Final answer:
In a commutative ring R with unity, if the factor ring R/N is finite, an ideal N is maximal if and only if it is prime. This equivalence holds because in a finite ring without zero divisors (finite integral domain), the concepts of a field and a simple ring coincide, and maximal ideals correspond to fields while prime ideals correspond to integral domains.
Step-by-step explanation:
The content provided seems to be coming from physics, but your question actually concerns a concept in abstract algebra, specifically within ring theory. In abstract algebra, an ideal N of a commutative ring R with unity is said to be prime if, whenever ab is in N for some elements a and b of R, then either a is in N or b is in N. Similarly, an ideal is maximal if there is no larger ideal containing it apart from R itself.
The problem statement mentions that the factor ring R/N is finite. To show that in this context maximal implies prime and vice versa, we can start by assuming N is maximal. If N is maximal, R/N would be a simple ring without any proper non-trivial ideals. Because R/N is finite and therefore a finite integral domain (no zero divisors), it becomes a field (every non-zero element has a multiplicative inverse). This fact implies that N must necessarily be prime, since in a field any product that results in zero means one or both of the multiplicands must be zero.