# Is $\mathbb Z[[X]]\otimes \mathbb Q$ isomorphic to $\mathbb Q[[X]]$?

Is $\mathbb Z[[X]]\otimes \mathbb Q$ isomorphic to $\mathbb Q[[X]]$?

Here tensor product is over the ring $\mathbb Z$ and $\mathbb Z[[X]]$ denotes formal power series over $\mathbb Z$. I think this is true if we take polynomial rings instead of power series. Any help in this regards will be appreciated.

Consider the natural homomorphism ${\mathbb Z}[[x]]\otimes_{\mathbb Z}{\mathbb Q}\to{\mathbb Q}[[x]]$. It is injective but not an isomorphism since $1+\frac{1}{2}x+\frac{1}{4}x^2 + ...$ does not belong to the image.

What about other 'strange' isomorphisms? If there was some isomorphism ${\mathbb Z}[[x]]\otimes_{\mathbb Z} {\mathbb Q}\cong{\mathbb Q}[[x]]$, then ${\mathbb Z}[[x]]\otimes_{\mathbb Z} {\mathbb Q}$ was a discrete valuation ring, i.e. a principal ideal domain with a unique prime element $\pi$.

Consider now the elements $x$ and $x-2$ in ${\mathbb Z}[[x]]\otimes_{\mathbb Z} {\mathbb Q}$. They are both non-invertible in ${\mathbb Z}[[x]]\otimes_{\mathbb Z} {\mathbb Q}$: for $x$, it is not even invertible in ${\mathbb Q}[[x]]$, and for $2-x$, it is invertible in ${\mathbb Q}[[x]]$, but its inverse $\frac{1}{2}+\frac{1}{4}x+\frac{1}{8}x^2 + ...$ does not come from ${\mathbb Z}[[x]]\otimes_{\mathbb Z}{\mathbb Q}$. Hence $x$ and $2-x$ are of the form $\pi^k \varepsilon$ and $\pi^l\eta$ for $k,l\geq 1$ and units $\varepsilon,\eta$. This however would force $x^l$ to be associate to $(2-x)^k$, which is a contradiction since this is not even true in ${\mathbb Q}[[x]]$ as $(2-x)^k$ is a unit there but $x^l$ is not.

Nice question! First let me make the weaker claim that there is a natural map $\mathbb{Z}[[x]] \otimes \mathbb{Q} \to \mathbb{Q}[[x]]$ and that it is not an isomorphism. This is the one induced by the natural inclusion $\mathbb{Z}[[x]] \to \mathbb{Q}[[x]]$. In terms of this inclusion $\mathbb{Z}[[x]] \otimes \mathbb{Q}$ is the subring of $\mathbb{Q}[[x]]$ consisting of rational formal power series whose coefficients have a common denominator (because the tensor product consists of finite linear combinations). So, for example, the formal power series

$$e^x = \sum_{n \ge 0} \frac{x^n}{n!}$$

lies in $\mathbb{Q}[[x]]$ but not in $\mathbb{Z}[[x]] \otimes \mathbb{Q}$ because its denominators get arbitrarily large.

Conceptually, the problem is that power series are a limit but tensor products of commutative rings are a colimit. In general it's not formal to verify that a limit commutes with a colimit; that usually isn't true, and when it is it usually requires work to verify.

Now let's show that they aren't isomorphic at all. (Edit: There was a mistake here which is handled correctly in Hanno's answer.) $\mathbb{Q}[[x]]$ is a local ring, and in particular it has a unique maximal ideal $(x)$, and any element not in $(x)$ (a power series with nonzero constant term) is invertible. But $\mathbb{Z}[[x]] \otimes \mathbb{Q}$ is not a local ring: it has $(x)$ as a maximal ideal, but (as in Hanno's answer) the element $x - 2$ does not lie in this maximal ideal but is also not invertible.

