partial derivative of a polynomial belongs to a maximal ideal

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.

• @baharampuri The $f$ seems not contained in $\mathfrak m$. – user8891548 Nov 24 '18 at 7:09
• Sorry missed that. – baharampuri Nov 24 '18 at 7:22
• One clarifying question: which field are you considering $T_x\Bbb A^n_K$ as a vector space over when you ask for its dimension? If you ask over the residue field at $x$, the answer is always $n$, whereas if you ask over the field $K$, the dimension is $n$ times the degree of the residue field as an extension over $K$. – KReiser Nov 24 '18 at 8:06
• @KReiser $T_x\mathbb A_K^n$ considered as a $\kappa(x)$-vector space. May I ask how to prove that the dimension is $n$? – user8891548 Nov 24 '18 at 8:45