Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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 $S^{-1}(\mathbb Z[[X]])$.
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 with $\mathbb{Q}$ is right exact and 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.
