Prove the Equality for Hypersurface
Let \(V = V(F)\) be a hypersurface, and \(F\) be an irreducible polynomial. We want to show that for any point \(P \in V\), the tangent space \(T_P(V)\) is given by the set \(\{(v_1, \dots, v_n) \mid \sum F_{X_i}(P)v_i = 0\}\).
Since \(V\) is a hypersurface, it is defined by a single polynomial, and the ideal of the variety is \(I(V) = (F)\). Now, consider any \(G \in I(V)\). Since \(G\) belongs to the ideal generated by \(F\), we can write \(G = aF\) for some polynomial \(a\).
To show that the tangent spaces are equal, we need to prove two inclusions:
Claim: \(\{(v_1, \dots, v_n) \mid \sum G_{X_i}(P)v_i = 0\} \subset \{(v_1, \dots, v_n) \mid \sum F_{X_i}(P)v_i = 0\}\)
Take an arbitrary vector \((v_1, \dots, v_n)\) from the left-hand set, i.e., \(\sum G_{X_i}(P)v_i = 0\). Since \(G = aF\), we have \(\sum (aF)_{X_i}(P)v_i = 0\). Differentiating the product \(aF\) with respect to \(X_i\), we obtain:
$$(aF)_{X_i} = a_{X_i}F + aF_{X_i}$$
Evaluating this expression at point \(P\), we get:
$$\sum (a_{X_i}(P)F(P) + a(P)F_{X_i}(P))v_i = 0$$
However, since \(P \in V\), we have \(F(P) = 0\) and thus:
$$\sum a(P)F_{X_i}(P)v_i = 0$$
Thus, the vector \((v_1, \dots, v_n)\) belongs to the right-hand set as well, proving the first inclusion.
Claim: \(\{(v_1, \dots, v_n) \mid \sum F_{X_i}(P)v_i = 0\} \subset \{(v_1, \dots, v_n) \mid \sum G_{X_i}(P)v_i = 0\}\)
Since \(G \in I(V)\), and \(I(V) = (F)\), the reverse inclusion can be proven similarly to the first inclusion, and thus both sets are equal.