# Compute tangent cone from a given ideal?

I have learned how to compute a tangent cone if we have a given polynomial in the following way: consider the part of the polynomial that is of minimal homogeneous degree. For example if $f=x^2+xy-y^3$, then consider $f_m=x^2+xy$=$x(x+y)$ and then the tangent cone is given by setting $f_m=0$, so we have the lines $x=0$ and $y=-x$.

Now, my question is, given an ideal of a variety, $I(V)$, how does one recognize the ideal of a tangent cone? For example, if $I(V)=(xy,yz,zx)$ a subset of $k[x,y,z]$ (which the variety is the 3 coordinate axis in $R^3$), what is the tangent cone of the variety $V$? Then, how would one determine what the ideal of that cone is?

Thank you!

## 1 Answer

Take the ideal $\hat{I}$ of minimal degree homogeneous parts. The tangent cone of $V$ at the origin $O$ is $Z(\hat{I})$ i.e the closed set defined by the ideal $\hat{I}$.
To obtain the tangent cone at some point $P$ you need first to move $P$ to the origin.

In your case $I(V)=(xy,yz,zx)$ so $\hat{I}=I$ and the tangent cone at the origin is just $V$.
Indeed the tangents at the origin in $V$ are the three axes.