Setting x to 0 is pointless, since [0:x] becomes R, and there are no prime ideals containing R. In contrast, set x to a unit, whence [0:x] = 0. Thus d(R) includes the minimal primes containing 0, which are the minimal primes of R.
If you want a minimal prime ideal below a given prime ideal, that contains some other ideal H, mod out by H and apply the above to the quotient ring. This gives a minimal prime ideal, which pulls back to a minimal prime ideal in R containing H.
By the way, all this works in the noncommutative world. If the intersection contains xRy, each prime in the chain contains xRy, each prime contains x or y, and either x or y will survive in the intersection. The same proof works for semiprime ideals; and you can apply it to the union of an ascending chain of ideals. But I digress.
If there is a prime ideal P containing [0:x] for some x, P descends to a minimal prime ideal containing [0:x], a member of d(R).
Conversely, let y belong to P in d(R). Let P be minimal over [0:x]. Mod out by the annihilator and P maps to a minimal prime ideal containing 0. If y maps to 0 then y is in the annihilator, and y becomes a zero divisor. If y maps to a nonzero element in the quotient ring, it is a zero divisor. Pull this back to R and yz lies in the annihilator, whence xyz = 0. This makes y a zero divisor.
In summary, each zero divisor is contained in a prime in d(R), and each prime in d(R) consists entirely of zero divisors.
Map prime ideals and annihilators over to R/S. If P intersects S its image becomes all of R/S, so assume P misses S. With this constraint in hand, prime ideals in R and in R/S correspond.
If one ideal contains another in R, the same is true in R/S. Thus the minimal prime containing an annihilator implies a corresponding minimal prime in R/S containing the image of the annihilator. We only need ask whether the image of an annihilator is in fact an annihilator.
If y is in [0:x] then all the fractions of y kill x/1 in R/S. The image of [0:x] lies in the annihilator of x/1, relative to R/S.
Conversely, if y/w times x/1 = 0, then uyx = 0 for some u in S. If y does not kill x then uy kills x, and lies in P. Since P misses S, u cannot lie in P, hence P contains y. Now y might not be in [0:x], but it lies in P, and when we move to R/S, the image of P contains the fractions of y, which are part of the annihilator in the fraction ring. Prime ideals containing annihilators corespond to prime ideals containing annihilators.
Put this all together and d(R/S) equals d(R)/S, discarding any primes that intersect S.
In an earlier section we showed these prime ideals are in fact the associates of 0, denoted σ(0). Each associate is the radical of an annihilator, rad([0:x]). Each associate is prime, and when a radical is prime, it is the minimal prime ideal containing [0:x]. Therefore σ(0) is contained in d(R).
Conversely, let Q be a prime in d(R), the smallest prime over an annihilator [0:x]. Remember that 0 is the intersection of primary ideals. Now y drives x into the intersection iff it drives x into each ideal. The annihilator becomes the intersection of [Hi:x] over all the primary ideals. This is contained in Q. Its radical must also be contained in Q. Remember that radical commutes with intersection. The radical of each conductor [Hi:x] is either Pi or R. Thus an intersection of prime ideals is contained in Q. A finite product of prime ideals is contained in Q, and at least one of these prime ideals is contained in Q. Let Q contain P, the radical of [H:x]. Certainly [0:x] is inside [H:x], hence P contains [0:x]. With Q minimal over [0:x], Q has to equal P. In other words, Q is an associate of 0.
If 0 has a primary decomposition, d(R) = σ(0).