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 $K(p, \alpha)$."
Best Answer
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":
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.