# How can I translate this to the polynomial setting?

Imagine I have these equations $$x^2 = x \cdot x \; \land \; 0 = x^2-x,$$ and I ask for a solution in $$\mathbb{Z}_q = \{0,1,\dots,q-1\}$$ where $$q$$ is a prime number. It is easy to see that the only solutions to these equations are $$0$$ and $$1$$.

However, if we ask for solutions in the polynomial ring $$\mathcal{R}_q = \mathbb{Z}_q[x]/\langle x^n+1 \rangle$$ where $$n=2^k$$ with $$k \in \mathbb{N}$$ (i.e., $$\mathcal{R}_q$$ is the ring of polynomials with coefficients in $$\mathbb{Z}_q$$ of degree at most $$n-1$$), there are different from $$0$$ or $$1$$ as we are not working in a field anymore.

I am wondering if there are some set of equations that are only satisfied by $$0$$ and $$1$$ in $$\mathcal{R}_q$$.

There are not, in general. The problem is that your ring can in general be written as a product of other rings (whenever $$x^n + 1$$ has a nontrivial factorization $$\bmod q$$), and the solutions to a system of polynomial equations in a product $$R \times S$$ have the following property:
$$(r, s)$$ is a solution iff $$r$$ is a solution to the system when projected down to $$R$$ and $$s$$ is a solution to the system when projected down to $$S$$.
It follows that if $$(0, 0)$$ and $$(1, 1)$$ are both solutions then $$0$$ and $$1$$ are both solutions to the system projected down to both $$R$$ and $$S$$, which means that necessarily $$(0, 1)$$ and $$(1, 0)$$ must also be solutions.
i.e., $$\mathcal{R}_q$$ is the ring of polynomials with coefficients in $$\mathbb{Z}_q$$ of degree at most $$n-1$$
is not an accurate description of $$R_q$$. "Polynomials of degree at most $$n-1$$" only forms a vector space and multiplication can be defined in many different ways.