[proofplan] We prove the theorem by reducing the projection $K^{n+m} \to K^n$ to a sequence of one-coordinate projections. For a one-coordinate projection, the key input is the generic image lemma: the image of a non-empty open subset of an irreducible affine variety contains a non-empty open subset of the closure of the image. Noetherian induction then handles the remaining lower-dimensional closed part. Finally, constructible sets are finite unions of locally closed pieces, and projections preserve finite unions. [/proofplan]

[step:Reduce the projection to one-coordinate projections] For each integer $r \geq 0$, let $\mathbb{A}^r_K$ denote the affine space $K^r$ with its Zariski topology. It is enough to prove the result for the coordinate projection \begin{align*} \rho: \mathbb{A}^{n+1}_K &\to \mathbb{A}^n_K \\ (a_1,\dots,a_n,t) &\mapsto (a_1,\dots,a_n). \end{align*} Indeed, the projection \begin{align*} \pi: \mathbb{A}^{n+m}_K &\to \mathbb{A}^n_K \end{align*} is the composition of the $m$ one-coordinate projections \begin{align*} \mathbb{A}^{n+m}_K \to \mathbb{A}^{n+m-1}_K \to \cdots \to \mathbb{A}^{n+1}_K \to \mathbb{A}^n_K. \end{align*} If each one-coordinate projection sends constructible sets to constructible sets, then their composition does the same. Thus, for the rest of the proof, fix the projection $\rho: \mathbb{A}^{n+1}_K \to \mathbb{A}^n_K$ defined above. [/step]

[step:Prove the generic image lemma for irreducible locally closed sets][claim:Generic image lemma] Let $Y \subset \mathbb{A}^{n+1}_K$ be an irreducible closed subset, and let $U \subset Y$ be a non-empty open subset in the [subspace topology](/page/Subspace%20Topology) on $Y$. Define \begin{align*} Z := \overline{\rho(Y)} \subset \mathbb{A}^n_K, \end{align*} where the closure is taken in the Zariski topology. Then $\rho(U)$ contains a non-empty open subset of $Z$. [/claim] [proof] Let $A := K[x_1,\dots,x_n]$ and $B := K[x_1,\dots,x_n,t]$. Let $P \subset B$ be the prime ideal defining $Y$, and let \begin{align*} \mathfrak{p} := P \cap A. \end{align*} Then $Z = V(\mathfrak{p})$, because $\mathfrak{p}$ is precisely the ideal of all polynomial functions on $\mathbb{A}^n_K$ that vanish on $\rho(Y)$. Since $U \subset Y$ is non-empty and open, there exists a polynomial $g \in B$ such that \begin{align*} \varnothing \neq Y \cap D(g) \subset U, \end{align*} where $D(g) := \{q \in \mathbb{A}^{n+1}_K : g(q) \neq 0\}$. Therefore it is enough to prove that $\rho(Y \cap D(g))$ contains a non-empty open subset of $Z$. Let \begin{align*} A_0 := A/\mathfrak{p}, \qquad B_0 := B/P, \end{align*} and let $\bar{g} \in B_0$ denote the image of $g$. Since $Y \cap D(g) \neq \varnothing$, the element $\bar{g}$ is non-zero in the domain $B_0$. The inclusion $A \hookrightarrow B$ induces an injective map $A_0 \hookrightarrow B_0$, because $\mathfrak{p}=P\cap A$. Let $F := \operatorname{Frac}(A_0)$ be the fraction field of $A_0$. The localized algebra \begin{align*} C := (B_0)_{\bar{g}} \otimes_{A_0} F \end{align*} is non-zero: tensoring with $F$ localizes $B_0$ at the multiplicative set $A_0\setminus\{0\}$, and then inverting the non-zero element $\bar{g}$ cannot kill the domain $B_0$. It is finitely generated over $F$ because $(B_0)_{\bar{g}}$ is finitely generated over $A_0$. Choose a maximal ideal $\mathfrak{m} \subset