Consider $Q_3$ field which is rational number $Q$ completed at $p=3$.(In other words, $Q_3$ is $3-$adic rational numbers.) Let $f=x^3-17\in Q_3[x]$. Let $O$ be the complete DVR associated to $Q_3$.

Clearly $f\in O[x]$. From Gauss lemma by $O$ PID, I see that if $f=gh,g,h\in Q_3[x]$,then $g,h\in O[x]$. Therefore, I can perform $3-$reduction by considering $\frac{O}{3}[x]$.

Now $\bar{f}\in Z_3[x]$ has $x^3+1=(x+1)^3\in Z_3[x]$. So I cannot apply hensel lemma here to determine factor of $f$ in $Q_3[x]$. The book says it has a degree 2 irreducible factor in $Q_3[x]$.

The other version of Hensel uses $f'(x)=3x^2$. And $x=-1$ yields $3^2\vert f(-1)$ and $3^2\not\vert f'(x)$ but $2=1+1$. The absolute value requires $|f(-1)|<|f'(-1)|^2$ where $|x|$ is the $\frac{1}{3^{v_3(x)}}$ and $v_3(x)$ is $3-$adic valuation of $x\in Q_3$. Both sides yields $\frac{1}{9}$ for absolute value. So I cannot apply this hensel lifting.

$\textbf{Q:}$ Can someone kindly provide hints to methods to determine reducibility for local fields? If I am lucky, I can proceed by hensel lemma but it requires factors after prime reduction being coprime. Have I done something wrong above?

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    $\begingroup$ Perhaps take a look at some of the solutions to this question regarding how to find the root in $\Bbb{Z}_3$. $\endgroup$ – sharding4 Jul 8 '18 at 0:12
  • $\begingroup$ @sharding4 So the trick is to look at the transformed polynomial as one normally plays Eisenstein criterion? $\endgroup$ – user45765 Jul 8 '18 at 0:17
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    $\begingroup$ Yes. I think you may be able to find a slicker approach along those lines. If not, you can slog through it using the method I used in the question referenced above. $x\equiv 2 \pmod 3$ lifts because $(3k+2)^3 \equiv 17 \pmod 9$ is identically true. The calculation just becomes messy and tedious. $\endgroup$ – sharding4 Jul 8 '18 at 0:21
  • $\begingroup$ @sharding4 Thanks for the info. $\endgroup$ – user45765 Jul 8 '18 at 0:24
  • $\begingroup$ No problem. With a little cheat on the computer I found $1/3 \sqrt[3]{17^2}-1/3\sqrt[3]{17}+1/3$ has minimal polynomial $g(x)=x^3-x^2+6x-12$. You can more easily find a root of that polynomial using Hensel's Lemma or some other method of your choice. Then $\sqrt[3]{17}=1/2\alpha^2-1/2\alpha+2$ where $\alpha$ is the root of $g$ that you found. $\endgroup$ – sharding4 Jul 8 '18 at 0:31

Here’s a method of using Strong Hensel, which you seem to know:

The cubic $X^3-17=f(X)$ has a root if and only if $f(X-1)=X^3-3X^2+3X-18$ has a root. Three-adically, the Newton polygon has a vertex at $(1,1)$, which already tells you that there’s a factorization.

If you’re not comfortable with Newton , go one step further, and substitute $3X$ for $X$, to get $f(3X-1)=27X^3-27X^2+9X-18=9\left(3X^3-3X^2+X-2\right)$. Modulo $3$, we get $3X^3-3X^2+X-2\equiv X-2\pmod3$, and this has the nice factorization $(X-2)\cdot1$ into two relatively prime factors. You can lift one of them to characteristic zero, by Strong Hensel, preserving its degree. Of course you choose the $X-1$. And there you have it.

In case you’re nervous about using $1$ as one of the factors, let me reassure you. This freedom comes in handy surprisingly often.

EDIT: Addendum

You asked about the factorization of $f(X)=X^3-17$ modulo $3$ and why that didn’t tell the splitting of $(3)$ in $\Bbb Q(w)$, where $\text{Irr}(w,\Bbb Q[X])=X^3-17=f(X)$. The theorem should read that the splitting of the irreducible modulo $p$ tells the splitting of $(p)$ only when the root $r$ generates the integers of $\Bbb Q(r)$ over $\Bbb Z$. And of course $r=w$ doesn’t do that.

  • $\begingroup$ This might be a dumb follow up question. For local field $Q_3$, the factorization of $x^3-17$ indicates existence of two primes above $(3)$ via extension and one of which is ramified. However, for number field $Q((17)^{1/3})$, the ramification of $(3)$ in number ring of $Q((17)^{1/3})$ shows that there is only 1 prime as $x^3-17$ reduction by $(3)$ becomes $(x+1)^3$ and this indicates prime total ramification. Should I have local prime and global prime correspondence? Or did I make wrong observation here. Thanks. $\endgroup$ – user45765 Jul 8 '18 at 3:17
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    $\begingroup$ You’re probably misapplying the relevant theorem. The factorization, calling $w=\sqrt[3]{17}$, is $(X-w)(X^2+wX+w^2)$, which does reduce, modulo $3$, to $X^3-1$, it’s true; but this factorization also shows that the other prime above $3$ has to do with a primitive cube root of $1$. In fact, over $\Bbb Q$, the splitting of $(3)$ in $\Bbb Q(w)$ is: a prime of degree one (unramified) corresponding to the linear factor, and a ramified prime with $e=2$ for the quadratic factor. That’s the source of the ramification! (Maybe I’ll make an addition to the Answer, when I get time.) $\endgroup$ – Lubin Jul 8 '18 at 14:07
  • $\begingroup$ Ah, Thanks a lot for the clarification. I got a different answer from my friend who used global approach yielding a totally ramified prime and I was checking the ramification from local field approach which I should get one ramified and one unramified prime. That is why I was trying to look at the local field reducibility of $x^3-17$. $\endgroup$ – user45765 Jul 8 '18 at 14:23
  • $\begingroup$ One final comment: an integral basis is $\{1,w,(w^2-w+1)/3\}$, and this basis shows the (global) discriminant to be $3\cdot17^2$. $\endgroup$ – Lubin Jul 8 '18 at 19:49
  • $\begingroup$ Yes. That is given at the end of Marcus Number fields, Chapter 2. That discriminant gives $pq^2$ factorization as well by Chapter 3, exercise 21b). $\endgroup$ – user45765 Jul 8 '18 at 20:38

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