I have been using the fact that if I have a single equation given in the affine space \mathbb{A}^{n}, then taking the closure in \mathbb{P}^{n} can be given by homogenizing the equation. However, I have also realized that I never thought about why! We will only discuss for the case n = 2, but the proof should work for any n \geq 1. The references for this posting are Vakil's book and Kidwell's answer to this MSE question.
Computing projective closure. Let g(x, y) \in k[x, y] be any polynomial and d be the maximum total degree of the monomials of g(x, y). Then \overline{V_{\mathbb{A}^{2}}(g(x, y))} = V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)), where the closure is taken in \mathbb{P}^{2}.
Proof. Since V_{\mathbb{A}^{2}}(g(x, y)) = V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)) \cap D_{\mathbb{P}^{2}}(z) \subset V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)), we have \overline{V_{\mathbb{A}^{2}}(g(x, y))} \subset V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)). To show the reverse inclusion, first, note that there is a homogeneous ideal J \subset k[x, y, z] such that V_{\mathbb{P}^{2}}(J) = \overline{V_{\mathbb{A}^{2}}(g(x, y))}. We want to show that V_{\mathbb{P}^{2}}(J) \supset V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)), and to do so, it is enough to show that J \subset \sqrt{(z^{d}g(x/z, y/z))} in k[x, y, z]. Since V_{\mathbb{A}^{2}}(g(x, y)) \subset V_{\mathbb{P}^{2}}(J), we see that J must be contained in the homogeneous ideal I(V_{\mathbb{A}^{2}}(g(x, y))) generated by all the homogeneous elements of k[x, y, z] contained in homogeneous prime ideals in V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)) \cap D_{\mathbb{P}^{2}}(z) = V_{\mathbb{A}^{2}}(g(x, y)). Thus, it is enough to show that I(V_{\mathbb{A}^{2}}(g(x, y))) \subset (z^{d}g(x/z, y/z)) in k[x, y, z]. Fix any homogenous h(x, y, z) \in I(V_{\mathbb{A}^{2}}(g(x, y))). By definition, we have V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)) \cap D_{\mathbb{P}^{2}}(z) = V_{\mathbb{A}^{2}}(g(x, y)) \subset V_{\mathbb{P}^{2}}(h(x,y,z)), so V_{\mathbb{P}^{2}}(z^{d}g(x/z, y/z)) \subset V_{\mathbb{P}^{2}}(zh(x,y,z)). This implies that (e.g., 4.5.H. (a) of Vakil) z^{N}h(x, y, z)^{N} \in (z^{d}g(x/z, y/z)) for some N \gg 0. Due to the definition of d, we see that z^{d}g(x/z, y/z) is not divisible by z in k[x, y, z]. Thus, if we denote by z^{d}g(x/z, y/z) = p_{1}(x, y, z)^{e_{1}} \cdots p_{r}(x, y, z)^{e_{r}} the irreducible factorization in k[x, y, z], then each p_{i}(x, y, z)^{e_{i}} divides h(x, y, z)^{N}. Hence, we must have h(x, y, z)^{N} \in (z^{d}g(x/z, y/z)). This implies that h(x, y, z) \in \sqrt{(z^{d}g(x/z, y/z))}, as desired. \Box
What's going on? At first, I was wondering why Vakil's book is missing this important piece of information, but I realized that it was quite clear that this follows from 4.5.H. This is very general, say we have a graded ring S := S_{0} \oplus S_{1} \oplus S_{2} \oplus \cdots. As usual, we write S_{+} := S_{1} \oplus S_{2} \oplus \cdots the ideal generated by homogenous elements. Recall that we can consider \mathrm{Proj}(S) as the set of homogeneous prime ideals that do not contain S_{+}. We use the notation V_{+}(I) := \{\mathfrak{p} \in \mathrm{Proj}(S) : \mathfrak{p} \supset I\} for a homogeneous ideal I \subset S and D_{+}(f) := \mathrm{Proj}(S) \smallsetminus V_{+}(f). Given any subset T \subset \mathrm{Proj}(S), we denote by I_{+}(T) to be the ideal generated by homogenous elements that are contained in any (homogeneous) prime ideals in T. In Vakil, 4.5.H says:
(a) V_{+}(I) \subset V_{+}(f) if and only if f \in \sqrt{I}.
(b) V_{+}(I_{+}(T)) = \overline{T} in \mathrm{Proj}(S) for any subset T \subset \mathrm{Proj}(S).
The statement (b) is basically proven in the proof above. To see (a), note that the only nontrivial direction is to assume V_{+}(I) \subset V_{+}(f) and show f \in \sqrt{I}. This easily follows once we show the following.
Lemma. Suppose that V_{+}(I) \neq \emptyset. Then \sqrt{I} is the intersection (in S) of all \mathfrak{p} \in \mathrm{Proj}(S) that contains I.
Proof. We start with the fact that \sqrt{I} is the intersection of all prime ideals of S containing I. Since we are assuming that V_{+}(I) is nonempty, we do not need to consider the possibility of S_{+} being the only prime ideal for this intersection. Hence, it remains to show that we can keep this intersection only using homogenous prime ideals of S containing I. Given any (not necessarily homogenous) ideal \mathfrak{p} of S, we can construct \mathfrak{p}^{h} := \bigoplus_{d \geq 0}(\mathfrak{p} \cap S_{d}), which is a homogeneous ideal. What's great about this ideal is that it is prime if \mathfrak{p} is prime, and because I is homogenous, we have I \subset \mathfrak{p} if and only if I \subset \mathfrak{p}^{h}. This lets us get rid of all the primes \mathfrak{p} in the intersection that are not homogeneous, which finishes the proof. \Box
Remark. Let J be any ideal of a \mathbb{Z}_{\geq 0}-graded ring S. We always have J^{h} \subset J. and J = J^{h} if and only if J is homogeneous. We do not have this equality when J is not homogenous. For instance, take S = k[x, y] with the usual grading and J = (x + y^{2}). Then J^{h} = (0).
More information. Having V_{+}(g) \subset V_{+}(f) is equivalent to D_{+}(f) \subset D_{+}(g), so we must have \mathrm{Spec}(S_{f})_{0} \hookrightarrow \mathrm{Spec}(S_{g})_{0}, which is given by (S_{g})_{0} \rightarrow (S_{f})_{0} coming from the further localization may S_{g} \rightarrow S_{f}. This seems to suggest that D(f) \subset D(g), which is true, but this actually uses the argument in Lemma given above, so we could not finish the proof this way.
Remark. As an application, if H \subset \mathbb{P}^{n} (over k) is a reduced closed subscheme of codimension 1, then we may find a homogeneous polynomial F \in k[x_{0}, \dots, x_{n}] such that H \simeq \mathrm{Proj}\left(\frac{k[x_{0}, \dots, x_{n}]}{\sqrt{(F)}}\right). I am not so sure if one can say more.
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