otimes \overline{r_i} = \left( \sum_{i=1}^n r_i m_i \right) \otimes 1. 
$$
Thus addition in $\mathfrak m \otimes_R k$ behaves exactly like addition in $\mathfrak m$. Let $m \in \mathfrak m$. If $m \equiv 0 \pmod{\mathfrak m^2}$, then $m \otimes 1 = 0$ since $m = \sum_i m'_im''_i$ for some $m'_i,m''_i \in \mathfrak m$, hence $m \otimes 1 = \sum_i m'_i \otimes m''_i = 0$. Conversely, the multiplication map $\mathfrak m \otimes_R R/\mathfrak m \to \mathfrak m/\mathfrak m^2$ maps $m \otimes 1$ to a non-zero element if $m \not\equiv 0 \pmod{\mathfrak m^2}$ (it is well-defined since the multiplication map $\mathfrak m \otimes_R \mathfrak m \to \mathfrak m/\mathfrak m^2$ is the zero map), thus $m \otimes 1 \neq 0$ if $m \in R \setminus \mathfrak m^2$. Therefore, the multiplication map gives an isomorphism $\mathfrak m \otimes_R R/\mathfrak m \simeq \mathfrak m / \mathfrak m^2$. 
More generally, for a ring $R$ and an $R$-module $M$ together with an ideal $\mathfrak a \trianglelefteq R$, you always have the isomorphism $M/\mathfrak a M \simeq M \otimes_R R/\mathfrak a$. I usually like to prove this using Snake's lemma. This is not hard : the cokernel of the map $\varphi : R/\mathfrak a \otimes_R M \to M/\mathfrak a M$ is obviously zero (because $\overline 1 \otimes m \mapsto \overline m$) and since the map $R \otimes_R M \to M$ is an isomorphism, its kernel is zero. Since the map $\mathfrak a \otimes_R M \to \mathfrak aM$ is surjective, its cokernel is zero ; since the Snake exact sequence is exact at $\ker \varphi$, we have an exact sequence of the form $0 \to \ker \varphi \to 0$, which means $\ker \varphi = 0$. (I suggest you draw the diagram.)
Hint : Put consider the two exact sequences
$$
\mathfrak a \otimes_R M \to R \otimes_R M \to (R/\mathfrak a) \otimes_R M \to 0
$$ 
and
$$
0 \to \mathfrak a M \to M \to M/\mathfrak a M \to 0.
$$
There is an obvious way to map the top part to the bottom part in a way that coincides with the diagram arising in Snake's lemma. Compute kernels and cokernels to deduce that the canonical map $(R/\mathfrak a) \otimes_R M \to M/\mathfrak aM$ is an isomorphism. 
Hope that helps,
