[proofplan] The Zariski closure of any subset $S \subset \mathbb{P}^n_k$ is the smallest closed set containing $S$, which by the projective Galois correspondence equals $V(I(S))$, where $I(S)$ is the homogeneous ideal of homogeneous polynomials vanishing on $S$. So the strategy is: identify the homogeneous ideal of polynomials vanishing on $\iota_0(V^{\mathrm{aff}})$, and show its vanishing locus equals $V(I^h)$. The bridge between affine and projective ideals is the [Homogenisation and Dehomogenisation](/theorems/2140) lemma, which writes any homogeneous polynomial $f$ as $X_0^m \cdot g^h$ for $g$ its dehomogenisation. Vanishing on $\iota_0(V^{\mathrm{aff}})$ on the chart $X_0 = 1$ then translates into vanishing of $g$ on $V^{\mathrm{aff}}$, which by [Hilbert's Nullstellensatz](/theorems/2124) (applicable because $k$ is algebraically closed) means $g \in \sqrt{I}$. Care is needed on the locus $\{X_0 = 0\}$ (the "boundary at infinity"); we show that taking the projective vanishing locus collapses these subtleties so that $V(I^h)$ is exactly the closure. [/proofplan] [step:Express the projective closure via the homogeneous ideal of $\iota_0(V^{\mathrm{aff}})$] For a subset $S \subset \mathbb{P}^n_k$, define the homogeneous ideal \begin{align*} I(S) := \langle f \in k[X_0, \ldots, X_n] : f \text{ homogeneous}, \ f(p) = 0 \text{ for all } [p] \in S \rangle \end{align*} — the ideal generated by all homogeneous polynomials vanishing on $S$. By the [Characterisation of Homogeneous Ideals](/theorems/2132), $I(S)$ is a homogeneous ideal in the standard sense, and by the projective [Algebraic-Geometric Duality](/theorems/2127), \begin{align*} \overline{S}^{\,\mathbb{P}^n_k} = V(I(S)). \end{align*} (A closed set $V(J) \supset S$ satisfies $J \subset I(S)$, so $V(I(S)) \subset V(J)$; conversely $V(I(S))$ is closed and contains $S$. Hence $V(I(S))$ is the smallest closed set containing $S$.) Apply this with $S := \iota_0(V^{\mathrm{aff}})$ and write $J := I(\iota_0(V^{\mathrm{aff}}))$. To prove the theorem it suffices to show \begin{align*} V(J) = V(I^h). \end{align*} Since the vanishing locus of a homogeneous ideal depends only on its radical by [Vanishing is Ideal-Independent](/theorems/2133), this follows from the radical equality \begin{align*} \sqrt{J} = \sqrt{I^h}. \end{align*} [/step] [step:Characterise homogeneous polynomials vanishing on $\iota_0(V^{\mathrm{aff}})$] A homogeneous polynomial $f \in k[X_0, \ldots, X_n]$ vanishes on $\iota_0(V^{\mathrm{aff}})$ if and only if for every $(a_1, \ldots, a_n) \in V^{\mathrm{aff}}$, \begin{align*} f(1, a_1, \ldots, a_n) = 0, \end{align*} i.e. the dehomogenisation $g(Y_1, \ldots, Y_n) := f(1, Y_1, \ldots, Y_n) \in k[Y_1, \ldots, Y_n]$ vanishes on $V^{\mathrm{aff}} = V_{\mathrm{aff}}(I)$. By [Hilbert's Nullstellensatz](/theorems/2124) (applied to $k$ algebraically closed and the ideal $I \subset k[Y_1, \ldots, Y_n]$), \begin{align*} g \in \sqrt{I}, \end{align*} i.e. some power $g^N$ lies in $I$. By [Homogenisation and Dehomogenisation](/theorems/2140), every homogeneous $f \in k[X_0, \ldots, X_n]$ factors uniquely as \begin{align*} f(X_0, \ldots, X_n) = X_0^m \cdot g^h(X_0, \ldots, X_n) \end{align*} where $m = \max\{k \ge 0 : X_0^k \mid f\}$ and $g^h$ is the homogenisation of $g = f(1, Y)$. We have thus shown: \begin{align*} \{f \text{ homogeneous} : f \text{ vanishes on } \iota_0(V^{\mathrm{aff}})\} = \{X_0^m \cdot g^h : m \ge 0,\ g \in \sqrt{I}\}. \end{align*} [guided] Two ingredients combine here. \textbf{Vanishing on $\iota_0(V^{\mathrm{aff}})$ as a condition on dehomogenisation.