If we consider an affine space $\mathbb{A}_K^n=\mathrm{Spec}\,K[T_1,\cdots,T_m]$ over a field $K$. It's easy to show that $T_x\mathbb{A}_K^n\simeq K^n$ where $x$ is a $K$-point corresponding to the maximal ideal of the form $(T_1-x_1,\cdots,T_n-x_n)$. But I wonder how to show that $\dim T_x\mathbb{A}_K^n=n$ (or maybe fail to equal) for a general closed point correspond to a general maximal ideal $\mathfrak m$.
I tried to consider the map $\mathfrak m\to \kappa(x)^n,\,g\mapsto(\frac{\partial g}{\partial T_1}(x),\cdots,\frac{\partial g}{\partial T_n}(x))$. If $x$ is $K$-point, it's easy to show that the kernel is $\mathfrak m^2$, and induced a bijection: $\mathfrak m/\mathfrak m^2\to \kappa(x)^n=K^n$. But in the general case, is that right? I think it's just a injection. This is equivalent to prove the following:
Conjecture: If $g\in\mathfrak m$ and we have $\dfrac{\partial g}{\partial T_i}\in\mathfrak m$ for all $i$, then $g\in\mathfrak m^2$.
If $\mathfrak m=(T_1-x_1,\cdots,T_n-x_n)$, it is verified by Taylor expansion. But for general maximal ideal, I don't know how to do it.