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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