Prime elements and $p$-adic completions In my efforts to translate Hensel's Theorie der algebraischen Zahlen (which first introduced the world to $p$-adic numbers), I've come across a statement of his which I struggle to make sense of, since the only way I can conceivably make sense of it has it say something which I feel must be false, and which if true is something that I just cannot see how it follows from what has been mentioned and proven up till that point.
In an effort to get to the bottom of it, let me ask the community this question:
Let's say that $\mathbb{Q}(\alpha)$ is some algebraic number field, and let $K_p$ be the $p$-adic completion thereof for an arbitrary real prime $p$ (apologies if the notation is non-standard). Is it then always a given that if $\beta \in \mathbb{Q}(\alpha)$ is a prime element of $\mathbb{Q}(\alpha)$, it is also a prime element of $K_p$?
If so, why, and if not, please provide me with a good counter-example.
In addition, does there exist any element $\gamma \in \mathbb{Q}(\alpha)$ which is a unit in $K_p$ but which is not a unit in $\mathbb{Q}(\alpha)$?
As always, I look forward to what you have to teach me!
EDIT: To give a more complete reference in the hope that it helps by providing context, see p. 146 of Theorie der algebraischen Zahlen, under p. 146, under §7: Die konjugierten Körper und die konjugierten Entwicklungen für den Bereich von p. Specifically:

Eine ganze Zahl $\pi_1$ in dem Körper $K(\alpha_1)$ is ja dadurch als Primzahl charakterisiert, daß ihre Norm
$$n (\pi) = \pi_1 \pi_2 \dots \pi_{\lambda} = p^f E$$
keine Einheit und von möglicht niedriger Ordnung in $p$ ist. Ebenso ist eine Einheit $\varepsilon_1$ von $K(\alpha_1)$ charakterisiert, daß ihre Norm eine Einheit modulo $p$ ist.

As I write later down in answer to comments by Torsten Schoeneberg, in as far as I could tell, Hensel spends Sec. 6.4 explaining why the notion that the norm should be a unit modulo $p$ characterizes all units in $K(p, \alpha)$, and he spends Sec. 6.5 explaining why the notion that the norm should not be a unit and should be of lowest possible positive order in $p$ characterizises all primes in $K(p,\alpha)$. Then all of a sudden, at the beginning of Sec. 6.7, he now says that the properties which I thought only applied to units and primes in $K(p, \alpha)$ also applies to units and primes in $K(\alpha)$, from which all I could assume was "Well, $K(\alpha)$ is a subdomain of $K(p,\alpha)$, so I would assume that what he's saying is equivalent to saying that all primes of $K(\alpha)$ are primes of $K(p, \alpha)$, and all units of $K(\alpha)$ are units of $K(p, \alpha)$."
 A: As Bourbaki noted in the historical notes of its volume on commutative algebra, Hensel struggled to write a clear account of his work, making various missteps, in part because the conceptual language of abstract algebra and general topology was unavailable at that time. He confused himself about $e^p = \sum p^n/n!$ as a real number and as a $p$-adic number, since he could not fully appreciate at first the meaning of two different topologies on the same field $\mathbf Q$ (or more accurately, that the notion of an infinite series depends on the notion of convergence).
Hensel also mixed up inequalities involving valuations since small $p$-adic numbers have large $p$-adic valuations; this is something that initially confuses many students even today.
When Hensel refers to the completion of a number field $\mathbf Q(\alpha)$ at a "real prime" $p$ (meaning an ordinary prime number like $2, 3, 5, \ldots$), he is making a bad mistake, because the $p$-adic completions of a number field are related to the prime ideals in it lying over $p$, and completions at different prime ideals can be genuinely different fields! They don't even need to have the same degree over $\mathbf Q_p$.
