[proofplan] We first show that every affine algebraic subset of $K^n$ is definable by a quantifier-free formula in the ring language with constants from $K$. The only algebraic input is that every ideal of $K[x_1,\dots,x_n]$ is finitely generated, so even algebraic sets defined by arbitrary families of polynomials can be defined by finitely many equations. We then use the fact that quantifier-free formulas are closed under Boolean connectives, matching exactly the finite Boolean operations used to build constructible sets. [/proofplan]

[step:Express each affine algebraic set by finitely many polynomial equations] Let $S \subset K[x_1,\dots,x_n]$ be a set of polynomials, and define its affine algebraic set \begin{align*} V_K(S) := \{a \in K^n : f(a)=0 \text{ for every } f \in S\}. \end{align*} Let $I := (S) \subset K[x_1,\dots,x_n]$ be the ideal generated by $S$. By the [Hilbert Basis Theorem](/theorems/860) (citing a result not yet in the wiki: Hilbert Basis Theorem), the ideal $I$ is finitely generated. Choose polynomials $f_1,\dots,f_r \in K[x_1,\dots,x_n]$ such that \begin{align*} I = (f_1,\dots,f_r). \end{align*} We claim that \begin{align*} V_K(S)=V_K(f_1,\dots,f_r). \end{align*} If $a \in V_K(S)$ and $g \in I$, then $g$ is a finite $K[x_1,\dots,x_n]$-linear combination of elements of $S$, so evaluating at $a$ gives $g(a)=0$. In particular $f_i(a)=0$ for every $i \in \{1,\dots,r\}$, hence $a \in V_K(f_1,\dots,f_r)$. Conversely, if $a \in V_K(f_1,\dots,f_r)$ and $h \in S$, then $h \in I=(f_1,\dots,f_r)$, so \begin{align*} h = \sum_{i=1}^r q_i f_i \end{align*} for some $q_1,\dots,q_r \in K[x_1,\dots,x_n]$. Evaluating at $a$ gives \begin{align*} h(a)=\sum_{i=1}^r q_i(a) f_i(a)=0, \end{align*} so $a \in V_K(S)$. Therefore the two algebraic sets are equal. [/step]

[step:Translate finitely many polynomial equations into a quantifier-free formula] For each polynomial $f_i \in K[x_1,\dots,x_n]$, let $t_{f_i}(x_1,\dots,x_n)$ denote the corresponding $\mathcal{L}_{\mathrm{ring}}(K)$-term obtained by using addition, multiplication, $0$, $1$, and constant symbols for coefficients in $K$. Define the quantifier-free formula \begin{align*} \psi_{V}(x_1,\dots,x_n) := \bigwedge_{i=1}^r \bigl(t_{f_i}(x_1,\dots,x_n)=0\bigr). \end{align*} For every $a=(a_1,\dots,a_n)\in K^n$, interpretation of polynomial terms in the structure $K$ gives \begin{align*} K \models t_{f_i}(a)=0 \quad \iff \quad f_i(a)=0. \end{align*} Hence \begin{align*} K \models \psi_V(a) \quad \iff \quad a \in V_K(f_1,\dots,f_r) \quad \iff \quad a \in V_K(S). \end{align*} Thus every affine algebraic subset of $K^n$ is quantifier-free $\mathcal{L}_{\mathrm{ring}}(K)$-definable. [/step]

[step:Use Boolean connectives to define finite Boolean combinations] By hypothesis, $X$ is constructible, so there exist affine algebraic subsets $V_1,\dots,V_m \subset K^n$ and a finite Boolean expression $B$ in $m$ variables such that \begin{align*} X = B(V_1,\dots,V_m), \end{align*} where the Boolean operations are finite union, finite intersection, and complement in $K^n$. For each $j \in \{1,\dots,m\}$, the previous step gives a quantifier-free $\mathcal{L}_{\mathrm{ring}}(K)$-formula $\psi_j(x_1,\dots,x_n)$ defining $V_j$. Form a new formula $\varphi(x_1,\dots,x_n)$ by replacing, in the Boolean expression $B$, each set variable $V_j$ by the formula $\psi_j$, union by disjunction $\vee$, intersection by conjunction $\wedge$, and complement by negation $\neg$. The formula $\varphi$ is quantifier-free because it is built from quantifier-free formulas using only Boolean connectives. For every $a \in K^n$, satisfaction of Boolean combinations of formulas agrees with membership in the corresponding Boolean combinations of their definable sets. Therefore \begin{align*} K \models \varphi(a) \quad \iff \quad a \in B(V_1,\dots,V_m) \quad \iff \quad a \in X. \end{align*} Thus \begin{align*} X = \{a \in K^n : K \models \varphi(a)\}, \end{align*} which is the desired quantifier-free definition. [/step]