Oracle results on P vs BPP Let $A$ be any EXP complete problem. Then, $P^A = NP^A$.

Let $B$ be some oracle that takes into accounts the queries that $M$ (a TM in P) will make, and we can get $P^B \neq NP^B$.

Question: Do we have similar oracle results for P vs BPP?
I had a vague recollection that I knew an excellent reference for such oracle separations. I finally found it.

A great reference for oracle separations (for classes between P and PSPACE) is the following paper:


  Vereshchagin, N K (1994), "RELATIVIZABLE AND NONRELATIVIZABLE THEOREMS IN THE POLYNOMIAL THEORY OF ALGORITHMS", Russian Academy of Sciences. Izvestiya Mathematics 42 (2): 261


The paper shows (or gives a citation for) an oracle separation between almost every pair of classes that you might care about between P and PSPACE (e.g., it has classes like P, RP, BPP, UP, FewP, NP, MA, AM, other levels of PH, PH, IP, PSPACE, etc.).

For example, Theorem 8 shows an oracle problem in coRP that is not in NP. Since (relative to all oracles) coRP is in BPP and NP contains P, we get an oracle problem in BPP that is not in P.

As I mentioned in my comment, showing an oracle for which $\text{P}^A = \text{BPP}^A$ is easy. Let A be a EXP-complete language or a PSPACE-complete language.