For $X$ an affine variety and $p \in X$ define $T_p X = \mathrm{Der}(k[X], \mathrm{ev}_p)$.

Claim: If $Y = X \setminus Z(f)$ is some Zariski open affine subvariety of $X$ and $p \in Y$, then $T_p X \cong T_pY$.

Definition: Let $X$ be an arbitrary variety, $p \in X$. Define $T_p X = T_p U$ for any open affine neighbourhood $U$ of $p$.

Apparently the above claim tells us this definition makes sense, but I can't see why. I'd appreciate an explanation (note that I'm only really concerned with projective varieties, so if a simpler explanation can be afforded by restricting to this case then please do so!)


  • 1
    $\begingroup$ math.stackexchange.com/questions/144202/… $\endgroup$
    – Myshkin
    Commented Jun 2, 2012 at 10:40
  • $\begingroup$ Let $U$ and $V$ be two open affine neighborhoods of $p$ ... and therefore $T_pU = T_pV$. $\endgroup$
    – user14972
    Commented Jun 2, 2012 at 10:55
  • $\begingroup$ @Hurkyl I must be missing something. Why? $\endgroup$
    – Matt
    Commented Jun 2, 2012 at 11:02
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    $\begingroup$ Every open affine neighbourhood contains an open affine neighbourhood of that form. $\endgroup$
    – Zhen Lin
    Commented Jun 2, 2012 at 12:22
  • 2
    $\begingroup$ Distinguished open subsets are a base for the Zariski topology. $\endgroup$ Commented Jun 2, 2012 at 18:14

1 Answer 1


There are two typical topological ideas being used here.

The first is the idea of "localizing" at a point $P$ and performing a calculation $f(U)$ that involves an open neighborhood $U$ of $P$ that is independent of the choice of neighborhood. You can reduce the case of showing $f(U) = f(V)$ to the case of $V \subseteq U$, by the argument $f(U) = f(U \cap V) = f(V)$.

The second idea is that when you have a basis for a topology, you can typically rephrase any idea using open sets into one using the basis sets, because every open set is a union of basis sets.

The sets $D(f)$ (the complements of $Z(f)$) are a basis for the topology of an affine scheme.

$U$ and $V$ both contain a basis set $D(f)$ and $D(g)$, and the their intersection $D(fg)$ is a basis set contained in $U \cap V$. Therefore

$$ T_p U \cong T_p D(fg) \cong T_p V.$$

Incidentally, you really want to show a stronger claim: that the isomorphisms involved are 'coherent': e.g. that if you have a chain of inclusions $W \subseteq V \subseteq U$, that the isomorphisms

$$ T_p U \cong T_p V \cong T_p W $$

gives the same isomorphism as $T_p U \cong T_p W$ from invoking the theorem directly on $U$ and $W$. More precisely, you want to construct a functor between categories:

Open neighborhoods of p and inclusions --> Vector spaces and isomorphisms

If such a functor didn't exist, then even while the tangent space is well-defined as an abstract group, it would be rather difficult to actually use due to subtle technical issues.


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