Is $\mathbb{Z}[[x]]/(x-6) \cong \widehat{\mathbb{Z}}_2 \times \widehat{\mathbb{Z}}_3$?

It seems intuitive that $\mathbb{Z}[[x]]/(x-p)$ is the p-adic numbers, and I think this is not too hard to show formally. Really, it doesn't seem too hard to show this is true for any number p, not just primes. This gives a nice explicit description of $\mathbb{Z}[[x]]/(x-n)$, when combined with the well known (and hopefully well remembered) facts that $p^k$-adics is just a fancy name for p-adics, and that n-adics in general are the direct product of the p-adics for the distinct primes p dividing n.

Everything was fine and good until I tried to show this direct product decomposition directly in the ring $\mathbb{Z}[[x]]/(x-6)$. I want to use Chinese remainder theorem on the ideals $(x-2)$, $(x-3)$, and $(x-6)$. Clearly $(x-2)+(x-3) = (1)$, but I think it is also pretty clear that $(x-6) \neq (x-2) \cap (x-3)$.

Indeed, I don't think $x-6$ is even an element of $(x-2)$, since $$\frac{x-6}{x-2} = 3 + x + \tfrac12 x^2 + \ldots + \tfrac1{2^n} x^{n+1} + \ldots \notin \mathbb{Z}[[x]]$$

What has gone wrong?

  • $\begingroup$ I don't understand what you're doing. $\mathbb{Z}[[x]]/(x - 6)$ mod $(x - 2)$ can't be $\mathbb{Z}_2$ because $4 = 0$ in the quotient. The map $\mathbb{Z}[[x]]/(x - 6) \to \mathbb{Z}_6 \to \mathbb{Z}_2$ (where the second map is the projection) doesn't have $x - 2$ in its kernel; $x$ maps to $6 = 2 + 4$ like it ought to. $\endgroup$ Jun 13, 2011 at 20:09
  • $\begingroup$ Sorry about asking such a silly question, but what does $\widehat{\mathbb{Z}}_2$ denote? My first instinct is to read it as $\widehat{\mathbb{Z}}$ localized at the set $\{1,2,4,\ldots\}$, but I could also understand if it meant $\widehat{\mathbb{Z}}$ localized at the prime ideal $(2)$. Or is it a construction involving $\mathbb{Z}_2$? Since $\mathbb{Z}_2$ is already the completion of $\mathbb{Z}$ at $(2)$, I'm not sure what $\widehat{\,\,\,\,\,\mathbb{Z}_2\,\,}$ would mean. $\endgroup$ Jun 13, 2011 at 20:10
  • $\begingroup$ I think you should be considering the homomorphisms $x\mapsto 2x$ and $x\mapsto 3x$ in $\mathbb{Z}[[x]]$. $\endgroup$ Jun 13, 2011 at 20:11
  • $\begingroup$ Is $\mathbb{Z}[[x]]/(x-p)$ isomorphic to the $p$-adics? In the $p$-adics you have the "infinite expansions" $a_0 + a_1p + a_2p^2 + \cdots$, but we usually require $0\leq a_i\leq p-1$ (e.g., see Gouvea's p-adic numbers, Definition 1.2.1 (coherent sequence). $\endgroup$ Jun 13, 2011 at 20:12
  • 2
    $\begingroup$ @Jack: Yes, I think you're right on that. But now look at the constitutent maps. You go from $\mathbb{Z}[[x]]$ to $\mathbb{Z}/6^n\mathbb{Z}$ by moding out by $(x-6,x^n)$; then you go from $\mathbb{Z}/6^n\mathbb{Z}\to(\mathbb{Z}/2^n\mathbb{Z}) \times (\mathbb{Z}/3^n\mathbb{Z})$ in the obvious way, but what happens to $x$ under the compositions? If you are mapping $x$ to $6$ under the first, it cannot map to $2$ when you get to $\mathbb{Z}/2^n\mathbb{Z}$. So the kernel is not what you think it is. $\endgroup$ Jun 13, 2011 at 20:26

4 Answers 4


The isomorphism of $\widehat{\mathbb{Z}_n}$ with $\mathbb{Z}[[x]]/(x-n)$ is realized by mapping $\mathbb{Z}[[x]]$ onto $\mathbb{Z}/n^k\mathbb{Z}$ by sending $1$ to $1$, $x$ to $n$, and $x^k$ to $0$, that is, as you note, the map with kernel $(x-n,x^k)$. These maps are consistent, so by the universal property of the inverse limit you get a map $$\mathbb{Z}[[x]] \to \lim_{\leftarrow}\mathbb{Z}/n^k\mathbb{Z} =\widehat{\mathbb{Z}_n}.$$ The map is onto, and the kernel is indeed $$\cap_{k=1}^{\infty}(x-n,x^k) = (x-n).$$

But when you consider the natural decomposition of $\widehat{\mathbb{Z}_6}$ into $\widehat{\mathbb{Z}_2}\times\widehat{\mathbb{Z}_3}$ by factoring out the cyclic groups along the inverse limit, you don't get the corresponding maps. That is, the maps $$\mathbb{Z}[[x]]\to\mathbb{Z}/2^k\mathbb{Z}\qquad\text{and}\qquad\mathbb{Z}[[x]]\to\mathbb{Z}/3^k\mathbb{Z}$$ that we just defined will naturally give you map into the product, which is isomorphic to $\mathbb{Z}/6^k\mathbb{Z}$; but the map you get via this composition is not the map we defined above into $\mathbb{Z}/6^k\mathbb{Z}$: note that $(a+2^k\mathbb{Z},b+3^k\mathbb{Z})$ maps to the unique residue class $m$ modulo $6^k$ such that $m\equiv a\pmod{2^k}$ and $m\equiv b\pmod{3^k}$. Since the image of $x$ under the product is $(2+2^k\mathbb{Z},3+3^k\mathbb{Z})$, this is not equal to $6+6^k\mathbb{Z}$ except in the case $k=1$; but for the isomorphism to work in the "intuitive way" that you are trying you would need $x$ to map to both $6+6^k\mathbb{Z}$ and to $(2+2^k\mathbb{Z},3+3^k\mathbb{Z})$.

