So I've been going through Kurt Hensel's old German classic Theorie der algebraischen Zahlen, and I've reached the point where he begins discussing $p$-adic algebraic number fields $\mathbb{Q}(p , \alpha)$ (though Hensel uses $K$ instead of $\mathbb{Q}$ for the rationals), the $p$-adic extension of $\mathbb{Q}(\alpha)$, where $\alpha$ is some algebraic integer.
Hensel proves that just as how there is, up to multiplication by units, a single prime in $\mathbb{Q}_p$, there is only a single prime, up to multiplication by units, in $\mathbb{Q}(p , \alpha)$, though one that, in general, is a divisor of $p$. He even provides an algorithm to find it. Excellent, I figure, let's construct ourselves an example.
I take $\alpha$ to be the real root of the polynomial equation $x^3 – x – 2 = 0$, and set about constructing the $2$-adic extension of $\mathbb{Q}(\alpha)$. I pick the $2$-adic extension, because $(1+\alpha)(-\alpha+\alpha^2)=2$ in $\mathbb{Q}(\alpha)$, which makes this situation interesting. And, indeed, by Hensel's algorithm, I am able to determine that both $\alpha$ and $1+\alpha$ are primes.
Then cometh the problem: they don't appear to differ by a multiplicative factor of a unit! Specifically,
$$(1+\alpha) \cdot \frac{1}{2} (2 + \alpha – \alpha^2) = \alpha$$
and
$$\alpha \cdot \frac{1}{2} (1 + \alpha^2) = (1+\alpha),$$
but neither $\frac{1}{2} (2 + \alpha – \alpha^2)$ nor $\frac{1}{2} (1 + \alpha^2) = (1+\alpha)$ appear to be integral, something which itself goes against what Hensel has written earlier, that in the $p$-adic extension of a number field, every element is either integral, or its inverse is integral!
At this point I've been going through my calculations a number of times, and I just find myself unable to see the error, if this comes down to computation. Perhaps the problem is in my understanding of the theory. Anyway, I'm at my wit's end, and could much appreciate someone more well-wandered in algebraic number theory have a look at this and tell me where I'm going wrong.
Look forward to what you have to offer!
Best Answer
I haven’t read Hensel’s book so I’m not entirely sure what’s the content that you are referring to. It however is very reasonable to require that the polynomial be irreducible (rather than “indecomposable”, that’s the modern word, unless Hensel means something different?) over $\mathbb{Q}_2$ to perform any operations in the extension generated by one of its roots.
So here, we want to show that $P(x)=x^3-x-2$ actually has a root over $\mathbb{Q}_2$.
For this, we use a result which is called Hensel’s lemma. The coefficients of $P$ are in $\mathbb{Z}_2$, that $P(0) \in 2\mathbb{Z}_2$ but $P’(0) \notin 2\mathbb{Z}_2$. Hensel’s lemma states, under these assumptions (more general ones, in fact), that $P$ has a unique root in $\mathbb{Z}_2$ congruent to $0$ mod $2\mathbb{Z}_2$.
How does the proof go? The idea is to construct a Cauchy sequence $z_n \in \mathbb{Z}_2$ congruent to $0$ mod $2$ such that $P(z_n) \rightarrow 0$.
If $u$ is congruent to $0$ mod $2$, then you can check $P’(u)\notin 2\mathbb{Z}_2$, and we know that $P(u+h)\in h^2 \mathbb{Z}_2 + P(u)+hP’(u)$. So take $z_{n+1}=z_n-P(z_n)/P’(z_n)$: if $P(z_n)\in 2\mathbb{Z}_2$, then you can see that $P(z_{n+1})$ has greater $2$-adic valuation than $P(z_n)$ and $z_n$ and $z_{n+1}$ are congruent mod $2$. We start with $z_0=0$, and that ends the proof.