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I wonder whether there is a connection between the decomposition of an irreducible integer (monic?) polynomial $f(x)$ over $\mathbb F_p$ into irreducible factors and the decomposition of $(p)\mathcal O_K$ for $K=\mathbb Q[x]/(f(x))$ into prime ideals. It feels like everybody knows something about it and tacitly supposes it) The problem for me is that $\mathcal O_K$ is in general not generated by a root $\alpha$ of $f(x)$, but is something rather monstrous. Any results, including obvious counterexamples, are welcome!

evgeny
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2 Answers2

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There is a general result about this. You are correct, in that it cannot always be applied, but it is widely applicable. It can be found in Neukirch's "Algebraic Number Theory" as proposition 8.3 (with slightly different notation). Let $\mathcal{O}$ be any Dedekind domain, and let $\mathcal{R}$ be the integral closure of $\mathcal{O}$ in a finite extension $L/\operatorname{Frac}\mathcal{O}$. Let $\theta\in\mathcal{R}$, and let $p\in\mathcal{O}[X]$ be the minimal polynomial of $\theta$. Recall that the conductor $\mathfrak{F}$ of $\mathcal{O}[\theta]$ in $\mathcal{R}$ is defined to be $$ \mathfrak{F} := \{\alpha\in\mathcal{R}\mid \alpha\mathcal{R}\subseteq\mathcal{O}[\theta]\}. $$

Your scenario is precisely the case $\mathcal{O} = \Bbb Z$ and $\mathcal{R} = \mathcal{O}_K$ for a number field $K/\Bbb Q$. The result becomes a bit simpler (you can avoid this description of the conductor), and you may find it stated explicitly in this case (along with a proof) in the excellent notes of Keith Conrad I have linked at the end of the post.

Proposition (Dedekind): Let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}$ which is relatively prime to the conductor $\mathfrak{F}$ of $\mathcal{O}[\theta]$, and let $$ \overline{p}(X) = \overline{p}_1(X)^{e_1}\cdots\overline{p}_r(X)^{e_r} $$ be the factorization of the polynomial $\overline{p}(X) = p(X)\pmod{\mathfrak{p}}$ into irreducibles $\overline{p}_i(X) = p_i(X)\pmod{\mathfrak{p}}$ over the residue class field $\mathcal{O}/\mathfrak{p}$, with all $p_i(X)\in\mathcal{O}[X]$ monic. Then $$ \mathfrak{P}_i = \mathfrak{p}\mathcal{R} + p_i(\theta)\mathcal{R},\qquad i = 1,\dots, r, $$ are the different primes above $\mathfrak{p}$. The inertia degree $f_i$ of $\mathfrak{P}_i$ is the degree of $\overline{p}_i(X)$, and one has $$ \mathfrak{p} = \mathfrak{P}_1^{e_1}\cdots\mathfrak{P}_r^{e_r}. $$

In particular, note that if $\mathcal{R} = \mathcal{O}[\theta]$, then $\mathfrak{F} = \mathcal{R} = (1)$, and so any prime $\mathfrak{p}$ is relatively prime to $\mathfrak{F}$. In your case, you can state the theorem by saying that (when it applies to $(q)$) the primes over $(q)$ are all of the form $(q,p_i(\theta))$, and $(q,p_i(\theta))$ divides $(q)$ with multiplicity equal to the multiplicity of $\overline{p}_i$ in the factorization of $\overline{p}$.

As a (non)example, let $\mathcal{O} = \Bbb{Z}$, and let $\mathcal{R} = \mathcal{O}_{\Bbb Q(i)} = \Bbb Z[i]$. You have $$ (2) = (1 + i)^2, $$ (to see this, you can use the theorem with $\theta = i$, or check that $(1 + i)^2 = (2)$ and that $(1 + i)$ is prime by taking norms). But if you try to apply the theorem with $\theta = 2i$, you can check that $\mathfrak{F} = 2\Bbb Z[i]$. Then $(2)$ is not relatively prime to $\mathfrak{F}$. You find that $p(X) = X^2 + 4$ is the minimal polynomial of $\theta = 2i,$ and $$ \overline{p}(X) = X^2. $$ Then as $X\in\Bbb Z[X]$ is a monic lift of $X\in\Bbb F_2[X]$, you would want $$ \mathfrak{P} = (2)\Bbb{Z}[i] + (2i)\Bbb{Z}[i] = 2\Bbb{Z}[i], $$ to be a prime of multiplicity $2$ over $(2)$ in $\Bbb{Z}[i]$, which would imply that $(2)$ factors in $\Bbb{Z}[i]$ as $$ 2\Bbb{Z}[i] = 4\Bbb{Z}[i], $$ which is clearly not the case. You can find a nice extended discussion of this with more examples here.

