Let $\Omega$ be an open set of $\mathbb R^n$ and $\mathcal F(\Omega)$ be the unital commutative ring of infinitely differentiable real functions over $\Omega$. Given $a\in\Omega$, consider $\mathfrak m_a$ as the ideal given by: $$\mathfrak m_a=\big\{f\in\mathcal F(M)\text{ such that }f(a)=0\big\}$$ I want prove to that $\dim (\mathfrak m_a/\mathfrak m_a^2)=n$.
For do so, I am going to suppose that $\Omega$ is open convex. By Hadamard's lemma each $f\in \mathfrak m_a$ can be written as: $$f(x)=\sum_{\mu=1}^n f_\mu(x)(x^\mu-a^\mu),$$ where $f_1,\ldots,f_n$ are smooth real functions on $\Omega$ such that $f_\mu(a)=(\partial_\mu f)(a)$ for each $\mu$.
This implies that two functions $f,g\in\mathfrak m_a$ lies in the same equivalence class if and only if: $$(\partial_\mu f)(a)=(\partial_\mu g)(a),\qquad\qquad(\star)$$ for each $1\leq\mu\leq n$
Now, denote $T_a^\star\mathbb R^n$ as the $\mathbb R$-vector space of linear maps $T_a\mathbb R^n\to \mathbb R$. Then using ($\star$) it is not difficult to show that: $$\begin{array}{rcll} \Phi: &\mathfrak m_a/\mathfrak m_a^2&\longrightarrow &T_a^\star\mathbb R^n\\ &f+\mathfrak m_a^2&\longrightarrow & \Phi(f+\mathfrak m_a^2)=d_af \end{array}$$ is a linear isomorphism.
My question is: How can I show that $\Phi$ is in fact an isomorphism without supposing that $\Omega$ is convex?