How can I find the tangent cone and blowing up of $\mathbb{A}^2$ at $0$ in these curves How can I find the tangent cone and blowing up of $\mathbb{A}^2$ at $0$ in these curves:


*

*$x=y^2$

*$x^2=y^3$

*$(x+y)(x-y)+y^3 = 0$


I'm studying algebraic geometry from Kempf's Algebraic Varieties. This is an exercise in the book, namely 6.2.5. It has covered the theory behind tangent cones but according to the book, it has given the tangent cone the wrong scheme structure because schemes are not covered in his book. Yet, I see no relation between the theory and calculating the tanget cone and blowing up of these curves in $\mathbb{A}^2$. Can someone help me proceed with calculating these things?
 A: The exceptional divisor in the blow-up $Bl_p(X)$ of a scheme $X$ at a point $p \in X$ is the projectivized tangent cone ${\Bbb P}TC_p(X)$ to $X$ at $p$. See e.g. The Geometry of Schemes by David Eisenbud and Joe Harris.
The affine tangent cones at the origin are just the lowest degree homogeneous terms 1. $x$ 2. $x^2$ and 3. $(x+y)(x−y)$
A: For blow ups - recall that the blow up $\mathbb{A}^2$ at $0$, is given by the zero locus of $xS-yT = 0$ in $\mathbb{A}^2 \times \mathbb{P}^1$, where $x,y$ are coordinates on $\mathbb{A}^2$ and $S,T$ are homogeneous coordinates on $\mathbb{P}^1$, 
Let's denote this by $\pi:Bl_0\mathbb{A}^2 \rightarrow \mathbb{A}^2$.
Now if you want to blow up $C:x=y^2$, this is the same as taking the closure of $\pi^{-1}(C-\{(0,0)\})$. In terms of equations, you have that the preimage $\pi^{-1}(C)$ is cut out by $xS-yT=0$ and $x=y^2$ in $\mathbb{A}^2 \times \mathbb{P}^1$, i.e. you have 
$$x=y^2$$
$$y(yS-T)=0$$
but notice that if you just take the zero locus cut out by 
$$x=y^2$$
and 
$$yS-T=0$$
Then this is a closed subset of $\mathbb{A}^2\times \mathbb{P}^1$ which contains $\pi^{-1}(C-\{(0,0)\})$ as an open dense subset, in particular this is the closure of $\pi^{-1}(C-\{(0,0)\})$, hence this is the blow up of $C$ (well - depending on how you defined blow ups you may need to justify why this is the blow up). You see that we removed the $y=0$ component, which actually corresponds to the exceptional divisor of $\pi$. In terms of pictures, if you don't remove the $y=0$ term you'll see that upstairs you have a curve (which is isomorphic to $y=x^2$) with a $\mathbb{P}^1$ attached to it - removing the $y=0$ term removes this copy of $\mathbb{P}^1$. 
I'll leave you to figure out how to do the other two cases. You should be able to find more details on how to compute these things in Shafarevich's Basic Algebraic Geometry 1. 
