Computation of normal cones I am reading Fulton's intersection theory but I have a poor intuition about the normal cone. I know the cone $C_{Y/X}$ is the normal bundle if $Y\subset X$ is smooth. I would appreciate it if someone could show what the normal cone of the following schemes of $\mathbb{A}^2$. 


*

*An intersection of two lines $k[x,y]/(xy)$.

*A fat line $k[x,y]/(x^k)$

*A fat point $k[x,y]/(x,y)^k$

 A: I'll take a stab at a partial answer, though I'm not well-acquainted with the normal cone.
By definition, $C_{Y/X}=\operatorname{Spec}\left(\oplus_l I^l/I^{l+1}\right),$ where $Y\subseteq X$ is defined by the ideal $I\subseteq k[X]$ in the affine coordinate ring of $X.$ In the first case, we have $I=(xy)\subseteq k[x,y]=k[X].$ Thus, $C_{Y/X}$ has coordinate ring $\oplus_l(xy)^l/(xy)^{l+1}.$ 
Noticing that $(xy)^0/(xy)^1=k[x,y]/(xy),$ we see that there is a canonical surjection $\left(k[x,y]/(xy)\right)[t]\to\oplus_l(xy)^l/(xy)^{l+1}$ taking $t\mapsto xy\pmod {(xy)^2}$. Moreover, according to Fulton p.12-13, this map is an isomorphism, since $xy$ is a nonzerodivisor in $k[x,y]$ and $k[x,y]/(xy)$ is nonzero (i.e. $xy$ is a regular sequence). Thus, there is an isomorphism $C_{Y/X}\to Y\times\Bbb A^1=\operatorname{Spec}\left(k[x,y,t]/(xy)\right)$ is a cartesian product of a union of two coordinate axes with a line.
For the next example, all this analysis should work, since $x^n$
 is a regular sequence of $k[x,y].$ Thus we should find that $C_{Y/X}=Y\times\Bbb A^1=\operatorname{Spec}\left(k[x,y,t]/(x^n)\right)$ is a cartesian product of a fat line with a reduced line.
For the third example, $I=(x,y)^n$ is not generated by a regular sequence, so it is not so obvious to me what the geometry is. We now have a surjection $k[x,y,t_1,\ldots, t_N]/(x,y)^n\to\oplus_l(x,y)^{nl}/(x,y)^{n(l+1)}$ taking $t_1\mapsto x^n,t_2\mapsto x^{n-1}y,\ldots,t_N\mapsto y^n,$ giving a closed embedding of $C_{Y/X}$ in $Y\times\Bbb A^N$ with $N={n+2\choose 2}.$ I would be happy to see further elaborations of the techniques to be pursued here.