So the problem is that the map you get by taking $\mathbb{Z}[[x]]/(x-6)$ is not the map induced by looking at $\mathbb{Z}[[x]]/(x-2)\times \mathbb{Z}[[x]]/(x-3)$.

  • $\begingroup$ Ah, that's a nice way to think about the isomorphism. $\endgroup$ Jun 13, 2011 at 20:46

It is true that, for $a,b$ coprime integers then $$ \mathbb{Z}[[x]]/(x-ab)\cong\left(\mathbb{Z}[[x]]/(x-a)\right)\times\left(\mathbb{Z}[[x]]/(x-b)\right) $$ which is what you would expect from $\mathbb{Z}_{ab}\cong\mathbb{Z}_a\times\mathbb{Z}_b$.

First, consider the homomorphism $$ \begin{align} &\theta_a\colon\mathbb{Z}[[x]]\to\mathbb{Z}[[x]],\\ &f\mapsto f(ax). \end{align} $$ As $(x-ab)\subseteq\theta^{-1}_a((x-b))$, this induces a homomorphism $\tilde\theta_a\colon\mathbb{Z}[[x]]/(x-ab)\to\mathbb{Z}[[x]]/(x-b)$. Defining $\theta_b,\tilde\theta_b$ similarly, you should be able to use the Chinese Remainder Theorem to check that $f\mapsto (\tilde\theta_b(f),\tilde\theta_a(f))$ gives the required isomorphism.

Alternatively, an element $c$ of $\mathbb{Z}_a$ can be written as $\sum_nc_na^n$ for $c_n\in\mathbb{Z}$. If $a,b$ are coprime then you can always take $c_n$ to be a multiple of $b^n$ (by the Chinese Remainder Theorem) so, in fact, $c$ can be written as $\sum_n (c_nb^n)a^n$. So, it is the image of $\sum_n c_n(ab)^n\in\mathbb{Z}_{ab}$ where the coefficient $c_n$ of $(ab)^n$ is multiplied by $b^n$. Hence why I considered the map $\theta_b$ taking $x^n$ to $b^nx^n$.

  • $\begingroup$ Thanks! I tried using your hint, but kept going in circles. This looks very nice to answer my question about more exactly what the maps are. Tomorrow I plan on calculating all the kernels exactly, since one should still have that (x-ab) is the intersection of the kernels of θa and θb (both of which are prime, and should have "degree" 1, dangit). The intersection of (x-a) and (x-b) is the principal ideal ((x-a)(x-b)), right? $\endgroup$ Jun 13, 2011 at 23:14
  • $\begingroup$ @Jack: For any $n\in\mathbb{Z}$ and $f\in\mathbb{Z}[[x]]$, you can write $f=(x-n)g+h$ where either $h=0$ or the leading coefficient of $h$ is not a multiple of $n$. Show that if $f$ is in the kernel of $\tilde\theta_a$ and $\tilde\theta_b$ then its leading coefficient is a multiple of $ab$. At least, that's what I did. $\endgroup$ Jun 14, 2011 at 0:14

I don't find the first statement as obvious as you do, but here is a proof. First, there is clearly a map $\mathbb{Z}[[x]] \to \mathbb{Z}_p$ which sends $x$ to $p$, and we take the unique continuous such map. The kernel consists of all power series $\sum a_i x^i$ such that $\sum a_i p^i = 0$ in $\mathbb{Z}_p$. Intuitively we want to write

$$\sum a_i x^i = (x - p) \left( \sum a_i \frac{x^i - p^i}{x - p} \right)$$

but the sum in brackets does not converge in $\mathbb{Z}[[x]]$. However, the coefficient of $x^n$, as an element of $\mathbb{Z}_p$, is

$$\sum_{i \ge n+1} a_i p^{i-n-1} = \frac{-a_0 - a_1 p - ... - a_n p^n}{p^{n+1}} \in \mathbb{Z}$$

so we can collect terms such that the resulting sum converges in $\mathbb{Z}[[x]]$. So the kernel is $(x - p)$ as desired.

Note that any map $\mathbb{Z}[[x]] \to \mathbb{Z}_p$ which sends $x$ to an element divisible by $p$ uniquely extends to a continuous map, and in your situation with $p = 2$, you set $x = 2$ to define one map and set $x = 6$ to define the other. There's no contradiction in the fact that the resulting maps don't have the same kernel.

So: yes, it is true that $\mathbb{Z}[[x]]/(x - 6) \cong \mathbb{Z}_6 \cong \mathbb{Z}_2 \times \mathbb{Z}_3$, and I don't know why you're taking the profinite completion of something which is its own profinite completion.

  • $\begingroup$ Jack mentions in the comments that he's using $\widehat{\mathbb{Z}}_2$ to denote $\mathbb{Z}_2$ to indicate that he doesn't mean $\mathbb{Z}/2\mathbb{Z}$. $\endgroup$ Jun 13, 2011 at 20:49

I left this as a comment above, but it is at the end of a long thread, so I will also note it here: the isomorphisms being discussed here are a special case of the more general isomorphism discussed in this question, which states that for any Noetherian ring $R$, if $I = (a_1,\ldots,a_n)$ is an ideal, then $R[[x_1,\ldots,x_n]]/(x_1-a_1,\ldots,x_n-a_n) =$ the $I$-adic completion of $R$.


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