} The chart embedding sends $(y_1, \ldots, y_n) \mapsto [1 : y_1 : \cdots : y_n]$, so the image lives in the chart $U_0 = \{X_0 \neq 0\}$. For a homogeneous polynomial $f$, the value at the projective point $[1 : a_1 : \cdots : a_n]$ is well-defined as a vanishing condition (since $f$ is homogeneous), and equals $f(1, a_1, \ldots, a_n) \in k$ — exactly the dehomogenisation $g$ evaluated at $(a_1, \ldots, a_n) \in \mathbb{A}^n_k$. So \begin{align*} f \text{ vanishes on } \iota_0(V^{\mathrm{aff}}) \iff g \text{ vanishes on } V^{\mathrm{aff}}. \end{align*} \textbf{Vanishing on $V^{\mathrm{aff}}$ as containment in $\sqrt{I}$.} We apply [Hilbert's Nullstellensatz](/theorems/2124). Its hypotheses require $k$ to be algebraically closed (stipulated in the theorem statement) and $I$ to be an ideal of $k[Y_1, \ldots, Y_n]$ (given). The conclusion is $I(V_{\mathrm{aff}}(I)) = \sqrt{I}$. The polynomial $g$ vanishes on $V_{\mathrm{aff}}(I)$ iff $g \in I(V_{\mathrm{aff}}(I)) = \sqrt{I}$, i.e. $g^N \in I$ for some $N \ge 1$. \textbf{Recovering $f$ from $g$.} The dehomogenisation map $f \mapsto f(1, Y) = g$ has kernel $(X_0)$ (the principal ideal generated by $X_0$): if $f = X_0 \cdot \tilde f$ then $g = 1 \cdot \tilde f(1, Y) = \tilde f(1, Y)$, so $X_0$-multiples are not detected. [Homogenisation and Dehomogenisation](/theorems/2140) quantifies this: every homogeneous $f$ is uniquely $X_0^m \cdot g^h$. So the homogeneous polynomials vanishing on $\iota_0(V^{\mathrm{aff}})$ are exactly $\{X_0^m g^h : m \ge 0, g \in \sqrt{I}\}$. The free factor $X_0^m$ accounts for vanishing at the "boundary" $\{X_0 = 0\} \subset \mathbb{P}^n_k$, which is automatic (the chart misses this hyperplane) and does not constrain the closure on $U_0$. [/guided] [/step] [step:Identify the radical of the ideal $J$ with $\sqrt{I^h}$] By Step 2, the homogeneous ideal $J = I(\iota_0(V^{\mathrm{aff}}))$ is generated by elements of the form $X_0^m g^h$ with $g \in \sqrt{I}$ and $m \ge 0$. We claim \begin{align*} \sqrt{J} = \sqrt{I^h}. \end{align*} \textbf{Inclusion $\sqrt{I^h} \subset \sqrt{J}$.} Take a generator $g^h$ of $I^h$ with $g \in I$. Then $g \in \sqrt{I}$, so by Step 2 the polynomial $g^h = X_0^0 \cdot g^h$ vanishes on $\iota_0(V^{\mathrm{aff}})$, hence $g^h \in J$. Since the generators of $I^h$ all lie in $J$, we have $I^h \subset J$, hence $\sqrt{I^h} \subset \sqrt{J}$. \textbf{Inclusion $\sqrt{J} \subset \sqrt{I^h}$.} It suffices to show that every generator $X_0^m g^h$ with $g \in \sqrt{I}$ lies in $\sqrt{I^h}$. Choose $N \ge 1$ with $g^N \in I$. Then $(g^N)^h \in I^h$ by definition. We compare $(g^h)^N$ with $(g^N)^h$ by their dehomogenisations: \begin{align*} ((g^h)^N)(1, Y) = (g^h(1, Y))^N = g(Y)^N = (g^N)(Y) = ((g^N)^h)(1, Y). \end{align*} By [Homogenisation and Dehomogenisation](/theorems/2140), a homogeneous polynomial $F$ with dehomogenisation $F(1, Y) = G$ equals $X_0^{\deg F - \deg G} \cdot G^h$. Both $(g^h)^N$ and $(g^N)^h$ are homogeneous with dehomogenisation $g^N$, and both are not divisible by $X_0$ (as powers/homogenisations are not — this is