There's a lot of algebra in algebraic number theory. Much more than I knew. For the first time ever I saw an application of the following:
Theorem (Algebra - Chapter 0, pg 142) Let $I$ be an ideal of a ring $R$, let $J$ be an ideal of $R$ containing $I$. Then $J/I$ is an ideal of $R/I$ and $$\frac{R/I}{J/I} \simeq \frac{R}{J}.$$
It was used in showing that the quotient of a specific ring of integers by a prime ideal was $\mathbb Z/(2)$.
What is the relationship between non-principal prime ideals and failures of unique factorization? If we have a failure $AB=CD$ then I would expect $(A,C)$ to be a non-principal prime ideal. Note: The different between primes and irreducibles is very important in non-UFDs.
There are lists of quadratic rings (easy to work with) that aren't UFDs here:
For a concrete example $\mathbb Q(\sqrt{-15})$ has ring of integers $\mathbb Z[\alpha]$ (where $\alpha = \frac{1 + \sqrt{-15}}{2}$ and $I = (2, \alpha)$ is a prime ideal (Note: Neither of $(2)$, $(\alpha)$ are prime ideals).
We can prove that it's a prime ideal by showing that $\mathbb Z[\alpha]/I$ is an integral domain. The minimum polynomial of $\alpha$ is $X^2 - X + 4$ so $$\mathbb Z[\alpha]/I = \frac{\mathbb Z[X]/(X^2 - X + 4)}{(2,X)/(X^2 - X + 4)} \simeq \mathbb Z[X]/(2,X) \simeq \mathbb Z/(2)$$ which is indeed an integral domain.
Examples from <http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/quadraticgrad.pdf>
In $\mathbb Z[\sqrt {-14}]$:
The ideal $(2, \sqrt{-14})$ is not principal because the previous equation is already factored into irreducibles, if it were principal, say $(\beta)$, then we would be able to factor it futher since $2$ and $\sqrt{-14}$ would be multiples of $\beta$.
Another nice application of algebra was the following:
Theorem (Algebra - Chapter 0, pg 124) Let $R$ be a finite commutative ring, then if $R$ is an integral domain it's a field.
this is used in another proof that prime ideals are maximal in a ring of integers.. assuming it can be shown that $\mathcal O_K/\mathfrak q$ is finite.
Intuitively, $\mathfrak q$ is a lattice inside $\mathcal O_K$ - they both have the same dimension - the residue classes fill up the fundamental domain and cover the space by translation. So it should be finite. There is a proof (along completely different lines, using norm rather than det) in the notes <http://www.math.uiuc.edu/~r-ash/Ant/AntChapter4.pdf>.
A notion of determinant of a lattice then would give $|\mathcal O_K/\mathfrak q| = \det \mathfrak q$ and prove the theorem (again, it was shown in a different way in the previous post).
The notes by r-ash say that if $N(I)$ is prime then $I$ is a prime ideal. This could be useful.
A prime ideal in a number field would be finitely generated (every ideal is f.g. in a noetherian ring) - so it will always be expressible as some finite list of generators $(a,b,c)$.
in fact, according to <https://en.wikipedia.org/wiki/Noetherian_ring> every ideal in a number field is generated by at most 2 elements! The proof uses some advanced theory I'm not doing just now.
I'm understanding the subject much better and improving my grounding in algebra. I had to pull back a bit though, I was trying to study the ramification theorem but I really need to rigorously work out the full details of unique factorization of ideals - and log all the algebra bits that come up in it. It must be done in some generality so that the proof applies to the discrete valuation rings that come from "localizing" a ring of integers too.
I should start to compile notes into a PDF or on paper that prove all the results sequentially. There are so many little facts and results that it would take a lot of editing to get it all straight.
The way I see this subject is that we have a bunch of algebra background material - and this should all come first in a very clear general setting. Then the fundamental constructions and results about number fields can be developed mixed in with concrete examples.
So for leading up to the first important theorem my mental map is a bit like this: