Decomposition of rational primes, why work over ring of algebraic integers?

Question

The proof of Thm. 6.1 in this article https://people.reed.edu/~jerry/361/lectures/iqclassno.pdf actually proves the following generalization of Thm. 6.1 beyond quadratic number fields (copied from the last half of pg. 13):

For a field $F$ over $\mathbb Q$ with ring of algebraic integers $\mathscr O_F$ equal to $\mathbb Z[r]$, for $r\in F$ (an algebraic integer) with monic minimal polynomial $f(x)= \operatorname{Irr}(r,\mathbb Q) \in \mathbb Z[x]$, then given a rational prime $p\in \mathbb Z$, the factorization of $\bar f(x)$ in $\mathbb F_p[x]$ dictates the decomposition of the ideal $p\mathscr O_F$: $$\bar f(x) = \prod_{i=1}^g \bar \varphi_i(x)^{e_i} \implies p \mathscr O_F = \prod_{i=1}^g \mathfrak p_i^{e_i} \text{ where } \mathfrak p_i:=(\varphi_i(r), p)^{e_i}, \text{N}(\mathfrak p_i) = p^{\deg(\bar \varphi_i)},$$ where moreover $\sum_{i=1}^g e_i \deg(\bar \varphi_i) = \deg(\bar f)= \deg (f) = [F:\mathbb Q]$. (In the case that $F$ is quadratic, then the factorization of $f$ in $\mathbb F_p[x]$ is completely determined by whether or not the discriminant of $f$ is a (non-zero) square modulo $p$.)

However, I don't see where in the proof the fact that $\mathbb Z[r]$ was a ring of algebraic integers in the field was used. It also seems that the only place $f$ being monic was used was in the last line of the theorem: $\deg(\bar f) = \deg(f)$ (while of course $f\in \mathbb Z[x]$ was very important for the whole proof).

So, I am wondering if the following is true:

For a field $F$ over $\mathbb Q$, and $r\in F$ with minimal polynomial $f(x)= \operatorname{Irr}(r,\mathbb Q)$ normalized (via scalar multiplication) to be in $\mathbb Z[x]$, then given a rational prime $p\in \mathbb Z$, the factorization of $\bar f(x)$ in $\mathbb F_p[x]$ dictates the decomposition of the ideal $p\mathbb Z[r]$: $$\bar f(x) = \prod_{i=1}^g \bar \varphi_i(x)^{e_i} \implies p \mathbb Z[r] = \prod_{i=1}^g \mathfrak p_i^{e_i} \text{ where } \mathfrak p_i:=(\varphi_i(r), p)^{e_i}, \text{N}(\mathfrak p_i) = p^{\deg(\bar \varphi_i)},$$ where moreover $\sum_{i=1}^g e_i \deg(\bar \varphi_i) = \deg(\bar f)$, which is equal to $\deg(f)=[F:\mathbb Q]$ in the case that the leading (integer) coefficient of $f$ is not divisible by $p$. (In the case that $F$ is quadratic, then the factorization of $f$ in $\mathbb F_p[x]$ is completely determined by whether or not the discriminant of $f$ is a (non-zero) square modulo $p$.)

If so, why is the ring of algebraic integers "stealing all the spotlight"? The first half of pg. 14 of the aforementioned Reed article mentions that in the example case $F = \mathbb Q(\sqrt{5})$, we have $\mathscr O_F \supsetneq \mathbb Z[\sqrt 5]$, and "the reduction $x^2-5=(x-1)^2$ modulo $2$ seems to suggest that $2$ ramifies in $\mathbb Q(\sqrt 5)$, but in fact $2$ is inert because $x^2-5x+5$ is irreducible modulo $2$". This seems to answer my question, but this answer already assumes that $\mathscr O_F$ is the center of attention, since "ramifies/inert in $\mathbb Q(\sqrt 5)$" is defined with respect to $\mathscr O_F$!

So what's stopping me from being happy with a nice result about rational prime decomposition in $\mathbb Z[\sqrt 5]$, and defining ramification/inertness w.r.t. $\mathbb Z[\sqrt 5]$, but instead suggesting that my attention should turn to $\mathbb Z[\frac{1+\sqrt 5}2]$?

