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 $f_\nu(x)$.

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    $\begingroup$ Maybe you could define $\varphi_{\alpha,\nu}$ on $\mathbf{Z}_p[[X]]$ instead ? Then you could try to use the Weierstrass preparation theorem. I don't know though if $\ker \varphi_{\alpha,\nu}$ is principal in your setting. $\endgroup$ Commented Feb 22, 2011 at 12:29
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    $\begingroup$ The ideal $\ker \varphi_{\alpha,\nu}$ is indeed principal. The proof is as follows : $\mathbf{Z}[[X]]$ is a Noetherian unique factorization domain, so every prime ideal of height $1$ is principal. But $\mathbf{Z}[[X]]$ has Krull dimension $2$ and the image of $\varphi_{\alpha,\nu}$ is not a field, so $\ker \varphi_{\alpha,\nu}$ is principal. $\endgroup$ Commented Feb 24, 2011 at 10:14
  • $\begingroup$ Nice, what are your feelings about the question then? Thanks. $\endgroup$ Commented Feb 24, 2011 at 10:56
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    $\begingroup$ @Tommaso : my feeling is that the equality you mention seems plausible. What you're asking seems to be a generalization of the isomorphism $\mathbf{Z}[[X]]/(X-p) \cong \mathbf{Z}_p$, but I don't see the proof in general... $\endgroup$ Commented Feb 24, 2011 at 12:09

4 Answers 4


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 $(f)=\cap_{i=1}^t (f_i)=(f_1 \cdots f_t)$.

  • $\begingroup$ This is precisely the way I was trying to prove the statement yesterday, as I was talking to a friend. In the chain of isomorphisms you describe above, probably it is not so essential that $\mathbf{Z}[\alpha]=\mathcal{O}_K$, I will check if the same argument works when $\mathbf{Z}[\alpha]$ is not even finite over $\mathbf{Z}$, i.e. when $\alpha$ is not an algebraic integer. Thanks!! $\endgroup$ Commented Feb 25, 2011 at 10:36

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 $1$.

  • $\begingroup$ Good point. However in this special case f(x) is a unit in $\mathbf{Z}[[x]]$ and, up to units, the empty product, being equal to 1, is also a unit. $\endgroup$ Commented Feb 22, 2011 at 14:19
  • $\begingroup$ Probably it would then be more correct to ask to the equality above expressing the factorization to hold up to units. $\endgroup$ Commented Feb 22, 2011 at 14:27
  • $\begingroup$ Tommaso: $f(x)=2x-2$ seems to have the same problem? $\endgroup$ Commented Feb 22, 2011 at 19:32
  • $\begingroup$ The problem with 2x-2 is it is not irreducible (you can factor a 2). $\endgroup$ Commented Feb 22, 2011 at 19:44

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.)

  • $\begingroup$ Are you sure there are no such places $\nu$? If I haven't made a mistake, I think that 2 splits into $\mathfrak{p}\tilde{\mathfrak{p}}$ and that we can label the roots $\alpha$ and $\tilde{\alpha}$ such that $v_\mathfrak{p}(\alpha)$=$v_{\tilde{\mathfrak{p}}}(\tilde{\alpha})=1$ and $v_\mathfrak{p}(\tilde{\alpha})=v_{\tilde{\mathfrak{p}}}(\alpha)=-1$. $\endgroup$ Commented Feb 22, 2011 at 23:32
  • $\begingroup$ I think Kevin is right. Notice more generally if the root $\alpha$ of an irreducible polynomial $f(x)$ has norm equal to one (i.e. leading term = constant term), then for every p-adic place $\nu$ of $K=Q(\alpha)$ for which $\alpha$ is not an integer there must be another p-adic place (same p!) $\nu'$ so that $\alpha$ belongs to the maximal ideal of the valuation ring of $\nu'$. This implies that if there is no place $\nu$ for which $\alpha$ belongs to the maximal ideal defined by $\nu$, then $\alpha$ is a unit and so is $f(x)$. $\endgroup$ Commented Feb 23, 2011 at 0:46
  • $\begingroup$ I do not understand your counterexamples. I think that in both cases, once you fix a root of your polynomial, you have only one place $\nu$ with the property above. Which, if my question has a positive answer, would mean that $f(x)$ stays irreducible in the power series ring. Which is the case in both your examples! $\endgroup$ Commented Feb 24, 2011 at 2:13
  • $\begingroup$ I changed my answer. Hopefully I understood the question enough to write something coherent. $\endgroup$ Commented Feb 24, 2011 at 20:23

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 $\mathbb{Z}[[x]]$.

  • $\begingroup$ David, I don't think $x-6$ is a counter-example (I tried working that example out myself). The issue is that x-6 factors in the power series ring over the integers. If you identify x with 6, then as a 2-adic integer x looks like $0*1+1*2+1*2^{2}+0*2^{3}+\cdots$. Thus, $x^{n}=0*1+0*2+\cdots + 0*2^{n-1}+1*2^{n}+\cdots$ has its first nonzero term as a coefficient of $2^{n}$. In particular, we can form a power series $2+x+x^3+x^5+x^6+...$ which maps to zero in $\mathbb{Q}_{2}$ under the identification $x=6$. (We choose powers of $x$ so that we can use $2$ to carry infinitely.) $\endgroup$ Commented Feb 24, 2011 at 20:06
  • $\begingroup$ I don't have time to think about this right now, but you raise a good point. Also, I wrote above that $\mathbb{Z}_p \cap \mathbb{Q} = \mathbb{Z}$ but of course that isn't true. So it looks like what I have shown is the required statement with $\mathbb{Z}$ replaced by $\mathbb{Z}_p \cap \mathbb{Q}$. I think there is something valuable here though; please feel free to improve this. $\endgroup$ Commented Feb 24, 2011 at 20:22
  • $\begingroup$ I think so too. You might be able to patch it up by inverting the primes in the constant coefficient different from the prime you are considering, which one should intuitively be able to do using elements in $\mathbb{Z}[[x]]$. But it gets complicated, because the image of $\varphi_{\alpha,\nu}$ can be weird. (Try $\alpha=3+3\sqrt{2}$.) $\endgroup$ Commented Feb 24, 2011 at 20:27
  • $\begingroup$ @ David : Letting $\mathbb{Z}_{(p)}=\mathbb{Z}_p \cap \mathbb{Q}$, it seems that you prove that $f = (\prod h_i) \cdot u$ for some $u \in \mathbf{Z}[[x]]$. The constant term of $h_i$ is necessarily a power of $p$ (quotient everything by $x$). So $u(0)$ is the prime-to-$p$-part of $f(0)$. It seems that doing this for all $p$ gives the result, or am I missing something? $\endgroup$ Commented Feb 25, 2011 at 2:26
  • $\begingroup$ Sorry, in the comment above I meant : $u \in \mathbb{Z}_{(p)}[[X]]^{\times} \cap \mathbb{Z}[[X]]$. $\endgroup$ Commented Feb 25, 2011 at 2:29

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