Final answer:
To prove R is a principal ideal domain (PID), we first show that every finitely-generated ideal is principal by using the property that sums of principal ideals are principal. Next, we argue that all ideals in R are finitely generated due to R being a UFD. Consequently, R is indeed a PID because all ideals are principal and finitely generated.
Step-by-step explanation:
To prove that a unique factorization domain (UFD) R, where the sum of two principal ideals is again a principal ideal, is a principal ideal domain (PID), we first consider any finitely-generated ideal I in R. Since I is finitely generated, there exist elements a_1, a_2, ..., a_n ∈ R such that I = ⟪ a_1 ⟫ + ⟪ a_2 ⟫ + ... + ⟪ a_n ⟫. Utilizing the hypothesis that R is a UFD with the sum of two principal ideals being a principal ideal, we can combine these into a single principal ideal through successive applications of this property. Starting with ⟪ a_1 ⟫ + ⟪ a_2 ⟫, we get a principal ideal ⟪ a ⟫. Adding ⟪ a_3 ⟫ to ⟪ a ⟫ maintains the principal ideal structure, and so on until all a_i are included, thus confirming that I is principal.
Next, we must show that every ideal in R is finitely generated. Take any ideal I ∈ R. In the case that I is not finitely generated, there must be elements that cannot be represented as finite combinations of other elements in I, contradicting the definition of R as a UFD where every element can be uniquely factored into a finite product of irreducibles. Hence, every ideal in R must be finitely generated.
From the above arguments, we deduce that R is indeed a principal ideal domain. By definition, a PID is a domain in which every ideal is principal, and we have just established both required properties: that finitely-generated ideals can be expressed as a principal ideal and that all ideals are finitely generated.