Let $f_1,\ldots,f_r\in\mathbb{C}[x_0,\ldots,x_n]$ be homogeneous polynomials such that their common zero set $X$ in $\mathbb{P}^n$ is zero dimensional and consists of simple points only. Is it true that $f_1,\ldots,f_r$ already generate the homogeneous vanishing ideal of $X$?
I can show the claim in the affine situation: $A=\mathbb{C}[x_1,\ldots,x_n]/(\tilde{f}_1,\ldots,\tilde{f}_r)$, where $\tilde{f}_i=f_i(1,x_1,\ldots,x_n)$, is artinian and therefore a product of artinian local rings. Since each point is simple these factors have to be fields. Therefore $A$ is reduced.
Back in the projective/homogeneous situation this implies that the saturation of $(f_1,\ldots,f_r)$ is the vanishing ideal of $X$. But is taking the saturation really necessary?
Best Answer
Usually this is false. Let me, for simplicity, work over $n=2$, the projective plane. Then, in your situation, if $I$ is the sheaf of ideal of the points, you have an exact sequence, $0\to E\to F\to I\to 0$, where $F$ is the direct sum of $r$ line bundles and the map from $F\to I$ is given by the $f_i$s. Then, the kernel $E$ is a vector bundle. The $f_i$s generate a saturated ideal is equivalent to $H^0(F(d))\to H^0(I(d))$ is onto for all $d$. This is equivalent to $H^1(E(d))=0$ for all $d$ and that is equivalent to saying $E$ is a direct sum of line bundles by Horrock's theorem.
So, to get examples, you can work backwards. Start with a vector bundle $E$ which is not a direct sum of line bundles (for example, the tangent bundle) and then by Bertini type theorems, we can find an exact sequence as above upto some twist of $E$.
Your title is misleading. If $r=n$, then you have a complete intersection and then of course, the ideal generated by the $f_i$s is indeed saturated, but the example above says, if $r>n$, things can go wrong.