In fact we want to prove that $$S^{-1}(\mathbb Z[[X]])\not\simeq(S^{-1}\mathbb Z)[[X]]$$, where $$S=\mathbb Z\setminus\{0\}$$.
This is more or less obvious: the second ring (is $$\mathbb Q[[X]]$$ and it) is local, as it was already noticed, while the first has plenty of maximal ideals: every ideal of $$\mathbb Z[[X]]$$ generated by $$X-p$$, with $$p\in\mathbb Z$$ a (non-negative) prime number, gives rise to a maximal ideal in $$S^{-1}(\mathbb Z[[X]])$$.

• Dear sir, can you kindly explain why the ideal $(X-p)$ is maximal in $\mathbb{Z}[[X]]$ for $p$ prime in $\mathbb{Z}.$ I tried to give an evaluation map from $\mathbb{Z}[[X]]$ to $\mathbb{Z}$ at the point $p,$ I think for this may be some division algorithm we need by a monic polynomial e.g., $(X-p).$ – user371231 Mar 13 at 17:43
• Where did I say that $(X-p)$ is maximal in $\mathbb Z[[X]]$? – user26857 Mar 13 at 22:18
• Then $(x-p)$ must be a prime ideal in $\mathbb{Z}[[Z]]$. What ring will $\mathbb{Z}[[X]]/(X-p)$ is ? – user371231 Mar 14 at 3:21
• The ring of $p$-adic integers. – user26857 Mar 14 at 12:15

That the morphism induced by the inclusion $\mathbb{Z}[[x]]\to\mathbb{Q}[[x]]$ is not an isomorphism can also be derived from group theoretical properties.

As abelian groups, $\mathbb{Z}[[x]]$ and $\mathbb{Q}[[x]]$ are just direct products of copies of $\mathbb{Z}$ and $\mathbb{Q}$. Consider the exact sequence $$0\to \mathbb{Z}^{\mathbb{N}}\to \mathbb{Q}^{\mathbb{N}} \to (\mathbb{Q}/\mathbb{Z})^{\mathbb{N}} \to0$$ Tensoring wit $\mathbb{Q}$ is flat, we get the exact sequence $$\mathbb{Z}^{\mathbb{N}}\otimes\mathbb{Q}\to \mathbb{Q}^{\mathbb{N}}\otimes\mathbb{Q} \to (\mathbb{Q}/\mathbb{Z})^{\mathbb{N}}\otimes\mathbb{Q} \to0$$ but the group $(\mathbb{Q}/\mathbb{Z})^{\mathbb{N}}$ is not torsion, so $(\mathbb{Q}/\mathbb{Z})^{\mathbb{N}}\otimes\mathbb{Q}\ne0$, which means $\mathbb{Z}^{\mathbb{N}}\otimes\mathbb{Q}\to \mathbb{Q}^{\mathbb{N}}\otimes\mathbb{Q}$ is not surjective.

(Note that the map $\mathbb{Z}^{\mathbb{N}}\otimes\mathbb{Q}\to \mathbb{Q}^{\mathbb{N}}\otimes\mathbb{Q}$ is injective because $\mathbb{Q}$ is flat, but it's irrelevant for the problem at hand.)

However, $\mathbb{Z}^{\mathbb{N}}\otimes\mathbb{Q}$ and $\mathbb{Q}^{\mathbb{N}}\otimes\mathbb{Q}$ are isomorphic (through a different map). Indeed they are both torsion free divisible groups (that is, $\mathbb{Q}$-vector spaces) with the same cardinality $2^{\aleph_0}$, so they have the same dimension $2^{\aleph_0}$ (assuming choice, of course). However, as shown in another answer, this group isomorphism cannot be a ring homomorphism.

• Well, I think the question deals with a ring isomorphism. – user26857 Feb 4 '15 at 21:46
• @user26857 Of course! What I showed is that tensoring the inclusion doesn't give a group isomorphism, so no ring isomorphism either. – egreg Feb 4 '15 at 21:48