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Here is a quote from @AlexYoucis who my other question about Zariski (co)tangent spaces:

"Let $X=k[x_1,\ldots,x_n]/(f_1,\ldots,f_r)$ be our affine finite type $k$-scheme and $p=(a_1,\ldots,a_n)=(x_1-a_1,\ldots,x_n-a_n)$ be our point. We obtain a linear map $J_p:k^n\to k^r$ defined by the Jacobian map:

$$J_p=\begin{pmatrix}\frac{\partial f_1}{\partial x_1}(p) & \cdots & \frac{\partial f_1}{\partial x_n}(p)\\ \vdots & \ddots & \vdots\\ \frac{\partial f_r}{\partial x_1}(p) & \cdots & \frac{\partial f_r}{\partial x_n}(p)\end{pmatrix}$$

Then, one can show that $T_{X,p}$ is isomorphic to $\ker J_p$."

For a discussion of how to prove this, follow the link above to my previous question. Apparently, we never use the hypothesis that $k$ is algebraically closed. However, Vakil states in a not very clear remark (12.2.5) that things don't work if $k$ is not algebraically closed, although he only introduces this hypothesis in exercise 12.2.D, which should follow as an easy consequence of what Youcis said above:

12.2.D. EASY EXERCISE. Suppose $k = \bar k$. Show that the singular closed points of the hypersurface $f(x_1, . . . , x_n) = 0$ in $\Bbb A^n_k$ are given by the equations $$f =\frac {∂f}{∂x_1}= · · · =\frac {∂f}{∂x_1n} =0$$

Even in this exercise I don't see where we need $k$ to be algebraically closed. The corank of the transpose of the Jacobian in this case is equal to $n-1$ unless all the derivatives vanish (in which case it will be equal to $n$). Where do we need algebraically closed here?

In these other notes from Vakil (which are a slightly modified excerpt of FOAG) Vakil does mention that to prove that the (co)tangent space of a finite type $k$-scheme can be obtained with the Jacobian, one first starts shows that it holds at the origin $o=(x_1-0,...x_n-0)$ (where this step doesn't require algebraically closedness) and then uses that $k=\bar k$ to translate any other point $p$ to $o$. How so? Why do we need algebraically closedness to translate "$p$ to the origin"?

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  • I just wanted to remark that the Hilbert Nullstellensatz only yields that all maximal ideals are of the form $(x_1-a_1,\ldots,x_n-a_n)$ over an algebraically closed field. Maybe that is the source of the problem, at least in the case where you need to go from the origin to any point. – Jesko Hüttenhain Feb 15 '14 at 07:30
  • The key is that the computation I laid out only works for $k$-rational point. The reason that $k=\overline{k}$ is so desirable, is then $k$-rational is equivalent to closed. – Alex Youcis Feb 15 '14 at 09:47
  • This holds for separable points. In general the jacobian counts the differential forms, and the tangent space is not the dual of the differential forms for inseparable points. For example, if $X$ is regular but not smooth at some closed point $x_0$, then the tangent space has dimension $d=\dim_{x_0} X$, but the Jacobian has rank $<d$. – Cantlog Feb 15 '14 at 18:48
  • @Cantlog I expressed this to the OP in the other post--I'm not precisely sure what he wants. – Alex Youcis Feb 16 '14 at 07:29

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