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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.

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  • $\begingroup$ Have you studied differential geometry already, and seen different ways of describing the tangent space to a point on a manifold? $\endgroup$ – KCd Nov 21 '18 at 23:58
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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 $(dG)_a= \sum_i H_i(a) \cdot (dF_i) \mid_a$

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