I was asked about the proof in the notes of Prop 6.4 (1) which says that if is maximal then is -primary. The proof in the notes is a little brief, so here is a longer version: Since is maximal in and contains , it follows that is maximal in (since there is an inclusion-preserving bijection between ideals of containing and ideals of : see page 1 of the notes, and use the definition of maximal ideal that says that there are no ideals strictly between and ). Hence is also a prime ideal of. On the other hand, is the nilradical of (see page 6, just after Definition 1.19) so is the intersection of all primes of (Proposition 1.10), so is contained in all primes of . But is maximal so is not contained in any primes except itself. So it is the only prime of, which is therefore a local ring. Finally, every element of which is not nilpotent is not in the nilradical,, hence is a unit (since is local with unique maximal ideal), hence is not a zero-divisor. So is primary (by the alternative definition of primary given in Definition 6.1). I hope putting in all this detail has not made this seem harder than it is. Proving this was question 5(e)(iii) in the 2012 exam, but you can not draw any conclusions at all from that fact!