Exercise 1.4.5.2 of Shafarevich's Basic Algebraic Geometry, Vol. 1, asks to prove that there is a one to one correspondence between affine closed sets in $\mathbb{A}^n _0$ and projective closed sets in $\mathbb{P}^n$ that have no component contained in the hyperplane $\{S_0 = 0\}$.
I defined the two maps as $U \mapsto \overline{U}$, that is, affine closed sets are being mapped to their projective closure, and $V \mapsto V \cap \mathbb{A}_0^n$, for $V$ a projective closed set (with no component contained in $\{S_0 = 0\}$).
Then if $U$ is affine, it is the very book who a few pages before states that $\overline{U} \cap \mathbb{A}_0^n = U$, so that, in that one direction, things work. (It is clear that a the projective closure of $U$ will not contain any component that is subset of $\{S_0 = 0\}$, because leaving out that component we would still have a closed projective set containing $U$, hence the set we started with was not the minimal projective closed set containing $U$.) I also worked that part of the book out to convince myself, so I think my thoughts are clear here.
In the other direction I encounter some difficulties. I made myself an example of a projective closed set, namely, a line and a point, outside the line and $\mathbb{A}_0^2$, all of this in the projective plane: $ X = V(S_1(S_1 - S_2), S_1 S_0) = \{S_1 = 0\} \cup \{(0:1:1)\}$. If I intersect $X$ with $\mathbb{A}_0^n$ and then take the projective closure, the point goes lost, as was obvious. If $X=V(G_1,\ldots,G_m)$ is the closed projective set, then, as the book says, $\mathbb{A}_0^n \cap X = V(F_1, \ldots, F_m)$, where $F_i(x_1, \ldots, x_n) = G(1, x_1, \ldots, x_n)$; this is readily checked. This process does not reverse properly, as is obvious from the counterexample: if I homogenise the $F_i$'s again, for instance, in the counterexample with the line and the point, I do not get my $G_i$'s back, at least not all of them: $S_1(S_1 - S_2)$ comes back, but $S_1 S_0$ becomes $S_1$. I thought this was the core of the problem, but I cannot figure out how the hypothesis of no component contained in $\{S_0 = 0\}$ could be used.
Anyway, is this the right path, or is there a much simpler way than that of looking at the equations? (Be it of the irreducible components only.) Or am I missing something obvious about how the equations are under this hypothesis? Thanks in advance.
Edit: Ok, eventually I think I figured it out. If I take $X \neq \varnothing$ irreducible projective not contained in $\{S_0 = 0\}$, then $X \setminus ( X \cap \mathbb{A}_0^n)$ is closed in $X$, and by hypothesis is not the whole $X$; also $\overline{X\cap \mathbb{A}_0^n}$ is closed in $X$, and by hypothesis is non empty; so the union of these two closed sets being $X$, which is irreducible, we get $X = \overline{X\cap \mathbb{A}_0^n}$. The general case easily follows from this case, using that the closure of the union is the union of the closures (finite union).
Does this work? Now I've got another question, actually...