Factorizing polynomials in $\mathbf{Z}[[x]]$ - MathOverflow

Factorizing polynomials in $\mathbf{Z}[[x]]$

2011-02-22T09:42:30Z

Let $f(x)\in\mathbf{Z}[x]$ be a non-constant, irreducible polynomial, and let $\alpha \in\mathbf{C}$ be a root of $f(x)$. Denote by $\varphi_\alpha:\mathbf{Z}[x]\rightarrow\mathbf{C}$ the ring homomorphism sending $x$ to $\alpha$. Let now $\nu$ be a $p$-adic place of the number field $K=\mathbf{Q}(\alpha)$ such that $\alpha$ has strictly positive valuation with respect to $\nu$, i.e. such that $\alpha$ belongs to the maximal ideal of the integers of $K$ defined by $\nu$. There are only a finite number of such places $\nu$, they are precisely those occurring with positive exponent in the prime factorization of the principal fractional ideal of $K$ generated by $\alpha$.

Then $\varphi_\alpha$ can be extended uniquely to a continuous map $$\varphi_{\alpha,\nu}:\mathbf{Z}[[x]]\rightarrow K_\nu,$$ where $K_\nu$ denotes the completion of $K$ at the place $\nu$ (the topology considered here on $\mathbf{Z}[[x]]$ is the $x$--adic one). Let $f_\nu(x)\in\mathbf{Z}[[x]]$ be a power series generating the kernel of $\varphi_{\alpha,\nu}$ (such ideal should indeed be principal, right?).

Is it true that the $f_\nu(x)$ can be chosen so that $$f(x)=\prod_{\nu}f_\nu(x),$$ where $\nu$ ranges through the places of $K$ considered above?

EDIT: As suggested below, it would be more correct to ask that $f(x)$ be equal to the product $\prod_{\nu}f_\nu(x)$ only up to units, and for any choices of the $f_\nu(x)$.

Answer by Laurent Moret-Bailly for Factorizing polynomials in $\mathbf{Z}[[x]]$

2011-02-22T11:53:12Z

Assume $f$ is monic with constant term $\pm1$ (e.g. $f(x)=x-1$). Then there are no places $\nu$ as in the question, because every root is an algebraic unit. Therefore the displayed product is $1$.

Answer by Pace Nielsen for Factorizing polynomials in $\mathbf{Z}[[x]]$

2011-02-22T19:43:51Z

Edited due to mistakes pointed out in the comments:

I think the answer to the problem might be yes. Here are some preliminary thoughts.

First, you know $f_{\nu}(x)$ divides $f(x)$, since $f(x)$ is always in the kernel, the kernel is principal, and $\mathbb{Z}[[x]]$ is a UFD.

Second, you know $f_{\nu}(x)$ is irreducible, else you don't map into a domain.

Third, I think you can probably prove that these polynomials (the $f_{\nu}(x)$) are distinct. If so continue:

Fourth, this all says that $\prod_{\nu}f_{\nu}(x)$ divides $f(x)$ in $\mathbb{Z}[[x]]$. So, to get the result you just have to prove that the constant coefficients agree up to a unit (i.e. up to $\pm 1$). This is equivalent to showing that the constant coefficient of $f_{\nu}(x)$ is $p^{f_{P}\nu_{P}(\alpha)}$ where $P$ is the prime (above $p$) associated to $\nu$ and $f_{P}$ is the inertial degree of $P$ in the ring of integers over $K$. (Sorry for the double use of $f$--for the polynomial and for the inertial degree.)

Answer by David Speyer for Factorizing polynomials in $\mathbf{Z}[[x]]$

2011-02-24T19:58:21Z

I assume you want to fix the prime $p$, and I assume you want $f$ to be the minimal polynomial of $\alpha$ over $\mathbb{Q}$. If you don't fix $p$, then $x-6$ is a counter-example (there are two valuations for which $v(6)>0$, but $x-6 \neq (x-6)^2$, even up to units.) If $f$ is not the minimimal polynomial then $(x-2)(x-4)$ is a counterexample.

Subject to these caveats, you are right.

I find it really confusing to multiply over all places as you are doing. What I would rather do is the following: Fix a normal extension $L$ of $\mathbb{Q}$ in which $f$ splits. Let $\alpha_1$, ..., $\alpha_N$ be the roots of $f$ in $L$. Factor $f$ over $\mathbb{Q}_p$ as $g_1 g_2 \cdots g_k$ and group the $\alpha_i$ together if they are roots of the same $g_j$. Fix an extension $v$ of the $p$-adic valuation to $L$. Then there are $k$ different $p$-adic valuations of $\mathbb{Q}(\alpha)$. They each arise as follows: For each $g_i$, choose a root $\alpha_i$ and let $\phi_i : \mathbb{Q}(\alpha) \to L$ be the map induced by $\alpha \mapsto \alpha_i$. Then your valuation is $v \circ \phi_i$.

Suppose that $v(\alpha_i)>0$ for the roots of $g_1$, $g_2$, ..., $g_s$ and $v(\alpha_i) \leq 0$ for $g_{s+1}$, \cdots, $g_r$. So you are interested in the first $s$ valuations.

Here are the lemmas you need; I'll leave them as exercises. I'm pretty sure they work.

Lemma 1: For $i>s$, the polynomial $g_i$ is a unit in $\mathbb{Z}_p[[x]]$

Lemma 2: For $i \leq s$, the polynomial $g_i$ can be written as $h_i u_i$ where $u_i$ is a unit in $\mathbb{Z}_p[[x]]$ and $h_i \in \mathbb{Z}[[x]]$.

Lemma 3: The kernel of $\mathbb{Z}[[x]] \mapsto L_v$, where $x$ goes to a root of $g_i$, is generated by $h_i$.

Proof of Lemma 3: If the map were from $\mathbb{Z}_p[[x]]$, then the kernel would be generated by $g_i$. Since $h_i = g_i u_i^{-1}$, we see that $h_i$ is in the kernel. If $q$ is in the kernel, then $q=g_i v = h_i u_i v$ in $\mathbb{Z}_p[[x]]$. Since $h_i$ and $q$ are both in $\mathbb{Z}[[x]]$, we see that the coefficients of $u_i v$ are in $\mathbb{Q}$, so they are in $\mathbb{Q} \cap \mathbb{Z}_p = \mathbb{Z}$. Thus, $q$ is a multiple of $h_i$ in $\mathbb{Z}[[x]]$.

We take $h_i$ as your representative of the kernel. Then $f$ is $\prod h_i$ times a unit $u$ of $\mathbb{Z}_p[[x]]$. But $f$ and $\prod h_i$ are both in $\mathbb{Z}[[x]]$, so $u$ and $u^{-1}$ are in $\mathbb{Q}[[x]]$ and we deduce that $u$ is a unit of $\mathbb{Z}[[x]]$, as desired.

Remark: Lemma 2 should be thought of as an analogue of the Weierstrass preparation theorem, although it is not the standard $p$-adic analogue. To see this, note that the (formal) Weierstrass preparation theorem is

If $g$ is a power series in $x$ and $y$ then $g=uh$ where $u$ is a unit of $k[[x,y]]$ and $h$ is in $k[[y]][x]$.

The standard $p$-adic analogue is: If $g \in \mathbb{Z}_p[[x]]$ then $g=uh$ with $u$ is a unit of $\mathbb{Z}_p[[x]]$ and $h$ is in $\mathbb{Z}_p[x]$.

Lemma 2 goes in the "other" direction: If $g \in \mathbb{Z}_p[[x]]$ then $g=uh$ with $u$ is a unit of $\mathbb{Z}_p[[x]]$ and $h$ is in $\mathbb{Z}[[x]]$.

Answer by François Brunault for Factorizing polynomials in $\mathbf{Z}[[x]]$

2011-02-25T01:39:23Z

I think the answer to the question is yes. Here is the idea of a proof.

Let us assume that $\mathcal{O}_K=\mathbf{Z}[\alpha]$. I hope this is not too restrictive for you (hopefully, someone can extend the argument).

Put $(\alpha) = \mathfrak{p}_1^{a_1} \cdots \mathfrak{p}_t^{a_t}$ where the $\mathfrak{p}_i$ are distinct maximal ideals of $\mathcal{O}_K$. Let $\nu_i$ be the place of $K$ associated to $\mathfrak{p}_i$. Let $K_i$ be the completion of $K$ with respect to $\mathfrak{p}_i$, and let $\mathcal{O}_i$ be the ring of integers of $K_i$. We have isomorphisms

\begin{equation*} \mathbf{Z}[[X]]/(f) \cong \varprojlim_{n \geq 1} \mathbf{Z}[X]/(X^n,f) \cong \varprojlim_{n \geq 1} \mathcal{O}_K/(\alpha^n) \end{equation*}

\begin{equation*} \cong \varprojlim_{n \geq 1} \prod_{i=1}^t \mathcal{O}_K/\mathfrak{p}_i^{a_i n} \cong \prod_{i=1}^t \mathcal{O}_{i}. \end{equation*} Now, convince yourself that the composition is just $(\varphi_{\alpha,\nu_1},\ldots,\varphi_{\alpha,\nu_t})$, using your notations (this is because $X$ is sent to $\alpha$).

Recall that $\mathbf{Z}[[X]]$ is a UFD. We already know (see my comment to the question) that $\ker(\varphi_{\alpha,\nu_i}) = (f_i)$ with $f_i \in \mathbf{Z}[[X]]$ irreducible. The ideals $(f_i)$ are pairwise distinct (they are the preimages of distinct ideals of $\prod_{i=1}^t \mathcal{O}_{\nu_i}$). So we get $(f)=\cap_{i=1}^t (f_i)=(f_1 \cdots f_t)$.