Tangent space of a point of an algebraic variety

Solution 1:

Yes, there is a canonical isomorphism of $\mathbb C$-vector spaces $$i:S_{X,p}\stackrel {\cong}{\to} T_{X,p}=Der(\mathcal O_{X,p}, \mathbb C)$$ If one starts with a vector $v=(v_1,...,v_n)\in S_p\subset \mathbb C^n$ (so that $\sum_k \frac{\partial f_i}{\partial x_k}(p)v_k = 0$), the isomorphism $i$ associates to it the derivation $i(v)=\partial _v$ defined on $\mathcal O_{X,p}$ by the formula $$ \partial_v(g)=\sum_k \frac{\partial g}{\partial x_k}(p)v_k $$ where $g\in \mathcal O_{X,p}$ is an arbitrary local function.
The inverse isomorphism is given by $$i^{-1}: Der(\mathcal O_{X,p},\mathbb C) \stackrel {\cong}{\to} S_{X,p} :\partial \mapsto (\partial x_1,...,\partial x_n) $$

Edit
The vector space of derivations is also isomorphic to Zariski's tangent space $(\frak m_p/\frak m^2_p)^*$, defined via the maximal ideal $\mathfrak m_p\subset \mathcal O_{X,p}$.
The isomorphism is $$ Der(\mathcal O_{X,p} ) \stackrel {\cong}{\to} (\frak m_p/\frak m^2_p)^*:\partial \mapsto \overline{\partial} $$ where $\overline{\partial} (g \;\text {mod} \;m^2_p)=\partial (g)$ for $g\in \frak m_p$.

Second Edit
The following remark may be of some interest, since it does not seem to be addressed in Algebraic Geometry books:
If $X\subset \mathbb A^n_k$ is an affine algebraic variey and if $p\in X$, we may consider the maximal ideal $M_p\subset \mathcal O(X)$ of global functions vanishing at $p$.
We may also consider, as we already did, the maximal ideal $\mathfrak m_p\subset \mathcal O_{X,p}$ of germs of functions regular ay $p$ and vanishing at $p$.
We then have a natural $k$-linear map $ M_p/M_p^2 \to \mathfrak m_p/\mathfrak m_p^2 $ and the slightly surprising but pleasant fact is that this linear map is an isomorphism.

Solution 2:

Lemma 1 Let $A$ be a local algebra over a field $k$. Let $\mathfrak{m}$ be the unique maximal ideal of $A$. Suppose the canonical homomorphism $k \rightarrow A/\mathfrak{m}$ is an isomorphism. Let $f \in A$. There exists a unique $c \in k$ such that $f \equiv c$ (mod $\mathfrak{m}$). We denote this $c$ by $f(0)$. We call a $k$-linear map $v\colon A \rightarrow k$ a derivation if $v(fg) = v(f)g(0) + f(0)v(g)$ for $f, g \in A$. Let $Der(A, k)$ be the set of derivations. We regard $Der(A, k)$ as a vector space over $k$ in the obvious way. Let $v \in Der(A, k)$. If $f, g \in \mathfrak{m}$, $v(fg) = v(f)g(0) + f(0)v(g) = 0$. Hence $v(\mathfrak{m}^2) = 0$. Hence $v$ induces a $k$-linear map $\bar v\colon \mathfrak{m}/\mathfrak{m}^2 \rightarrow k$. Hence we get a $k$-linear map $\psi\colon Der(A, k) \rightarrow (\mathfrak{m}/\mathfrak{m}^2)^*$. Then $\psi$ an isomorphism.

Proof: This is proved here

Lemma 2 Let $k$ be a field. Let $L_i = \sum_{k = 1}^{n} a_{ik} x_k, i = 1,\dots, r$ be linear polynomials in $k[x_1,\dots,x_n]$. Let $T$ be the vector subspace $\{v \in k^n| L_i(v) = 0$ for all $i\}$ of $k^n$. Let $W$ be the vector subspace of $k^n$ generated by $L_i(e_1,\dots,e_n)$ for all $i$, where $\{e_1,\dots,e_n\}$ is the canonical basis of $k^n$. Then $T$ is canonically isomorphic to $(k^n/W)^*$.

Proof: An element of $(k^n/W)^*$ is identified with a linear map $f\colon k^n \rightarrow k$ such that $f(W) = 0$. Such a map $f$ is uniquely determined by the condition $\sum_{k = 1}^{n} a_{ik} f(e_k) = 0, i = 1,\dots, r$. Hence the assertion follows. QED

Proposition Let $A = k[x_1,\dots,x_n]$ be the polynomial ring over $k$. Let $I$ be an ideal of $A$ generated by $F_1,\dots,F_r$. Let $B = A/I$. Let $p = (a_1,\dots, a_n)$ be a point of $k^n$ such that $F_i(p) = 0$ for all $i$. Let $M = (x_1 - a_1,\dots,x_n - a_n)$ be the maximal ideal of $A$. Let $\mathfrak{m} = M/I$. Let $L_i = \sum_k \frac{\partial F_i}{\partial x_k}(p)x_k$ for $i = 1, \dots, r$. Let $T$ be the vector subspace $\{v \in k^n| L_i(v) = 0$ for all $i\}$ of $k^n$. Then $T$ is canonically isomorphic to $Der(B_{\mathfrak{m}},k)$.

Proof: $\mathfrak{m}/\mathfrak{m}^2$ is canonically isomorphic to $M/(I + M^2)$. $M/M^2$ is a $k$-vector space with a basis $\bar x_1,\dots,\bar x_n$, where $\bar x_i = x_i$ (mod $M^2$). Let $F \in I$. Since $F(p) = 0$, $F = \sum_k \frac{\partial F}{\partial x_k}(p)(x_k - a_k) + \cdots$. Hence $(I + M^2)/M^2$ is a vector subspace of $M/M^2$ generated by $L_i(\bar x_1, \dots,\bar x_n)$ for all $i$. Hence the assertion follows from Lemma 1 and Lemma 2. QED