How to generalize the tangent cone at X at regular point p?

Let $\mathbb{A}^n$ be an affine space with $n$ coordinates. Here is some notations.

• $f^{(i)}$: homogenous polynomial with degree $i$.
• $f^{in}$ (an initial polynomial) : the nonzero $f^{(i)}$ with the smallest degree.
• $I(X)=\left\{f(x): f(p)=0, for ~all ~p \in X \subseteq \mathbb{A}^n\right\}$
• $I(X)^{in}=\left\{f^{in} : f \in I(X)\right\}$
• $Z(X)=\left\{p \in \mathbb{A}^n : f(p)=0, for ~all~ f \in X \subseteq \bar{k}[x_1,...x_n] \right\}$

Then, a tangent cone $X$ at origin point, denoted by $\mathcal{C}_{X,0}:= Z(I(X)^{in})$.

And here is common example : Consider the following (elliptic) curve $C=Z(y^2-x^3-x^2)$. and let $f(x,y)=y^2-x^3-x^2$. Then,

[Step 1] $f^{in}(x)=y^2-x^2$. (or, $I(X)=\left\{x^2-y^2\right\}$)

[Step 2] $Z(I(X)^{in})=Z(y=x) \cup Z(y=-x)$ (or, $\mathcal{C}_{C,(0,0)}=Z(y=x) \cup Z(y=-x)$).

From here , I have a question: how extend the concept of the tangent cone to the regular points? For example, when considering again the curve $C=Z(y^2-x^3-x^2)$, how to find the $\mathcal{C}_{C,(-1,0)}$=?, or arbitrary point $\mathcal{C}_{C,p}$=? When I follow above two steps, the point $p$ is fixed only at the origin point. In other words, the above step does not seem to give any option to pick the point $p$ to me. (Of course, I guess that $\mathcal{C}_{C,p}$ would be a tangent (line) at the point $p$ like the below figure, but how to justify based on the 'definition' of tangent cone at $\mathcal{C}_{C,p}$?

Solution 1:

Move the point to the origin by rewriting $y^2-x^3-x^2=0$ as $y^2-(x+1)^3+2(x+1)^2-(x+1)=0.$

Then the tangent cone is $-(x+1)=0$ or $x=-1.$