[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]
