# homogeneous ideal in $k[Z_0, Z_1, . . . , Z_n]$

Suppose that $I$ is a homogeneous ideal in $k[Z_0, Z_1, . . . , Z_n]$ such that the radical ideal $√I = (Z_0, . . . , Z_n)$.

Show that:
(a) There is an integer $N$ so that for any $d ≥ N$ and any homogeneous polynomial $f$ of degree $d$, then $f ∈ I$.
(b) Show conversely that if $I$ is a homogeneous ideal such that $V (I) = ∅∈ \mathbb{P}^ n$ then there exists an $N$ with the property from (a).

Since $√I = (Z_0, . . . , Z_n)$ by definition of radical ideal we can say that for each $j = 0, . . . , n$ there is a positive integer $n_j$ so that $Z_j^{n_j}∈ I$
Is $N=max({n_j})$? Because any polynomial of degree $≥ N$ would be included in the ideal. The variety in $I$, i.e. $V(I)$ is empty in $\mathbb{P}^n$ because it correponds to the point $(0,0,...,0)$ which is doesn't belong to $\mathbb{P}^n$. But how do we get (a) from that?

• No, the max is not enough. But the sum is enough (in fact, you can do a little better, but we don't need to be economical here). – quasi Mar 21 '17 at 3:05
• For example, suppose $I = (x^3,y^3,z^3)$. How would you generate $xyz$? Obviously, you can't. Note that as suggested in my previous comment, 9 is enough. But in fact, 7 is also enough (pigeonhole principle). – quasi Mar 21 '17 at 3:06
• @quasi thanks! I get that – user286826 Mar 21 '17 at 3:14
• I'm still stuck on part b though – user286826 Mar 21 '17 at 3:28
• For part b you can use the (affine) Nullstellensatz. – Daniele A Mar 21 '17 at 12:05