Do we know that LWE is harder than Ring LWE?

The plain, normal-form, decisional LWE problem over $$\mathbb{Z}/q\mathbb{Z}$$ is: given a uniformly random $$n\times n$$ matrix $$A$$ and vector $$b\in \mathbb{Z}/q\mathbb{Z}^n$$, decide if $$b=As+e$$ for small $$s$$ and small $$e$$.

The decisional ring LWE problem is, in $$R_q:=\mathbb{Z}[x]/(q,p(x))$$, given uniformly random $$a\in R_q$$ and $$b\in R_q$$, decide if $$b=as+e$$ for small $$s,e$$.

The intuition that LWE is at least as hard as ring LWE is that we can express the ring LWE problem as a plain LWE problem by expressing $$b$$, $$s$$, and $$e$$ as vectors, and $$a$$ as a matrix, and then it is exactly an LWE problem. $$a$$ produces a circulant matrix $$A$$, which feels like it should be easier.

But this is only a proof that ring-LWE reduces to worst-case LWE. I want a reduction to average-case LWE. The same technique does not work: circulant matrices might be the hardest type of matrix for LWE. Or at least, I can't prove otherwise.

If there were an easy random self-reducibility proof for the randomness of the matrix $$A$$, that would give the result. But everything I know of goes through the heavy (and non-tight!) machinery of the lattice -> LWE reductions, rather than directly LWE -> LWE.

Does such a reduction exist?

Few things: Circulant LWE is easy --- it corresponds to RLWE in $$\mathbb{Z}_q[x]/(x^n-1)$$, which (for $$n = 2^k$$) is (ring) isomorphic to $$\mathbb{Z}_q[x]/(x-1)\times \prod_i \mathbb{Z}_q[x]/(x^{2^i}+1)$$. In particular, one can project down onto the $$\mathbb{Z}_q[x]/(x-1)$$ component (corresponding to evaluation of the polynomial at 1) to reduce to a 1D plain LWE instance. To avoid this we instead work in $$\mathbb{Z}_q[x]/(x^{2^{k-1}}+1)$$, e.g. the largest-degree irreducible factor of $$\mathbb{Z}_q[x]/(x^{2^k}-1)$$, which corresponds to negacyclic (rather than cyclic matrices). In general, $$\mathbb{Z}_q[x]/(f(x))$$ need not have any worst-case hardness. There are some papers on this (generally with titles that include some combination of "Provably" and "Weak RLWE").
Second, my understanding is that there is a fairly straightforward way to rerandomize the $$A$$ component of $$(A, u)$$, though the technique is not sample-preserving. This is done by letting $$r$$ be suitably random (generally uniform in $$\{0,1\}^m$$, or discrete Gaussian), and outputting $$(r^t A, r^t u)$$. As written this only makes sense for plain LWE, but for RLWE one can instead take random linear combinations of RLWE instances. There are some nuances here (plain LWE shows $$r^t A$$ is uniform by the leftover hash lemma, which is false for RLWE. Instead one needs a "regularity lemma"), but is this the kind of thing you are hoping existed?