I've started reading Matsumura's Commutative Algebra book. It's great. Very difficult but a good book.
A prime ideal is when $ab \in P$ implies $a \in P$ or $b \in P$. A multiplicative set is sort of the opposite, if $a,b \in S$ then $ab \in S$.
Theorem Every ideal is contained in a maximal ideal.
proof: Take the set of all the ideals and apply zorns lemma.
Theorem Every ideal disjoint from a multiplicative set is contained in a maximal ideal disjoint from that multiplicative set.
proof: similar.
Now the ideal of the previous theorem must be prime too,
proof: Suppose $x,y \not\in P$ then $P + x R$ meets $S$ (otherwise we'd have a larger than maximal ideal) so does $P + y R$, so their product does. The product is contained in $P + xy R$ hence $xy \not\in P$ and $P$ is prime.
Corollary Every ideal disjoint from a mult. set is contained in a prime ideal also disjoint from that set.
The radical of an ideal is the elements for which some power lies in the ideal: $$\sqrt{I} = \{ x \in R | \exists n > 0, x^n \in I \}.$$ The nilradical is the nilpotent elements $\sqrt{(0)}$.
Theorem $\sqrt{I} = \bigcap_{P \supset I} P$.
proof: If $P \supset I$ then $P \supset \sqrt I$ (simple exercise). So we have containment in the equality we want to prove. Now for equality suppose $x \not\in \sqrt{I}$ then there is a prime a ideal containing $I$ and not $x$ (use the multiplicative set of powers of $x$).. so $x$ is not in the intersection.
Definition A local ring is a ring with a unique maximal ideal.
Theorem A ring whose non-units form an ideal is local.
proof: (That the ideal is maximal) Suppose $I = \{ x \in R \setminus R^\times \}$ is an ideal then $R/I$ is a field since its elements may be written $u + I$ with $u \in R^\times$.
(That it is the only maximal ideal) Let $J$ be an ideal and suppose for contradiction that there is an element $j \in J \setminus I$, then $j$ is invertible so $J = (1)$.
I'm working on a hard problem from the book:
(3.3) Let $R$ be a local ring with a principal maximal ideal $\mathfrak m$ and the condition $$\bigcap_{r > 0} m^r = (0).$$ Show that $R$ is Noetherian and every idea is a power of $m$.
I'm sure it will require Nakayamas lemma but I haven't figured out how to apply that yet.
I did an easier exercise: If $M,N$ are submodules of some module and $M + N, M \cap N$ are f.g. then $M$ and $N$ are f.g.
proof: I came up with a lemma $(M+N)/M \simeq N/(M \cap N)$ from which you easily deduce the theorem using that quotients of a f.g. module are f.g., then break out the quotient. The lemma itself was easy just produce the homomorphism and check it.
No it doens't use NAK at all! Let $m = (\pi)$.
Step 1 finiteness: Let $I \not = (0)$ then $m^r \supset I \supset m^{r+1}$ since otherwise $I$ would be contained in the intersection of them all - impossible since it's zero.
Step 2 form of the elements: Each $x \in I \subset m^r$ is of the form $k \pi^r$ with $k \not \in m$, since otherwise $x \in m^{r+1}$ - but then $k$ is a unit. So $I = (\pi^r)$.
Great notes on nakayamas lemma, really easy to understand <http://www.maths.usyd.edu.au/u/de/AGR/CommutativeAlgebra/pp327-344.pdf>
Saw a nice web page about posts problem: <https://davidlazar.github.io/PCPL/>