Is the hilbert function a complete invariant for embeddings of zero dimensional (possibly reduced) schemes into P^n? I computed the Hilbert functions of several of these today. Its clear that the computation of the Hilbert function depended only "linear configuration" (made up word - what is the right one?) of these points up to automorphism, but its not obvious to me that these configurations are determined by the Hilbert function.
I'm assuming this is classical if it is true, it would be nice to see a reference or explanation or something. Interested in the non reduced case too, of course.
Of course I want the scheme to be defined over the same field as the projective space. And maybe ask that it has finite length too. I really only care about varieties, but if something weird happens when reasonable assumptions are dropped I would like to hear about that too.
 A: I think the answer is no. We have these two configurations of 4 points in $\mathbb{P}^2_{\mathbb{C}}$:


*

*Four points  in linearly general position (i.e. no three are collinear).

*Three points lying on a line and another point lying outside the line. 


Then these two configuration are clearly not linearly equivalent, but their Hilbert function is: 
$$ H(0)=1, \,\, H(1)=3, \,\, H(n)= 4 \,\,\text{ for all } n\geq 2  $$
To compute the Hilbert function, one can do everything explicitly with Macaulay2 for example, but we can also reason as follows.

First, let's denote the four points $p_1,p_2,p_3,p_4$ regardless of the configuration: in the second case, $p_1,p_2,p_3$ are collinear. Then, $H(0)=1$ is obvious and $H(1)=3$ is like saying that there is no line that contains all the points. To say that $H(n)=4$ for each $n\geq 2$ is like saying that these points impose $4$ independent conditions on the forms of degree $n$. 
More precisely, let $S=\mathbb{C}[X_0,X_1,X_2,X_3]$ be the coordinate ring of $\mathbb{P}^3_{\mathbb{C}}$ and let $S_2$ be its degree $2$ part. Evaluating at $p_1$ (fix arbitrarily some coordinates) we get a map 
$$ev_{p_1}\colon S_2 \to \mathbb{C}$$ 
whose kernel is the set of forms of degree $2$ vanishing at $p_1$. We denote this by $S_2(-p_1)$. If this map is surjective,then we know that $\dim S_2(-p_1)=\dim S_2 -1$, and to see that it is surjective, it is enough to prove that it is not zero, i.e. that not every form of degree $2$ vanishes on $p_1$.  But this is clear since we can just take the union of two lines  not passing through $p_1$.
Now, to compute the set $S_2(-p_1-p_2)$ of forms vanishing at $p_1$ and $p_2$ we need to compute the kernel of the map
$$ ev_{p_2}\colon S_2(-p_1) \to \mathbb{C} $$
Again, if we show the map is nonzero then it is surjective, so that it is enough to find a form of degree $2$ that vanishes on $p_1$ but not on $p_2$. We can take the union of two lines passing through $p_1$ and not $p_2$. This proves that
$$ \dim S_2(-p_1-p_2)= \dim S_2(-p_1) -1  = \dim S_2 -2 $$
i.e. that two points impose two independent conditions on forms of degree $2$.
Going on like this, it is easy to prove that (in both configurations) there is a form of degree $2$ vanishing on $p_1,p_2$ and not $p_3$: just take the union of a line passing through $p_1$ and not $p_3$ and another line passing through $p_2$ and not $p_3$. This implies, by the above reasoning, that
$$ \dim S_2(-p_1-p_2-p_3) = \dim S_2 - 3 $$
For the last point we need to find a form of degree $2$ vanishing on $p_1,p_2,p_3$ and not $p_4$, but this can be done again in both configurations: for the first one, take the union of the line from $p_1$ to $p_2$ and the line from $p_1$ to $p_3$, for the second take the line that contains $p_1,p_2,p_3$ and another line that does not contain $p_4$. This shows that
$$ \dim S_2(-p_1-p_2-p_3-p_4) = \dim S_2 - 4 $$
but, if $I\subseteq S$ is the ideal of $p_1,\dots,p_4$ then by definition $$ S_2(-p_1-p_2-p_3-p_4) = I_2$$
so that the above means exactly that $H(2)=4$.
To conclude, we need to prove that $H(n)=4$ for every $n\geq 3$, but to do this it is enough to show that if $p_1,\dots,p_4$ impose $4$ linearly independent conditions on the forms of degree $n$, then they impose again $4$ conditions on the forms of degree $n+1$. Indeed, suppose that for $1\leq i\leq 4$ we have a form of degree $n$ vanishing on $p_1,\dots,p_{i-1}$ and not $p_i$. Then to get a form of degree $n+1$ that does the same, it is enough to add to the previous one a line that does not pass through any of the points.