the defining property of the homogenisation operation). Hence $\deg(g^h)^N = N \deg g = \deg(g^N)^h$, and \begin{align*} (g^h)^N = (g^N)^h. \end{align*} Thus \begin{align*} (g^h)^N = (g^N)^h \in I^h, \end{align*} which means $g^h \in \sqrt{I^h}$. For the generator $X_0^m g^h$, we have $X_0^m g^h \in \sqrt{I^h}$ since $g^h \in \sqrt{I^h}$ and $\sqrt{I^h}$ is closed under multiplication by ring elements. As the generators of $J$ all lie in $\sqrt{I^h}$, we conclude $J \subset \sqrt{I^h}$, hence $\sqrt{J} \subset \sqrt{\sqrt{I^h}} = \sqrt{I^h}$. Combining both inclusions, $\sqrt{J} = \sqrt{I^h}$. [guided] We pin down the radical of the homogeneous ideal of the image, and show it coincides with the radical of the homogenised affine ideal. \textbf{Why radicals?} By [Vanishing is Ideal-Independent](/theorems/2133), $V(K)$ is a function of $\sqrt{K}$. Two ideals with the same radical have the same vanishing locus. So we lose no information by passing to radicals, and we gain the freedom to compare $J$ (which is messy — built out of all homogeneous polynomials vanishing on the image) with $I^h$ (which is cleanly defined as homogenisation of the affine ideal). \textbf{Inclusion $\sqrt{I^h} \subset \sqrt{J}$.} For each $g \in I \subset \sqrt{I}$, Step 2 gives $g^h \in J$ directly. So the generators of $I^h$ lie in $J$, hence $I^h \subset J$. \textbf{Inclusion $\sqrt{J} \subset \sqrt{I^h}$.} A generator of $J$ has the form $X_0^m g^h$ with $g \in \sqrt{I}$. Pick $N \ge 1$ with $g^N \in I$; then $(g^N)^h \in I^h$. The question is how $(g^N)^h$ relates to $(g^h)^N$. Both are homogeneous polynomials; both have dehomogenisation $g^N$ (direct substitution of $X_0 = 1$ commutes with taking powers and homogenisation). By [Homogenisation and Dehomogenisation](/theorems/2140), a homogeneous polynomial is determined by its dehomogenisation together with its degree (or equivalently the power of $X_0$ factored out). Now: $(g^h)^N$ is a power of a homogenisation, so it is not divisible by $X_0$; $(g^N)^h$ is a homogenisation, so it is not divisible by $X_0$. Their degrees must agree because $(\deg(g^h)^N) = N \deg g = \deg((g^N)^h)$. Hence they are equal: \begin{align*} (g^h)^N = (g^N)^h \in I^h. \end{align*} So $g^h \in \sqrt{I^h}$, and multiplying by $X_0^m$ keeps us in $\sqrt{I^h}$. \textbf{Why this is clean: the boundary at infinity.} A geometric point not in the image of $\iota_0$ corresponds to a point of the hyperplane at infinity $\{X_0 = 0\}$. The factor $X_0^m$ in elements of $J$ measures vanishing along this hyperplane, but the closure does not depend on this factor (since $V(\cdot)$ is determined by the radical, and $X_0^m$ is absorbed into the radical). The closure $V(I^h)$ adds exactly the points "at infinity" where the homogenisations of elements of $I$ vanish — no more, no less. [/guided] [/step] [step:Conclude $\overline{\iota_0(V^{\mathrm{aff}})} = V(I^h)$] Combining Step 1 and Step 3, and using that $V(\cdot)$ depends only on the radical of the input ideal by [Vanishing is Ideal-Independent](/theorems/2133): \begin{align*} \overline{\iota_0(V^{\mathrm{aff}})}^{\,\mathbb{P}^n_k} = V(J) = V(\sqrt{J}) = V(\sqrt{I^h}) = V(I^h). \end{align*} This is the desired equality. The analogous statement holds for any other standard chart $U_i$, with $X_0$ replaced by $X_i$ throughout. [/step]