I am self studying the notes here. The problem is exercise 2.18 on page 9 (solutions provided there as well).
Let R be a ring, p a prime ideal, R[X] the polynomial ring, pR[x] the product ideal and (x) the ideal generated by x. We want to show that pR[x] + (x) is a prime ideal.
In the solutions they state that (pR[x] + (x))/(pR[x]) = (x) ⊂ (R/p)[x]. This is the part I do not understand (I believe I follow the rest of the solution). Presumably this is obvious since there is no additional justification provided...
Define
$$\phi: pR[x]+(x)\to (x)\;,\;\;\phi(pf(x)+xg(x)):=xg(x)$$
Show the above is a surjective ring homomorphism (of rings without unity) and its kernel is $\;pR[x]\;$ , and now apply the first isomorphism theorem