Stahl
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  • @user1952009 As an extension of number fields and their rings of integers are examples of Dedekind domains, you may feel free to read the proposition in that light (take $\mathcal{O} = \mathcal{O}_K$ and $\mathcal{R} = \mathcal{O}_L$ for an extension $L/K$ of number fields). You will notice that I have also provided an example of this in the number field case which OP was curious about, and the link at the end includes even more examples of this proposition in that case, even stating the proposition for that case specifically. – Stahl Dec 15 '16 at 20:34
  • And clearly the OP might want a sketch proof (in the simplest cases) – reuns Dec 15 '16 at 20:35
  • @user1952009 The proof can be found in Neukirch in the general setting I've stated, and the proof in the number field situation (for extensions of $\Bbb Q$, at least) is in the link at the end. As OP did not ask for a proof (only results and examples, which I have provided), it would only serve to make a rather long answer even longer. If OP would like a proof, they may ask for it themselves and I could provide it at that point. – Stahl Dec 15 '16 at 20:42
  • If $\overline{p}(X) = \overline{p}_1(X)\overline{p}_2(X)$ I have the feeling that $(\mathfrak{p}) = (\mathfrak{p}, p_1(\theta)),(\mathfrak{p}, p_2(\theta))$ mainly because $gcd(\overline{p}_1(X),\overline{p}_2(X)) = 1$. Is it the good approach ? – reuns Dec 15 '16 at 23:54
  • @user1952009 The main idea is to say that $\mathcal{R}/\mathfrak{p}\mathcal{R}\cong(\mathcal{O}/\mathfrak{p})[X]/(\overline{p}(X))$ when $(\mathfrak{F},\mathfrak{p}) = 1$. Then you factor $\overline{p}$ into relatively prime $\overline{p}_i$'s and apply the Chinese remainder theorem, and finally do some analysis on the rings to conclude the results about actual primes in $\mathcal{R}$ over $\mathfrak{p}$ (find the corresponding primes in the other ring, use the correspondence between primes in a quotient and primes in the original ring, etc.). – Stahl Dec 16 '16 at 00:03
  • Yes it is obvious that $\mathcal{R}/\mathfrak{p}\mathcal{R}\cong(\mathcal{O}/\mathfrak{p})[X]/(\overline{p}(X))$ but how does it come to $\prod_i (\mathfrak{p}\mathcal{R} + p_i(\theta)\mathcal{R}) = \mathfrak{p}\mathcal{R}$ ? (that is the LHS contains $ \mathfrak{p}\mathcal{R}$) – reuns Dec 16 '16 at 16:04
  • Can you take a look at the answer I tried – reuns Dec 17 '16 at 02:15
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Let me try something following the lines given by Stahl

Assume $\mathbb{Z}[\alpha] \simeq \mathbb{Z}[X]/(f(X))$ where $f \in \mathbb{Z}[X]$ is the minimal polynomial of $\alpha$, so we have the isomorphism $$\varphi : \mathbb{Z}[X]/(f(X)) \to \mathbb{Z}[\alpha], \qquad \varphi(g(X)) = g(\alpha)$$

then for a prime number $p $ unramified in $\mathbb{Z}[\alpha]$ : $$A = \mathbb{Z}[\alpha]/(p) \simeq \mathbb{Z}[X]/(f(X))/(p) \simeq \mathbb{F}_p[X]/(f(X)) = B$$ (where the isomorphism in the middle is trivial)

in $B$, you have the product in zero-divisors : $$0 = f(X) = \prod_{j=1}^g f_j(X)$$ which is unique in the sense that when removing a factor the product is not zero anymore, and $B/(f_j)$ is a field

and use the lifted isomorphism $$\overline{\varphi}:B \to A, \qquad \overline{\varphi}(g(X)) = g(\alpha)$$ so that $$0 = \overline{\varphi}^{-1}(0) =\overline{\varphi}^{-1}(\prod_{j=1}^g f_j(X)) = \prod_j f_j(\alpha))$$ and again $A/(f_j(\alpha)) = \mathbb{Z}[\alpha]/(p)/(f_j(\alpha))= \mathbb{Z}[\alpha]/(p,f_j(\alpha))$ is a field,

and in $\mathbb{Z}[\alpha]$ : $$(p) = \prod_{j=1}^g (p,f_j(\alpha))$$

reuns
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