Geometric intuition of the tangent cone.

Question

In my algebraic geometry course, we are studying tangent cones and I have some troubles to intuitively understand the definition of the latter. Let $X$ be an affine variety that contains $0$. We define the tangent cone at $0$ as $$C_0 X = V(f^{in}; f \in J)$$ for $X = V(J)$ and $f^{in}$ the lowest degree term. If we take for example $X = V(y^2 -x^2 - x^3)$, we easily find that $$C_0X = \{y = x\} \cup \{y = -x\}$$ which are indeed the two tangent lines to $X$ at the origin. However, I don't have any geometric intuition of why, in general, this definition with the initial term should give us the tangent line to the origin. Any explanation ?

Answer 1

As suggested in the comments, the terms of lowest degree are dominant near the origin, so we may neglect all but the lowest-degree terms without changing which curves are tangent to the variety. The two arguments below rely on analytic tools and idioms (limits, order of vanishing, the chain rule) rather than being purely algebro-geometric.

For each argument, let $p(x, y) = \sum_{k} p_{k}(x, y)$ be a polynomial, with each $p_{k}$ non-zero and homogeneous of degree $k$ (i.e., "collect the degree-$k$ terms of the given polynomial"). Let $p_{m}$ denote the summand of smallest degree.

First, imagine zooming in on the plane at the origin. This amounts to the coordinate transformation $(x, y) \mapsto (tx, ty)$, whose effect on the variety is to replace $p(x, y) = 0$ with $$ 0 = p(tx, ty) = \sum_{k} p_{k}(tx, ty) = \sum_{k} t^{k} p_{k}(x, y) = t^{m} \sum_{k} t^{k-m} p_{k}(x, y). $$ Since $t \neq 0$, this equation is equivalent to $0 = p_{m}(x, y) + O(t)$. In the limit as $t \to 0$, this becomes $0 = p_{m}(x, y)$.

Second, if $c(t) = (x(t), y(t))$ is a smooth path in the zero set satisfying $c(0) = (0, 0)$ and $c'(0) \neq (0, 0)$, then $c(t) = c'(0)t + o(t) = t(c'(0) + o(1))$. Since $|p_{k}(c(t))| = |t|^{k}(C + o(1))$ near $t = 0$, $$ 0 = p(c(t)) = p_{m}(c(t)) + O(|t|^{m+1}). $$ Differentiating at $t = 0$ tells us $$ 0 = mt^{m-1} Dp_{m}(0, 0)c'(0) + O(|t|^{m}), $$ and therefore $0 = Dp_{m}(0, 0)c'(0)$, i.e., $c'(0)$ is tangent to the variety $p_{m}(x, y) = 0$.

The argument extends in the obvious way to varieties in more than two variables.