Example.  Consider $\mathbf Q(\sqrt[3]{2})$ and the prime $p = 5$. What is the $5$-adic completion of this field?  Intuitively, we're asking what $\mathbf Q_5(\sqrt[3]{2})$ means.  Well, that notation $\sqrt[3]{2}$ in the $5$-adics is ambiguous, because $x^3 - 2$ is reducible over $\mathbf Q_5$:
$$
x^3 - 2 = (x + 2 + 4\cdot 5 + 2 \cdot 5^2 + \cdots)g(x)
$$
where $g(x)$ is quadratic irreducible.  If $\sqrt[3]{2}$ is the cube root of $2$ in $\mathbf Q_5$ then $\mathbf Q_5(\sqrt[3]{2}) = \mathbf Q_5$. If $\sqrt[3]{2}$ is a cube root of $2$ not in $\mathbf Q_5$, then $\mathbf Q_5(\sqrt[3]{2})$ is a quadratic extension of $\mathbf Q_5$.  This is the same kind of ambiguity as we see in the notation $\mathbf R(\sqrt[3]{2})$, whose meaning changes if $\sqrt[3]{2}$ is a real or non-real cube root of $2$. (In contrast, there is no ambiguity in the meaning of $\mathbf  Q(\sqrt[3]{2})$ as an abstract field since $x^3 - 2$ is irreducible over $\mathbf Q$.) So $\mathbf Q(\sqrt[3]{2})$ has not one, but two, $5$-adic completions!
This ambiguity is related to what happens if we try to extend $|\cdot|_5$ from $\mathbf Q$ to $\mathbf Q(\sqrt[3]{2})$. These extension are related to the prime ideals in the factorization of $5$. Since
$x^3 - 2 \equiv (x+2)(x^2+3x+4) \bmod 5$, we have
$$
(5) = \mathfrak p \mathfrak q
$$
where $\mathfrak p = (5,\sqrt[3]{2}+2)$ and $\mathfrak q = (5,\sqrt[3]{4} + 3\sqrt[3]{2}+4)$.  Since
$f(\mathfrak p|5) = 1$ and $f(\mathfrak q|5) = 2$, the completions $\mathbf Q(\sqrt[3]{2})_{\mathfrak p}$ and $\mathbf Q(\sqrt[3]{2})_{\mathfrak q}$ have degree $1$ and $2$ over $\mathbf Q_5$, which matches what we found above about the two ways to interpret $\mathbf Q_5(\sqrt[3]{2})$: one way is $\mathbf Q_5$ (in which $|\sqrt[3]{2} + 2| < 1$) and one way is a quadratic extension of $\mathbf Q_5$ (in which $|\sqrt[3]{2} - n| = 1$ for all integers $n$).
You asked if an element $\beta$ of $\mathbf Q(\alpha)$ that is prime in $\mathbf Q(\alpha)$ is also prime in the $p$-adic completion. Because the notion of a $p$-adic completion is ambiguous, as shown in the above example, your question is ambiguous. You need to be more careful about describing which $p$-adic completion you have in mind. Here is a counterexample to show what can go wrong.
Example. There are elements $\beta$ of $\mathbf Q(\sqrt[3]{2})$ such that $|\beta|_\mathfrak p < 1$ and $|\beta|_\mathfrak q = 1$. For instance, let $\beta = \sqrt[3]{2} + 2$.  Then $\beta \equiv 0 \bmod \mathfrak p$, $\beta \not\equiv 0 \bmod \mathfrak q$,
and $\beta$ has minimal polynomial $x^3 - 6x^2 + 12x - 10$ over $\mathbf Q$. The constant term is divisible by $5$ once, so $\beta$ is a $\mathfrak p$-adic prime and a $\mathfrak q$-adic unit: $|\beta|_\mathfrak p = 1/5$ and $|\beta|_\mathfrak q = 1$.
So depending on what you want the notation $\mathbf Q_5(\sqrt[3]{2})$ to mean, $\beta$ might be prime in this field (if it denotes the $\mathfrak p$-adic completion) or it might be a unit in this field (if it denotes the $\mathfrak q$-adic completion).
There is also a sense in which the answer to your question is yes.  For a prime ideal $\mathfrak p$ in the integers of a number field, an element of the localization of the number field at $\mathfrak p$ is prime in that localization if and only if it is prime in the $\mathfrak p$-adic completion, because the meaning of being prime in the localization and the completion can be described in the same way: its $\mathfrak p$-adic valuation is $1$.
Remark. You indicate in a comment that the source of your interest in reading Hensel is to understand the source of the term "divisor" in algebraic geometry, which goes back to commutative algebra, which goes back to number theory. This terminology did not come from Hensel, but from Kronecker. On an algebraic curve (over $\mathbf C$), a divisor is a formal sum of finitely points with integer coefficients. This includes the principal divisors of a nonzero function by using as coefficients of points the order of vanishing of the function at all points (which are always integers and mostly $0$). Divisors on a curve are an additive analogue of the prime ideal factorization of a fractional ideal $\mathfrak a$ in a number field:
$$
\mathfrak a = \prod_\mathfrak p \mathfrak p^{e_\mathfrak p}
$$
where each $e_\mathfrak p$ is an integer and all but finitely many $e_\mathfrak p$ are $0$.  Taking $\mathfrak a = (\alpha)$ to be a principal fractional ideal is like taking a divisor to be a principal divisor. Multiplying fractional ideals is adding divisors.
