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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.

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  • $\begingroup$ @baharampuri The $f$ seems not contained in $\mathfrak m$. $\endgroup$ – user8891548 Nov 24 '18 at 7:09
  • $\begingroup$ Sorry missed that. $\endgroup$ – baharampuri Nov 24 '18 at 7:22
  • $\begingroup$ 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 $K$. $\endgroup$ – KReiser Nov 24 '18 at 8:06
  • $\begingroup$ @KReiser $T_x\mathbb A_K^n$ considered as a $\kappa(x)$-vector space. May I ask how to prove that the dimension is $n$? $\endgroup$ – user8891548 Nov 24 '18 at 8:45

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