Is using the polynomials ring $\mathbb{Z}_{2}[x]/(x^8+1)$ in RLWE secure against brute force attack?

Given the ring $$R_2=\mathbb{Z}_{2}[x]/(x^8+1)$$, initiate a public key encryption based on RLWE $$(A,b=As+e)$$ where all the parameters are selected as polynomials over the given ring. Under the assumption of brute force attack only and if the vector $$s$$ is 248-byte long, I have the following questions:

• is this secure initiation?
• What should be the dimension of the matrix $$A$$?
• Can Gaussian elimination finds $$s$$ given $$A$$ and $$b$$?.

It is not clear how to create correct encryption from this, let alone secure encryption.

I'll show the easiest lattice-based scheme to make correct, namely secret key encryption. Here, we have

1. Key Gen samples $$s(x)$$ somehow, generally with independent coordinates. The only reasonable choice here seems to be uniform.
2. $$\mathsf{Enc}_{s}(m) = [a(x), a(x)s(x) + e(x) + \mathsf{encode}(m(x))]$$
3. $$\mathsf{Dec}_s([a(x), b(x)]) = \mathsf{decode}(b(x) - a(x)s(x))$$

Here, $$\mathsf{encode}$$ and $$\mathsf{decode}$$ must correct errors, in the sense that

$$\mathsf{decode}(\mathsf{encode}(m(x)) + e(x)) = m(x)$$

Typically this is done by having $$\mathsf{encode}(m(x)) = \Delta m(x)$$ for some scaling factor $$\Delta$$, and then $$\mathsf{decode}(c(x)) = \lfloor c(x)/\Delta\rceil$$. The issue with your scheme is that working over $$\mathbb{Z}_2$$ leaves no room for any scaling factor $$\Delta$$, so it's really not clear what you can do to make the scheme correct, let alone secure.

Note that for this scheme, if you assume the coordinates of $$e(x)$$ are from a Gaussian of standard deviation $$\sigma$$, then one expects something like $$\Pr[\lVert e(x)\rVert_\infty > k\sigma]\leq n\exp(-k^2)$$. Under the assumption that $$k\approx 10$$, $$\sigma \approx 3$$ (which are common), this means that we really need to be working over $$\mathbb{Z}_q$$ for $$q\geq 30$$ just to get a basic notion of correctness for private-key encryption. Public-key encryption would require larger $$q$$ even though.