C$. By [Zariski's lemma](/page/Zariski%27s%20Lemma) applied to the finitely generated algebra $C$ over the field $F$, the residue field $L := C/\mathfrak{m}$ is a finite algebraic extension of $F$. Let \begin{align*} q: C \to L \end{align*} denote the quotient homomorphism. Choose an $F$-basis $\ell_1,\dots,\ell_s$ of $L$ with $\ell_1 = 1$. Since $q(C)$ generates $L$ as a field over $F$, and since $C$ is generated over $F$ by the image of the finitely generated $A_0$-algebra $(B_0)_{\bar{g}}$, each generator has coordinates in the basis $\ell_1,\dots,\ell_s$ with coefficients in $F$. The multiplication table also has finitely many coefficients $c_{ijk} \in F$ defined by \begin{align*} \ell_i\ell_j = \sum_{k=1}^s c_{ijk}\ell_k. \end{align*} Choose finitely many elements of $(B_0)_{\bar{g}}$ whose images generate $q((B_0)_{\bar{g}})$ as an $A_0$-algebra after tensoring with $F$. Each of their images has finitely many coordinates in the basis $\ell_1,\dots,\ell_s$, and the multiplication table has finitely many coefficients $c_{ijk}$. Choose a non-zero element $h_0 \in A_0$ clearing the denominators of all these coordinates and all the coefficients $c_{ijk}$. Then the $A_0[h_0^{-1}]$-submodule \begin{align*} M := A_0[h_0^{-1}]\ell_1 + \cdots + A_0[h_0^{-1}]\ell_s \subset L \end{align*} is closed under multiplication, because every product $\ell_i\ell_j$ has coefficients $c_{ijk}\in A_0[h_0^{-1}]$. Hence $M$ is a finite $A_0[h_0^{-1}]$-algebra. The same denominator choice puts the images under $q$ of the chosen algebra generators in $M$, so the quotient map $q$ restricts to a well-defined $A_0[h_0^{-1}]$-algebra homomorphism \begin{align*} (B_0)_{\bar{g}}[h_0^{-1}] \to M. \end{align*} Let $R$ be the image of this homomorphism. Because $A_0[h_0^{-1}]$ is Noetherian and $M$ is a finite $A_0[h_0^{-1}]$-module, the submodule $R \subset M$ is finite over $A_0[h_0^{-1}]$. Moreover $R \otimes_{A_0[h_0^{-1}]} F \cong L$, because localizing the restricted map recovers the surjective map $q: C \to L$. We now apply [generic freeness](/page/Generic%20Freeness) to the finite $A_0[h_0^{-1}]$-module $R$. The ring $A_0[h_0^{-1}]$ is a finitely generated domain over the field $K$, hence Noetherian and reduced, so the hypotheses of generic freeness hold. Therefore there exists a non-zero element $h_1 \in A_0[h_0^{-1}]$ such that $R[h_1^{-1}]$ is a finite free module over $A_0[h_0^{-1}][h_1^{-1}]$. Since $R \otimes_{A_0[h_0^{-1}]} F \cong L \neq 0$, the localized module $R[h_1^{-1}]$ remains non-zero. Choose a representative $h_1' \in A_0$ for $h_1$ after clearing the denominator coming from powers of $h_0$, and define $h := h_0h_1' \in A_0$. Replacing $R$ by $R[h_1^{-1}]$, the original quotient map localizes to a quotient \begin{align*} (B_0)_{\bar{g}}[h^{-1}] \to R \end{align*} whose target is a non-zero finite free $A_0[h^{-1}]$-algebra. Let \begin{align*} \Omega := D(h) \cap Z \subset Z. \end{align*} This is a non-empty open subset of $Z$ because $h \neq 0$ in the domain $A_0$. Let $a \in \Omega$. Evaluation at $a$ gives a $K$-algebra homomorphism \begin{align*} A_0[h^{-1}] \to K. \end{align*} Since $R$ is non-zero finite free over $A_0[h^{-1}]$, the fiber algebra \begin{align*} R_a := R \otimes_{A_0[h^{-1}]} K \end{align*} is a non-zero finite-dimensional $K$-algebra. Hence $R_a$ has a maximal ideal, and because $K$ is algebraically closed, the residue field at that maximal ideal is $K$. Therefore the quotient map $(B_0)_{\bar{g}}[h^{-1}] \to R$ produces a $K$-point of $Y \cap D(g)$ [lying over](/theorems/2876) $a$. Thus every $a \in \Omega$ belongs to $\rho(Y \cap D(g))$. Hence \begin{align*} \Omega \subset \rho(Y \cap D(g)) \subset \rho(U), \end{align*} and $\Omega$ is a non-empty open subset of $Z$. This proves the claim. [/proof][/step]

