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 inclusionpreserving 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 zerodivisor. 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!
MA3G6 Commutative Algebra
MA3G6 Commutative Algebra
Proposition 6.4 proof  more detail
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