The cone over a projective variety I'm trying to prove that $I(C(Y))=I(Y)$, where $C(Y)=\pi^{-1}(Y)\cup \{(0,\ldots,0)\}$ the cone over $Y$ and $\pi:\mathbb A^{n+1}-\{0,\ldots,0\}\to \mathbb P^n$ the projection which sends the point with affine coordinates $(a_0,\ldots,a_n)$ to the point with homogeneous coordinates $(a_0,\ldots,a_n)$.
My attempt of solution
$I(C(Y))=I\bigg(\pi^{-1}(Y)\cup\{(0,\ldots,0)\}\bigg)=I\bigg(\pi^{-1}(Y)\bigg)\cap I\bigg(\{(0,\ldots,0)\}\bigg)=?$
The last term should be equal to $I(Y)$, but I don't know why.
I really need help.
Thanks a lot.
 A: Let $S=\Bbbk[x_0,\ldots,x_n]$. Note that a homogeneous polynomial vanishes on $\pi^{-1}(Y)$ if and only if it "vanishes" on $Y$, simply by the definition of "vanshing" for homogeneous polynomials on sets of projective coordinates. Thus, $I(C(Y))$ contains $I(Y)$ and it is contained in $S_+$. Let us show that $I(C(Y))\subseteq I(Y)$. Let $f\in I(C(Y))$. Then, $f=\sum_{d=0}^{r} f_d$ with all the $f_d$ homogeneous of degree $d$. Let $a\in C(Y)$, then $t a\in C(Y)$ for any $t\in\Bbbk^\times$. Hence,
$$0 = \sum_{d=0}^r f_d(t a) = \sum_{d=0}^r f_d(a) t^d $$
Assuming that your field is infinite, this means that the polynomial $\sum_{d=0}^r f_d(a)\cdot T^d\in \Bbbk[T]$ has too many zeros to not be the zero polynomial. In other words, $f_d(a)=0$ for all $d$, so $f_d\in I(C(Y))$. Hence, $f$ is a sum of elements of $I(Y)$, therefore $I(C(Y))\subseteq I(Y)$. 
Note that you get $(0,\ldots,0)$ for free because homogeneous polynomials vanish at the origin anyway.
A: I think it follows straight from the definitions:
$$I_p(V)=\{F:F(\lambda x)=0\forall[x]\in V\forall\lambda\neq 0\}$$
$$C(V)=\{(x):[x]\in V\}\cup\{0\}=\{(\lambda x):[x]\in V,\lambda\in\mathbb{k}\}$$
$$I_a(C(V))=\{F:F(\lambda x)=0\forall [x]\in V\forall\lambda\in\mathbb{k}\}$$
Only difference is the case $\lambda=0$. But notice how $I_p(V)$ is always homogeneous, which means the homogeneous parts of each $F\in I_p(V)$ must be in the ring too. This includes the constant terms. So the constant terms must be zero. Consequently, these polynomials vanish at $0$ as well.
Of course, the above is true provided the field is infinite.