Answer 1

That the factorization of $f(x) \bmod p$ tells you the factorization of $p\mathcal O_F$ (for a finite extension $F/\mathbf Q$) when $\mathcal O_F = \mathbf Z[r]$ is true with a weaker condition: $p \nmid [\mathcal O_F:\mathbf Z[r]]$. Note $\mathcal O_F = \mathbf Z[r]$ is equivalent to $[\mathcal O_F:\mathbf Z[r]] = 1$, so $p \nmid [\mathcal O_F:\mathbf Z[r]]$ for all $p$. When the index is bigger than $1$, there are finitely many $p$ for which you're not guaranteed that the way $f(x) \bmod p$ factors tells you how $p\mathcal O_F$ factors.

Example. Let $F = \mathbf Q(\sqrt{5})$, so $\mathcal O_F = \mathbf Z[(1+\sqrt{5})/2]$, which is bigger than $\mathbf Z[\sqrt{5}]$, a subring with index $2$ in $\mathcal O_F$. So for a prime $p \not= 2$, the way $x^2 - 5 \bmod p$ tells you how $p\mathcal O_F$ factors, but you are not guaranteed that this works when $p = 2$.

This can be applied to the primes $3$, $5$, and $7$: $x^2 - 5 \bmod 3$ is irreducible, $x^2 - 5 \equiv x^2 \bmod 5$, and $x^2 - 5 \bmod 7$ is irreducible, so $3\mathcal O_F$ and $7\mathcal O_F$ are irreducble while $5\mathcal O_F = (\sqrt{5}\mathfrak p_F)^2 = \mathfrak p_5^2$.

Although $x^2 - 5 \equiv (x-1)^2 \bmod 2$, this does not mean $2\mathcal O_F = \mathfrak p_2^2$. In fact, $\mathcal O_F = \mathbf Z[(1+\sqrt{5})/2] = \mathbf Z[x]/(x^2 - x - 1)$, so $\mathcal O_F/(2) \cong (\mathbf Z/2\mathbf Z)[x]/(x^2-x-1)$ is a field since $x^2 - x - 1$ is irreducible mod $2$. Hence $2\mathcal O_F$ is a prime ideal (of norm $4$).

(Edit: In the ring $\mathbf Z[\sqrt{5}]$, the ideal $(2)$ also is not $\mathfrak p^2$ for a prime ideal $\mathfrak p$. That can be shown by contradiction. The ideal $\mathfrak q = (2,1+\sqrt{5})$ in $\mathbf Z[\sqrt{5}]$ is prime since it has index $2$, and it contains $2$, so if $(2)=\mathfrak p^2$ for some prime ideal $\mathfrak p$ then $\mathfrak p^2\subset \mathfrak q$, so $\mathfrak p \subset \mathfrak q$. Nonzero prime ideals in $\mathbf Z[\sqrt{5}]$ are maximal, so $\mathfrak p = \mathfrak q$. Thus if $(2)$ is the square of a prime ideal then it must be $\mathfrak q^2 = (2,1+\sqrt{5})^2$. Now check that $(2,1+\sqrt{5})^2 = (4,2+2\sqrt{5})$ and that this ideal does not contain $2$. In fact, $(4,2+2\sqrt{5})$ has index $2$ in the ideal $(2)$. We have $\mathfrak q^2 \subset (2) \subset \mathfrak q \subset \mathbf Z[\sqrt{5}]$ where each containment has index $2$. That this tower of containments of ideals does not lead to a prime ideal factorization of $(2)$ is analogous to a tower of normal subgroups $\{(1)\} \subset A_3 \subset S_3$ not leading to a direct product decomposition of $S_3$: if $N \lhd G$, it is not usually true that $G \cong N \times G/N$ as groups.)

You ask what is wrong with using $\mathbf Z[\sqrt{5}]$. Historically, people like Kummer and Dirichlet did work with the rings $\mathbf Z[\alpha]$ because (i) there was no abstract definition of an algebraic integer and (ii) such rings for specific choices of $\alpha$ seemed like the "natural" rings for various problems. But inside a number field $F$, the full ring of integers $\mathcal O_F$ has unique factorization of ideals while its proper subrings of finite index do not have unique factorization of ideals without some restrictions (like focusing on ideals prime to the conductor). When Dedekind discovered the general concept of an algebraic integer, he found some cubic fields, such as $K = \mathbf Q(\alpha)$ where $\alpha^3-\alpha^2-2\alpha-8=0$, for which $\mathcal O_K$ is not of the form $\mathbf Z[\beta]$. Therefore it was essential to shift the focus of attention away from the concrete rings $\mathbf Z[\beta]$ to consider all algebrac integers in a number field in order to have nice properties like unique factorization of ideals without exceptions.