I'd like to prove the following:
If $\mathfrak{a} \subseteq k[x_0, \ldots, x_n]$ is a homogeneous ideal, and if $f \in k[x_0,\ldots,x_n]$ is a homogeneous polynomial with $\mathrm{deg} \ f > 0$, such that $f(P) = 0 $ for all $P \in Z(\mathfrak{a})$ in $\mathbb P^n$, then $f^q \in \mathfrak{a}$ for some $ q > 0$.
I've been given the hint: interpret the problem in terms of the affine ($n+1$)-space whose affine coordinate ring is $k[x_0,\ldots,x_n]$ and use the usual Nullstellensatz.
I'm not really sure what the hint means. We have the isomorphism $k[x_0,…,x_n] \cong k[x_0,…,x_n] / I(\mathbb A_k^{n+1})$ (since $I(\mathbb A^{n+1}) = I(Z(0)) = 0$). But I don't see how this is helpful at all, nor am I sure this is what the hint means.
Any help would be greatly appreciated. Thanks
Best Answer
Consider $V(\mathfrak a)\subset \mathbb A^{n+1}(k)=k^{n+1}$, the cone in affine $n+1$ space defined by the ideal $\mathfrak a$ .
Your polynomial $f$ will vanish on $V(\mathfrak a)$ because that's exactly what it means that it vanishes on $Z(\mathfrak a)$ (see below)
The usual Nullstellensatz then implies that $f^q\in \mathfrak a$ for some $q\gt 0 $ .
NB Given a point $P=[a_0:a_1:...:a_n]\in \mathbb P^n (k)$, you cannot define the value of $f$ at $P$ : the expression $f(P)$ doesn't make sense.
What makes sense is to say that $f=0 $ on the line $k\cdot (a_0,a_1,...,a_n)\subset k^{n+1}$ .
One then writes $f(P)=0$, although $f(P)$ is not defined!
So saying that $f$ vanishes on $Z(\mathfrak a)$ really means that $f$ vanishes on $V(\mathfrak a)$ by definition.