Let $X$ be a (classical) variety over an algebraically closed field $k$, and let $x \in X$. Let $\mathcal O_{X,x}$ be the stalk at $x$, with maximal ideal $\mathfrak m_x$. Here is an intuitive way to think about the definition of the tangent space $T_x(X)$ as $(\mathfrak m_x/\mathfrak m_x^2)^{\ast}$:

If we instead pretend that $X$ is a real analytic manifold of dimension $n$, then $\mathcal O_{X,x}$ is isomorphic to the $\mathbb{R}$-algebra of convergent power series in $n$ variables. Its maximal ideal $\mathfrak m_x$ consists of all such power series with zero constant term, and $\mathfrak m_x^2$ consists of all power series where each nonzero term $a X_1^{\alpha_1} \cdots X_n^{\alpha_n}$ satisfies $\alpha_1 + \cdots + \alpha_n \geq 2$. Therefore, $$\mathfrak m_x/\mathfrak m_x^2 = \{ a_1X_1 + \cdots + a_nX_n + \mathfrak m_x^2 : a_i \in \mathbb{R}\}$$

with basis $X_i + \mathfrak m_x^2$. Now if $f = \sum\limits_{\alpha} a_{\alpha}X_1^{\alpha_1} \cdots X_n^{\alpha_n}$ is an analytic function centered at $x$, then the directional derivative of $f$ in a given direction $(c_1, ... , c_n)$ is equal to $a_1c_1 + \cdots + a_nc_n$. So we define the tangent space as a vector space having an inherent duality with $\mathfrak m_x/\mathfrak m_x^2$.

Now, let $X$ be a scheme of finite type over a field $k$. What should the definition of the tangent space at a point $x \in X$ be, and why? Should $x$ be a closed point?

  • $\begingroup$ To me it does not become clear what you are asking? I mean you cannot ask how the tangent space of a finite type scheme is defined, since you can check this in any text on algebraic geometry ?! $\endgroup$ – MooS Jun 6 '17 at 5:07
  • 2
    $\begingroup$ Your question already contains your answer: $(m/m^2)^*$ is fine. Tangent spaces to non-closed points are fine as well- they're simply vector spaces over fields of positive transcendence degree. $\endgroup$ – KReiser Jun 6 '17 at 7:23
  • $\begingroup$ You can find a nice answer in Vakil's FOAG! $\endgroup$ – Armando j18eos Jun 6 '17 at 10:40

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