Decision LWE vs Search LWE: Which one is harder?

Question

• Sometimes if we have an attacker who's able to solve decision-LWE problem then we can use them (as a sub-routine) to solve (search) LWE problem, i.e., $\mathsf{sLWE} \leq \mathsf{dLWE}$.
• Conversely, sometimes if we have an attacker who's able to solve (search) LWE problem then we can use them (as a sub-routine) to solve decisional-LWE problem, i.e., $\mathsf{dLWE} \leq \mathsf{sLWE}$.

1.That's quite confusing. Finally, which one is harder? Are they equivalent?

2.If the above question is not precise enough, in which circumstances one is at least as hard as the other? (for example, $m$, number of equations in one of them must be more than the other)

Some (maybe useful) references:

• [Daniele Micciancio, CSE208: Advanced Cryptography, slides, UCSD, Fall 2020, pp. 60-62]
• [Leo Ducas, PhD Thesis, Lattice Based Signatures: Attacks, Analysis and Optimization, theorem 3.23, page 22]

I don't know there are any typos in them or not.

Answer 1

The standard answer to this is Micciancio Mol. In general, we normally assume that search LWE is hard (algorithms breaking LWE typically break search LWE), and then connect the hardness of decision LWE to this. Micciancio Mol shows that this can be done in a sample-preserving way, meaning the number of equations $m$ is the same for both the search and decision LWE instances. This is not obvious from standard search to decision reductions.