Remarks
A special case occurs for d = X.
The condition on R being a field is necessary.
The condition on the degree of d is necessary.
Then
j is a bijection, so the residue class ring is in fact
R itself. In mathematical jargon,
R is identified with
R[X]/(X).
If R = Z/4Z,
then the map R ->
R[X]/(d) is not always injective:
2 + 2(2X + 1) = 4X + 4 = 0.
Take d = 2X + 1. Then the image of 2 equals
If d has degree 0, then d is an invertible element of R
and so, for any a 
R,
we would have
j(a) = a + (d) =
(a/d) · d + (d) = (d) = 0.
R,
we would have
In other words, j would be the zero map (mapping every element onto 0).