Starting with a simple observation. If $x,y\in\{0,1\}$, then the arithmetic product $x\cdot y$ corresponds to the logical conjunction if we interpret $1$ as true and $0$ as false.
But then, for a generic propositional logic formula $\phi$ in $n$ variables, we can obtain a polynomial $f(\phi)$ as follows:
\begin{align} f(p) & = x_p \\ f(\neg \phi) & = 1 - f(\phi) \\ f(\phi_1\land\phi_2) &= f(\phi_1)\cdot f(\phi_2) \\ f(\phi_1\lor\phi_2) &= f(\neg(\neg\phi_1\land\neg\phi_2)) \\ &= 1-((1-f(\phi_1))(1-f(\phi_2))) \end{align}
Then if $f(\phi)=1$ has integer solutions in $[0,1]^n$, the formula is satisfiable.
For example, if $\phi\equiv p\to q\equiv\neg p \lor q$, then $f(\phi)=x_px_q-x_p+1$, and indeed $x_px_q-x_p+1=1$ holds for $x_p=0$ and for $x_q=1$.
So, finding this kind of roots of polynomials is NP-hard, as interesting as this might be.
My question risks to be too vague, but has this been used in NP-hardness proofs for some more interesting algebraic problem?
Is this connection between propositional formulas and polynomials (hence between SAT and algebra) useful for anything?