Question about a step in Prop 7.50 in Milne's Algebraic Number Theory

Question

This question talks about the correspondence between finite unramified extensions of $K$, a complete field with a discrete, nonarchimedean absolute value, that are contained in an algebraic extension $L$ of $K$ and finite subextensions of the tower of residue fields $k \subset l$.

So Milne starts by considering a finite subextension $k' \supset k$ of the residue field of $k$ and writes it as $k[a]$ for some $a$. Now he uses Newton's Lemma (which was proved along with Hensel's Lemma) to argue that there should be an $\alpha \in L$ such that $f(\alpha)=0$ and $\alpha \equiv a \mod \mathfrak{p}$.

My issue is this: why should such an $\alpha$ exist in L? Since L is an arbitrary algebraic extension (not necessarily finite), it doesn't have to be complete under the extension of the absolute value and so I'm not sure why Newton's Lemma applies. Also, since L is not necessarily algebraically closed, why should the extension $K[x]/(f)$ embed into L (where f is a lift of the minimal polynomial of the primitive generator k')?

Am I missing something obvious or is there a mistake here?

You can look at the proposition and its proof in page 127 in Milne's notes.

EDIT:

I also took a look at these notes from 18.785 (page 1-2) where they consider $K[X]/(f)$ but they say that by construction it is true that the valuation ring of this is $A[\alpha]$. I don't see why this should be so. Could someone help with this as well?

Answer 1

There is nothing wrong here when $L$ is infinite. You just need to think more carefully by focusing on the actual finite extensions you use to create the lifting of a root. An infinite algebraic extension of $K$ is a union of finite extensions of $K$, and each finite extension of $K$ is complete.

Even if $L/K$ is not finite, each particular element of $L$ generates a finite extension of $K$, and finite extensions of $K$ are complete. So if $f(x) \in A[x]$ and $f(x) \bmod \mathfrak p$ is separable, then if you lift a root of $f(x) \bmod \mathfrak p \in k[x]$ to $A$, it is a number $\gamma \in A$ such that $|f(\gamma)| < 1$ and $|f'(\gamma)| = 1$. The field $K(\gamma)$ in $L$ has to be complete since it is finite over $K$. Apply Hensel's lemma to the complete field $K(\gamma)$ to see that $f(x)$ has a root $\alpha$ in the integers of $K(\gamma)$ such that $f(\alpha) = 0$ and $|\alpha - \gamma| < 1$.

Suppose $f(x)$ has two roots $\alpha$ and $\beta$ in $L$ such that $|\alpha - \gamma| < 1$ and $|\beta - \gamma| < 1$. Even if $L$ is not complete, the field $K(\alpha,\beta)$ inside of $L$ is a finite extension of $K$, and thus is complete. The numbers $\alpha$ and $\beta$ are in the integers of $K(\alpha,\beta)$ and they reduce to the same root of $f(x) \bmod \mathfrak p$ in the residue field of $K(\alpha,\beta)$. By the uniqueness of the lifting of a root in Hensel's lemma, we get $\alpha = \beta$. Thus in $L$ there is only one lifting of a root of $f(x) \bmod \mathfrak p$ in the residue field of $L$.

Answer 2

My issue is this: why should such an $\alpha$ exist in L? Since L is an arbitrary algebraic extension (not necessarily finite), it doesn't have to be complete under the extension of the absolute value and so I'm not sure why Newton's Lemma applies.

I agreed with you that there seems to be something wrong when $L$ is infinite.

However, the argument for existence of unramified $K'\subset L$ whose residue is $k'=k(a)$ can be rewritten following MIT course notes: Pick a lift $\alpha$ of $a\in k'$ to $B$, valuation ring of $L$. Let $g\in A[x]$ be minimal polynomial of $\alpha$ over $L$ then $K(\alpha)\cong K[x]/(g)$ is a separable finite subextension of $L$. Then you proceed to show $K'=K(\alpha)$ is unramified as in the MIT notes.

Also, since L is not necessarily algebraically closed, why should the extension $K[x]/(f)$ embed into L (where f is a lift of the minimal polynomial of the primitive generator k')?

When people write it like this, $K[x]/(f)\subset L$ would mean $K(\alpha)\subset L$ where $\alpha\in L$ with minimal polynomial $f\in K[x]$. I think they do this in this situation because it is safe to do so.

I also took a look at these notes from 18.785 (page 1-2) where they consider $K[X]/(f)$ but they say that by construction it is true that the valuation ring of this is $A[\alpha]$. I don't see why this should be so. Could someone help with this as well?

Once we view $K[x]/(f)$ and $K(\alpha)$ to be the same object $K'$, as discussed above, then its valuation ring is $A[\alpha]$ or $A[x]/(f)$ because it is algebraic extension of $A$ in $K'$, and as it is of degree $n=[K':K]$, so it is precisely algebraic closure of $A$ in $K'$, i.e. valuation ring of $K'=K[x]/(f)$.