# If $Y = X \backslash Z(f)$ for some affine variety $X$ with $p \in Y$, then $T_p Y \cong T_p X$

Let $Y = X \backslash Z(f)$ be a quasi-affine variety for some affine variety $X$, and let $p \in Y$. I'd like to prove that $T_p Y \cong T_p X$.

I have the following definition of $T_p X$:

If $A$ is a $k$-algebra and $\phi : A \to k$ a $k$-algebra homomorphism, then a $k$-linear map $D : A \to k$ is a 'derivation centred at $\phi$' if $D(fg) = \phi(f)D(g) + D(f)\phi(g)$ for all $f,g \in A$. Let $\mathrm{Der}(A,\phi)$ be the vector space of derivations centred at $\phi$. Define $T_p X = \mathrm{Der}(k[X], \mathrm{ev}_p)$.

My thoughts so far:

A morphism $\psi : Y \to X$ of affine varieties gives rise to a $k$-algebra homomorphism $\psi^*: k[X] \to k[Y]$ which maps $g \mapsto g \circ \psi$. This in turn gives rise to a $k$-linear map $d_\psi : T_pY \to T_{\psi(p)}X$, by sending $D \mapsto D \circ \psi^*$.

So for the case in question, we have a morphism $i : Y \to X$ given by the inclusion map, under which $i(p) = p$, so this looks pretty promising. I need to show that the map $d_i$ is a bijection.

If $d_i(D) = 0$, then $D \circ i^* = 0$ - I think this can only happen if $D$ is the zero map, so we have injectivity. Surjectivity I'm unsure about, and I haven't used $f$ anywhere yet. Any hints would be appreciated

Thanks

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The variety $X$ has a local ring $\mathcal O_{X,p}$ at $p$.
If $\mathfrak m \subset\mathcal O_{X,p}$ is its maximal ideal we have $T^*_p(X)=\mathfrak m/\mathfrak m^2$, a $k$-vector space, and the dual of that vector space is the tangent space of $X$ at $p$ : $T_p(X)=(\mathfrak m/\mathfrak m^2)^*$.
Since the local ring $\mathcal O_{X,p}$ does not change if you replace $X$ by an open neighbourhood $U$ of $p$, you have $T_p(X)=T_p(U)$.
Apply to $U=X\setminus Z(f)=Y$

The tangent space I defined is called the Zariski tangent space and is easily seen to be isomorphic to your tangent space defined with derivations.
The basic idea is that given your derivation $D$, you can extend it to $\mathcal O_{X,p}$, then restrict it to $\mathfrak m$ and since this linear form is zero on $\frak m^2$ you finally obtain a linear form on $\mathfrak m/\mathfrak m^2$, a tangent vector in the Zariski sense.

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In your definition of the tangent space, $X = \text{Spec }A$? I would approach this a little differently. Namely, show that any such derivation arises from a derivation of the local ring at $p$. It will then follow that your definition of the tangent space is independent of the choice of affine neighborhood of $p$ that you choose.
Sorry, I was unclear. I'm only defining the tangent space here for affine varieties. My notes then use the answer to this question as a justification for the definition of the tangent space for an arbitrary variety: define $T_p X = T_p U$ where $U$ is any affine open neighbourhood of $p$ –  algeom Apr 26 '12 at 22:45
Got it. Again, if you show that any such derivation arises from a derivation of the local ring at $p$, then you will have shown that the definition does not depend on the choice of open affine neighborhood. –  user29743 Apr 26 '12 at 23:05