The notion of an ideal in a ring comes from Dedekind, who was making precise Kummer's more vague notion of an "ideal number": a principal fractional ideal is associated to an actual number (a generator of it), while a nonprincipal fractional ideal becomes principal in a larger number field but you can't see that generator in the original number field, so the original fractional ideal is not an actual number but only an "ideal number". Similarly, a divisor for Kronecker was "something that could be a common divisor" (so divisor = factor) even if it didn't come from an actual element.  This is the distinction between principal divisors and nonprincipal divisors.
There is a comment by Leo Alfonso here about why Kronecker introduced "divisors":

It is used for the same purpose as an ideal: a kind of (substitute of) a common divisor of several elements on a ring. You are not really interested on what a divisor is but on what you can do with it. The trouble is the difficulty in checking the accuracy of your arguments.

If your actual goal is to understand where the notion of divisor first arose, then reading Hensel is a big diversion that will not lead to an answer.
A: [An earlier version of this answer was based on misunderstanding the OP. This is a rewritten version.]
As KCd writes in his very good answer, Hensel made mistakes. But, as far as I can tell, he did not make the "bad mistake" he is accused of there. In fact, starting on page 130, Hensel assumes that
($\ast$) $\alpha$ is an algebraic element over $\mathbb Q$ such that its minimal polynomial stays irreducible over $\mathbb Q_p$.
This entails that in the integer ring of $\mathbb Q(\alpha)$ there is a unique prime $\mathfrak p$ over $p$, and the notion of "the $p$-adic completion of $\mathbb Q(\alpha)$" makes perfect sense! (In modern language, we might preferably call it the "$\mathfrak p$-adic completion", but under assumption ($\ast$) these are canonically isomorphic.)
Now Hensel indeed embeds $\mathbb Q(\alpha)$ (which he calls $K(\alpha)$) into this $p$- (or $\mathfrak p$-)adic completion which he calls $K(p, \alpha)$. This, by the way, might be a better notation than the tempting modern $\mathbb Q_p(\alpha)$, because for that, one would already have to choose an embedding of $\mathbb Q(\alpha)$ into some, say, algebraic closure of $\mathbb Q_p$.
Anyway, with assumption ($\ast$), Hensel spends paragraphs 4--6 of chapter 6 (pages 133-146) proving all kinds of neat structure of that completion $K(p, \alpha)$, first of all that it is an extension of $\mathbb Q_p$ of degree $\lambda := [\mathbb Q_p(\alpha) : \mathbb Q]$. He makes strong use of the norm map $n$ from $K(p, \alpha)$ to $\mathbb Q_p$ to show that it contains a unique complete DVR whose fraction field it is: namely, its integer ring, whose unique maximal ideal $\mathfrak p$ satisfies $\mathfrak p ^e = (p)$ for a certain number $e$, or in other words, all of whose prime elements $\pi$ satisfy $\pi^e= p \cdot \epsilon$ for some unit $\epsilon$ of that DVR. He also notes that $n(\pi)$ has some $p$-adic order $f$, and that $ef = \lambda$. All this is standard material today, which Hensel discusses, not in the most streamlined fashion but, as far as I can tell, without serious mistakes.
But now comes paragraph 7, where Hensel goes back to the un-completed, original number field $\mathbb Q(\alpha)$, and I finally understand OP. Indeed, Hensel is sloppy in wording and notation there, but I think once one fixes his bad wording, he still makes no mistake.
Hensel's bad wording is precisely what you put your finger on: He talks about elements of $\mathbb Q(\alpha)$ being primes / units ("Primzahlen" / "Einheiten"), but what he means is "primes / units when viewed as elements in (the integer ring of) the completion $K(p, \alpha)$". He absolutely must mean that, and not primes / units in the integer ring of that number field, for otherwise obviously all he states is nonsense.
