I'm thinking about following theorem.
For a finitaly algebraic theory $\mathbb{T}$, $\text{FP}\mathbb{T}$ denotes the full subcategory of $\mathbb{T}\text{-Alg(Set)}$ consisting of finitely presented algebras. $F:\mathbb{T} \rightarrow \text{FP}\mathbb{T}^{op}$ is $n\mapsto \text{hom}_{\mathbb{T}}(n,-)$.
Theorem
Given a left exact category $E$. Then, to any $\mathbb{T}$-algebra $R: \mathbb{T} \rightarrow E$, there exists a left exact functor $\overline{R}: \text{FP}\mathbb{T}^{\text{op}}\rightarrow E$ such that $\overline{R}\circ F = R$.
$\text{Spec}_R(B)$ denotes $\overline{R}(B)$ for $B \in \text{FP}\mathbb{T}$. For a field $k$, $T_k$ denotes the algebraic theory whose morphisms $n\rightarrow m$ are m-tuples of polynomials over $k$ in $n$ variables.
Assume that $k$ is a field, $\mathbb{T} = \mathbb{T}_k$, $E = \text{Set}$, $R=k:\mathbb{T}\rightarrow E ; n\mapsto k^n$, $B(m) = (k[X_1,...,X_n]/I)^m$ for all $m$, where $I$ is an ideal in $k[X_1, ..., X_n]$.
Then, I think $\text{Spec}_R(B)$ is expected to be the zero points of $I$, but how to prove this?