About a proof of the Dedekind-Kummer Theorem

I'm following the lecture notes from MIT's 18.785, and I'm having trouble understanding the proof of the Dedekind-Kummer theorem. (Theorem 6.14 here)

Theorem 6.14 Assume $$AKLB$$ with $$L = K(\alpha)$$ and $$\alpha \in B$$. Let $$f \in A[x]$$ be the minimal polynomial of $$\alpha$$, let $$\mathfrak{p}$$ be a prime of $$A$$, and let $$\overline{f} = \overline{g}_1^{e_1} \cdots \overline{g}_r^{e_r}$$ be its factorization into monic irreducibles in $$(A/\mathfrak{p})[x]$$. Let $$\mathfrak{q}_i := (\mathfrak{p}, g_i(\alpha))$$, where $$g_i \in A[x]$$ is any lift of $$\overline{g}_i$$ in $$(A/\mathfrak{p})[x]$$ under the reduction map $$A[x] \to (A/\mathfrak{p})[x]$$. If $$B = A[\alpha]$$ then $$\mathfrak{p}B = \mathfrak{q}_1^{e_1}\cdots\mathfrak{q}_r^{e_r},$$ is the prime factorization of $$\mathfrak{p}B$$ in $$B$$ and the residue field degree of $$\mathfrak{q}_i$$ is $$\deg \overline{g}_i$$.

In the proof, the author first shows that $$\frak{q}_i$$ are prime ideals by quotienting, and then shows that $$\prod_i \mathfrak{q}_i^{e_i}$$ is divisible by $$\mathfrak{p}B$$. I am okay with these. The part I don't understand is the one following these arguments, where he writes

The $$\overline{g}_i(x)$$ are distinct as elements of $$(A/\mathfrak{p})[x]/(f(x)) \cong A[x]/(\mathfrak{p}, f(x)) \cong A[\alpha]/\mathfrak{p}A[\alpha]$$, and it follows that the $$g_i(\alpha)$$ are distinct modulo $$\mathfrak{p}B$$. Therefore, the prime ideals $$\mathfrak{q}_i$$ are distinct ...

I think I understand the first sentence, because if $$\overline{g}_i(x)$$ and $$\overline{g}_j(x)$$ have the same image in $$(A/\mathfrak{p})[x]/(f(x))$$ for some $$i \ne j$$, then $$\overline{f}$$ must divide $$\overline{g}_i(x) - \overline{g}_j(x)$$ inside $$(A/\mathfrak{p})[x]$$. Since $$\overline{g}_i(x)$$ and $$\overline{g}_j(x)$$ are distinct elements of $$(A/\mathfrak{p})[x]$$, it follows that one of them must have degree greater than or equal to $$\overline{f}$$. This contradicts that fact that they are both divisors of $$\overline{f}$$.

However, I don't understand why the fact that $$g_i(\alpha)$$ are distinct modulo $$\mathfrak{p}B$$ implies that the prime ideals $$\mathfrak{q}_i = (\mathfrak{p}, g_i(\alpha))$$ are distinct. Can someone explain the deduction here? It'll be great if you can check the validity of my argument above as well. Thank you so much.

The class of $$g_i(\alpha)$$ is the image of $$\overline g_i(x)$$ under the isomorphism $$(A/\mathfrak{p})[x]/(f(x)) \cong A[x]/(\mathfrak{p}, f(x)) \cong A[\alpha]/\mathfrak{p}A[\alpha]$$. Since the $$\overline g_i(x)$$ are distinct, so are the $$g_i(\alpha)\mod {\mathfrak p} B$$.
To see why the ideals $${\mathfrak q_i}$$ must be distinct note that it suffices to show they are distinct mod $${\frak p}B$$. Under the isomorphism $$B/{\frak p}B\cong A[x]/(\mathfrak{p}, f(x))\cong (A/\mathfrak{p})[x]/(f(x))$$ the ideal $${\frak q}_i$$ corresponds to the ideal $$(\overline g_i(x))$$ mod $$(f(x))$$ which in turn corresponds to the ideal $$(\overline g_i(x))$$ in $$(A/{\frak p})[x]$$. By assumption these ideals are distinct, thus so are the $${\frak q}_i$$.
I would guess when they wrote that "the $$g_i(\alpha)$$ are distinct mod $${\frak p}B$$" they meant the ideals $$(g_i(\alpha))$$, since just the fact that the elements are distinct does not show that the ideals are distinct.
• Thank you! Just one more question: Is it the fact that $\overline{g}_i(x)$ are distinct irreducible polynomials in PID $(A/\mathfrak{p})[x]$ the reason why the ideals generated by them are all distinct? Or is there an easier way to see this connection? Commented May 23, 2023 at 19:45
• @KyawShinThant In a integral domain $R$ two elements $a,b$ generate the same ideal iff $a=ub$ for some unit $u$. In this case if $\overline g_i(x)=u\overline g_j(x)$ for some unit $u$ in the ring $(A/{\frak p})[x]$, we must have $u=1$ (and so $\overline g_i(x)=\overline g_j(x)$) by looking at the highest degree term (both polynomials are monic) Commented May 23, 2023 at 19:54