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.
Best Answer
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.