Let $V$ be the curve defined by $Y^2=X^3$ over an algebraically closed field $k$. Consider the regular map $\phi:\mathbb{A}^1\to V$ where $t\mapsto (t^2,t^3)$. This induces a map on $k-$algebras $k[X,Y]/(Y^2-X^3)\to k[T]$ where $[X]\mapsto T^2$ and $[Y]\mapsto T^3$, and this map is supposed to induce a map $C_0(V)\to C_0(\mathbb{A}^1)$ of tangent cones at the origin.
The definition of tangent cones that I use is the following. For an affine variety $V= \text{Specm }k[X_1,\ldots X_n]/\mathfrak{a}$ over an algebraically closed field $k$, we can define the ideal $\mathfrak{a}_* =\{f_*:f\in\mathfrak{a}\}$ where, for $f\in k[X_1,\ldots X_n]$, $f_*$ is the homogeneous part of $f$ of lowest degree. The tangent cone at the origin is defined as the $k$-algebra $k[X_1,\ldots,X_n]/\mathfrak{a}_*$.
I know that $C_0(V)=k[x,y]/(y^2)$ and $C_0(\mathbb{A}^1)=k[T]$. Apparently, the tangent cone map should send both $[x]$ and $[y]$ to $0$, but why is this? Any help is greatly appreciated!
First, the coordinate algebra for the tangent cone of $X$ at $x$ can be described as the associated graded of $\mathcal{O}_{X,x}$, i.e. $\operatorname{gr} \mathcal{O}_{X,x} = \bigoplus \mathfrak{m}_x^n/\mathfrak{m}_x^{n+1}$, and given a map $f:X\to Y$ sending $x\mapsto y$, we get the map $\operatorname{gr} \mathcal{O}_{Y,y} \to \operatorname{gr} \mathcal{O}_{X,x}$ from the collection of induced maps $\mathfrak{m}_y^n/\mathfrak{m}_y^{n+1}\to \mathfrak{m}_x^n/\mathfrak{m}_x^{n+1}$. (Why? Since $\mathcal{O}_{Y,y}\to\mathcal{O}_{X,x}$ is a local map of local rings, we have that $s\in\mathfrak{m}_y^n$ implies the image of $s$ is in $\mathfrak{m}_x^n$ by induction on $n$ where the base case is the assumption that we have a local map.)
With this definition your problem can be answered: the images of $x$ and $y$ land in $\mathfrak{m}_0^2$, so they're zero in $\mathfrak{m}_0/\mathfrak{m}_0^2$.