0
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.