[guided]The point of the lemma is to show that a dominant projection is not merely dense on the image: it is actually open on some non-empty part of the target. We encode the geometry algebraically. Let $A := K[x_1,\dots,x_n]$ be the coordinate ring of the target $\mathbb{A}^n_K$, and let $B := K[x_1,\dots,x_n,t]$ be the coordinate ring of the source $\mathbb{A}^{n+1}_K$. Since $Y$ is irreducible and closed, its defining ideal $P \subset B$ is prime. Define \begin{align*} \mathfrak{p} := P \cap A. \end{align*} Then $V(\mathfrak{p})$ is exactly the Zariski closure of $\rho(Y)$, because a polynomial in the target coordinates vanishes on $\rho(Y)$ exactly when its pullback to the source vanishes on $Y$. The [open set](/page/Open%20Set) $U \subset Y$ contains a principal open subset of $Y$: choose $g \in B$ such that \begin{align*} \varnothing \neq Y \cap D(g) \subset U. \end{align*} So it suffices to prove that $\rho(Y \cap D(g))$ contains a non-empty open subset of $Z := V(\mathfrak{p})$. Set \begin{align*} A_0 := A/\mathfrak{p}, \qquad B_0 := B/P. \end{align*} The ring $A_0$ is the coordinate ring of $Z$, and $B_0$ is the coordinate ring of $Y$. The image $\bar{g} \in B_0$ is non-zero because $Y \cap D(g)$ is non-empty. Let $F := \operatorname{Frac}(A_0)$ be the function field of $Z$. Passing to the generic point of $Z$ means tensoring with $F$, and inverting $\bar{g}$ means restricting to the open set $D(g)$. Thus \begin{align*} C := (B_0)_{\bar{g}} \otimes_{A_0} F \end{align*} is the coordinate algebra of the generic fiber of $Y \cap D(g) \to Z$. It is non-zero and finitely generated over $F$. Choose a maximal ideal $\mathfrak{m} \subset C$. [Zariski's lemma](/page/Zariski%27s%20Lemma) applies because $C$ is a finitely generated algebra over the field $F$; it gives that the residue field \begin{align*} L := C/\mathfrak{m} \end{align*} is a finite algebraic extension of $F$. Let $q: C \to L$ be the quotient map. This is a point of the generic fiber, possibly defined over $L$ rather than over $F$. We now spread this generic point over a non-empty open subset of $Z$. Choose an $F$-basis $\ell_1,\dots,\ell_s$ of $L$ with $\ell_1 = 1$. The multiplication table has coefficients $c_{ijk} \in F$ defined by \begin{align*} \ell_i\ell_j = \sum_{k=1}^s c_{ijk}\ell_k. \end{align*} Also, the images under $q$ of the finitely many algebra generators of $(B_0)_{\bar{g}}$ have coordinates in this basis with coefficients in $F$. Choose finitely many elements of $(B_0)_{\bar{g}}$ whose images generate the image algebra after tensoring with $F$. Their images under $q$ have only finitely many coordinates in the basis $\ell_1,\dots,\ell_s$, and the multiplication table has only finitely many constants $c_{ijk}$. Choose a non-zero element $h_0 \in A_0$ clearing the denominators of all these finitely many coefficients. Then \begin{align*} M := A_0[h_0^{-1}]\ell_1 + \cdots + A_0[h_0^{-1}]\ell_s \subset L \end{align*} is closed under multiplication: for each pair $(i,j)$, the equality \begin{align*} \ell_i\ell_j = \sum_{k=1}^s c_{ijk}\ell_k \end{align*} has all coefficients $c_{ijk}$ in $A_0[h_0^{-1}]$. Hence $M$ is a finite $A_0[h_0^{-1}]$-algebra. The same denominator choice ensures that the images under $q$ of the chosen algebra generators lie in $M$, so $q$ restricts to a well-defined $A_0[h_0^{-1}]$-algebra homomorphism \begin{align*} (B_0)_{\bar{g}}[h_0^{-1}] \to M. \end{align*} Let $R$ be the image of this homomorphism. Since $A_0[h_0^{-1}]$ is Noetherian and $M$ is a finite module over it, the submodule $R \subset M$ is finite. Localizing at the generic point recovers the surjection $q: C \to L$, so \begin{align*} R \otimes_{A_0[h_0^{-1}]} F \cong L. \end{align*} Thus $R$ is non-zero. Finally we use [generic freeness](/page/Generic%20Freeness). The ring $A_0[h_0^{-1}]$ is a finitely generated domain over $K$, so it is Noetherian and reduced; these are the hypotheses needed to apply generic freeness to the finite module $R$. Hence there is a non-zero element $h_1\in A_0[h_0^{-1}]$ such that $R[h_1^{-1}]$ is finite free over $A_0[h_0^{-1}][h_1^{-1}]$. The module is still non-zero after this localization because \begin{align*} R \otimes_{A_0[h_0^{-1}]} F \cong L \neq 0. \end{align*} After clearing the denominator of $h_1$ coming from powers of $h_0$, choose $h\in A_0\setminus\{0\}$ so that $A_0[h^{-1}]$ contains this localization. Localizing the restricted quotient map gives a quotient \begin{align*} (B_0)_{\bar{g}}[h^{-1}] \to R[h_1^{-1}], \end{align*} and, after replacing $R$ by $R[h_1^{-1}]$, the target is a non-zero finite free $A_0[h^{-1}]$-algebra. Now define the open subset \begin{align*} \Omega := D(h) \cap Z. \end{align*} It is non-empty because $h \neq 0$ in the domain $A_0$. For any $a \in \Omega$, evaluation at $a$ gives a homomorphism $A_0[h^{-1}] \to K$. The fiber \begin{align*} R_a := R \otimes_{A_0[h^{-1}]} K \end{align*} is a non-zero finite-dimensional $K$-algebra because $R$ is non-zero finite free. Hence $R_a$ has a maximal ideal. Since $K$ is algebraically closed, the residue field at that maximal ideal is $K$, so this maximal ideal gives a $K$-rational point of the fiber over $a$. Because $R$ is a quotient of $(B_0)_{\bar{g}}[h^{-1}]$, that point lies in $Y \cap D(g)$ and projects to $a$. Therefore every $a \in \Omega$ belongs to $\rho(Y \cap D(g))$, and hence \begin{align*} \Omega \subset \rho(Y \cap D(g)) \subset \rho(U). \end{align*} This proves that $\rho(U)$ contains a non-empty open subset of $Z$.[/guided]

[step:Prove the one-coordinate theorem by Noetherian induction] For every closed subset $W \subset \mathbb{A}^n_K$, let $\mathcal{P}(W)$ be the assertion that for every [constructible subset](/page/Constructible%20Set) \begin{align*} C \subset \rho^{-1}(W) \subset \mathbb{A}^{n+1}_K, \end{align*} the image $\rho(C)$ is constructible in $W$, with the subspace Zariski topology. Since affine space with the Zariski topology is a [Noetherian topological space](/page/Noetherian%20Topological%20Space), we prove $\mathcal{P}(W)$ for all closed $W$ by [Noetherian induction](/page/Noetherian%20Induction) on the closed subsets of $\mathbb{A}^n_K$. Fix a closed subset $W \subset \mathbb{A}^n_K$ and assume $\mathcal{P}(W')$ holds for every proper closed subset $W' \subsetneq W$. Let $C \subset \rho^{-1}(W)$ be constructible. Since $\rho^{-1}(W)$ is Noetherian, the standard locally closed decomposition for [constructible subsets](/page/Constructible%20Set) in a Noetherian topological space gives that $C$ is a finite union of subsets of the form \begin{align*} Y \cap U, \end{align*} where $Y \subset \rho^{-1}(W)$ is irreducible and closed in $\rho^{-1}(W)$, and $U \subset Y$ is open in the subspace topology on $Y$. Because $\rho$ preserves finite unions and finite unions of constructible sets are constructible, it is enough to prove that $\rho(U)$ is constructible in $W$ for one such pair $(Y,U)$. If $U = \varnothing$, then $\rho(U) = \varnothing$, which is constructible in $W$. Assume $U \neq \varnothing$. Since $Y$ is closed in $\rho^{-1}(W)$ and $\rho^{-1}(W)$ is closed in $\mathbb{A}^{n+1}_K$, the set $Y$ is also closed in $\mathbb{A}^{n+1}_K$. Define \begin{align*} Z := \overline{\rho(Y)} \subset W, \end{align*} where the closure is taken in $\mathbb{A}^n_K$. The inclusion $Z \subset W$ follows from $Y \subset \rho^{-1}(W)$ and the fact that $W$ is closed. By the generic image lemma, there exists a non-empty open subset $\Omega \subset Z$ such that \begin{align*} \Omega \subset \rho(U). \end{align*} Let \begin{align*} Z_1 := Z \setminus \Omega. \end{align*} Then $Z_1$ is a proper closed subset of $Z$, hence also a proper closed subset of $W$ unless $Z = W$ and still a closed subset of $W$ in all cases. Moreover \begin{align*} \rho(U) = \Omega \cup \rho\bigl(U \cap \rho^{-1}(Z_1)\bigr). \end{align*} The set $U \cap \rho^{-1}(Z_1)$ is constructible in $\rho^{-1}(Z_1)$, because it is the intersection of the constructible set $U \subset \mathbb{A}^{n+1}_K$ with the [closed set](/page/Closed%20Set) $\rho^{-1}(Z_1)$. Since $Z_1 \subsetneq Z \subset W$, the induction hypothesis $\mathcal{P}(Z_1)$ gives that \begin{align*} \rho\bigl(U \cap \rho^{-1}(Z_1)\bigr) \end{align*} is constructible in $Z_1$. Because constructibility is preserved when passing from a closed subspace to the ambient Noetherian space, it is constructible in $W$ and in $\mathbb{A}^n_K$. The set $\Omega$ is open in the closed set $Z$, so $\Omega$ is locally closed in $W$ and therefore constructible in $W$. Hence $\rho(U)$ is a finite union of constructible subsets of $W$, and is constructible in $W$. This proves $\mathcal{P}(W)$ by Noetherian induction. Taking $W = \mathbb{A}^n_K$ proves the one-coordinate case. [/step]

[step:Pass from locally closed pieces to an arbitrary constructible set] Let $X \subset \mathbb{A}^{n+m}_K$ be constructible. By definition, there exist finitely many locally closed subsets $X_1,\dots,X_r \subset \mathbb{A}^{n+m}_K$ such that \begin{align*} X = \bigcup_{i=1}^r X_i. \end{align*} Applying the one-coordinate result successively to each $X_i$ along the factorization of $\pi$ into one-coordinate projections, we obtain that each $\pi(X_i)$ is constructible in $\mathbb{A}^n_K$. Since projection commutes with finite unions, \begin{align*} \pi(X) = \pi\left(\bigcup_{i=1}^r X_i\right) = \bigcup_{i=1}^r \pi(X_i). \end{align*} A finite union of constructible subsets of $\mathbb{A}^n_K$ is constructible. Therefore $\pi(X)$ is constructible in $K^n$, as required. [/step]