He is also a bit sloppy in identifying the norm map from $\mathbb Q(\alpha)$ to $\mathbb Q$ with the one from $\mathbb K(p, \alpha)$ to $\mathbb Q_p$ -- of course the former is the restriction of the latter under the canonical embedding, but a modern text would point this out and prove it. He also uses notation like $p^{\frac{1}{e}}$ which we all should be very careful with. Finally, the way he talks about things it seems like he imagines "everything contained in some superfield". This, finally, is close to his notorious errors pointed out in KCd's answer, but it does not cause any mistakes here, as far as I can tell.
In fact, what he seems to say, in modern language could be phrased as:
Let $\sigma_1, ... \sigma_\lambda$ be all embeddings of $\mathbb Q(\alpha)$ into (say) an algebraic closure of $\mathbb Q$. (That is, if $\alpha_1:=\alpha, \alpha_2, ..., \alpha_\lambda$ are the roots of $\alpha$'s minimal polynomial, we can assume $\sigma_i(\alpha)=\alpha_i$.)
Then each $\sigma_i$ extends canonically to an isomorphism of $K(p, \alpha)$ to $K(p, \alpha_i)$, which further is isometric, as it sends units to units and primes to primes.
(In modern language this seems so obvious I guess it is rarely proved, but it is pretty important I guess, and Hensel's proof idea, using that $\sigma$-conjugates have the same norm, is neat.)
He uses this to prove (p. 151) that

If the discriminant of $\mathbb Q(\alpha)$ is not divisible by $p$, then $\mathfrak p =(p)$.

which is one of the most important results about ramification in number fields (although still only true in the situation of assumption $(\ast)$.)
A: Unfortunately, the format of Math Stackexchange, at least in as far as I have been able to divine, does not allow you to post a longer reply to your question in response to feedback given by other users except as mere character-limited comments if you don't wish to go for the "answer your own question"-option. So let me make clear that I don't consider this an actual answer to my question but merely my reply to feedback given by KCd and Torsen Schoeneberg, and I am grateful to both of them for taking the time to look deeper into this conundrum of mine.
Pertaining to KCd's comments: Learning that Hensel apparently made such serious mistakes gave me a lot of thought, and I still have yet to properly digest most of what you've said regarding the first half of your comment. Pertaining to the second half, I have been fortunate enough to read up on divisors such as they feature in Kummer's and Kronecker's works. Such as I can understand their works, given a domain, they partition their elements into units and non-units, and when it pertains to the non-units, well, you have the prime divisors, and the divisors are made up from prime divisors and units as products. From what I've been able to tell, neither Kummer nor Kronecker ever give a strict, clear definition of what a divisor on its own actually is, and looking into the works of Harold M. Edwards, who devoted a lot of time into tracing out a history of divisor theory in the "Kummerian" sense, he claims that neither Kummer nor Kronecker ever gave one. And certainly I've never come across a place where Hensel gives one either.
I have in the course of my reading from Weil and Zariski gotten all the way back to Kummer, and from thence made it to Kronecker, and from thence to Hensel. In this project, papers and books on the history of algebra written by Harold Edwards have been very helpful. So I know the very most origins of the term. Nevertheless, those origins haven't been sufficient to give me the "operational" meaning of the term divisor in this particular context, and from what I have been reading, the point of interest is at Hensel's work. Nevertheless, you have supplied me with plently of helpful information, and for that I am deeply grateful.
Pertaining to Torsten Schoeneberg's comments: If he doesn't claim that, then as confused as I may be, I am still happy you are pointing that out to me, since I was thoroughly convinced that that was precisely what he claimed. In as far as I could tell, he spends Sec. 6.4 explaining why the notion that the norm should be a unit modulo $p$ characterizes all units in $K(p, \alpha)$, and he spend Sec. 6.5 explaining why the notion that the norm should not be a unit and should be of lowest possible positive order in $p$ characterizises all primes in $K(p,\alpha)$. Then all of a sudden, at the beginning of Sec. 6.7, he says that the properties which I thought only applied to units and primes in $K(p, \alpha)$ also applies to units and primes in $K(\alpha)$, from which all I could assume was "Well, $K(\alpha)$ is a subdomain of $K(p,\alpha)$, so I would assume that what he's saying is equivalent to saying that all primes of $K(\alpha)$ are primes of $K(p, \alpha)$, and all units of $K(\alpha)$ are units of $K(p, \alpha)$."
I hope that helps explains my confusion. I look forward to your next replies. :)
