# Tangent Spaces (Algebraic Geometry)

I'm in my algebraic geometry class and I have the definition of the space tangent to some variety $$W=V(F_1,F_2,...)$$ as the degree-1 components of $$F_1,F_2,...$$ . We then introduce the differential at a point $$p$$, which sends $$g$$ to $$\sum_i \frac{dg}{dx_i}(p)(x_i-p_i)$$. We then show that this is independent of generators of $$I$$, and want to prove the theorem that $$T_pV=V(dpF_1...dpF_n)$$.

This is where I get confused: isn't this the exact same as the degree-1 components? So haven't we already proven this?

We then show that $$dp$$ sends $$k[x_1...x_n]$$ to the set of linear functionals on the tangent space. But then there is some move where we restrict to $$m_p=(x_1-p_1,...,x_n-p_n)$$ and want to quotient this out, so we consider some $$g$$ in the kernel of $$dp$$, and expand it to find that $$g$$ must have degree greater than 2.

I guess I'm confused as to exactly what we are proving and what the significance of these results are.

• Have you studied differential geometry already, and seen different ways of describing the tangent space to a point on a manifold? – KCd Nov 21 '18 at 23:58

You want to show that the subspace $$(dF) \mid_a=0$$ for all $$F \in I(W)$$ can actually be reduced to the calculation $$(dF) \mid_a=0$$ for $$F= F_1, \dots ,F_n$$ which are a set of generators for $$I(W)$$.
This amounts to showing that $$(dF_1) \mid_a, \dots, (dF_n) \mid_a$$ generate $$\{(dF) \mid_a =0 \mid F \in I(W)\}$$ as a vector space.
To do this, you really just need the product rule. If $$G= \sum H_i F_i$$, then apply the product rule to obtain that $$(dG)_a= \sum_i H_i(a) \cdot (dF_i) \mid_a$$