Prove that $T_pM^*=\mathfrak m_p/\mathfrak m_p^2$. Let $M$ a manifold of dimension $n$ and $T_pM$ its tangent space. Let denote $T_p^*M=\mathcal L(T_pM,\mathbb R)$ it's dual. Let also denote $$\mathfrak m_p=\{f\in\mathcal C^\infty (M)\mid f(p)=0\}\quad \text{and}\quad \mathfrak m_p^2=\left\{\sum_{i=1}^k f_ig_i\mid f_i,g_i\in\mathfrak m_p\right\}.$$
Show that $T_pM^*=\mathfrak m_p/\mathfrak m_p^2$.
Let $$\omega :\mathfrak m_p\longrightarrow T_p^*M$$
defined by $$\omega (f)=\omega (f)(X)=X(f)$$
for all derivation $X\in T_pM$. It's easy to prove that $\mathfrak m_p^2\subset \ker \omega $. Then I would like to use the first isomorphisme theorem, but I don't know if I can.  
Am I on the right way ? How to continue ?
By the way, what does mean that $T_p^*M=\mathfrak m_p/\mathfrak m_p^2$ ? I can't give a concret representation of it.

Choose local coordinates $x_i$ around $p$, with $x_i(p)=0$. Then $m_p$ is the ideal of functions which vanish at $p$. Such a function is of the form $\sum_{i=1}^na_ix_i+ o(x)$. Then $m_p^2$ is the ideal generated by function which vanishes at the order 2 i.e. $f(x)=o(x)$, and the quotient is just the vector space of linear function $x\to \sum_{i=1}^na_ix_i$, i.e. the cotangent space at $p$.
