Fdtxjr

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 $mathbb Z[[x]]otimes_mathbb Zmathbb Qtomathbb Q[[x]]$. It is injective but not an isomorphism since $1+frac12x+frac14x^2 + ...$ does not belong to the image.

What about other 'strange' isomorphisms? If there was some isomorphism $mathbb Z[[x]]otimes_mathbb Z mathbb Qcongmathbb Q[[x]]$, 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 $frac12+frac14x+frac18x^2 + ...$ does not come from $mathbb Z[[x]]otimes_mathbb Zmathbb Q$. Hence $x$ and $2-x$ are of the form $pi^k varepsilon$ and $pi^leta$ for $k,lgeq 1$ 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 $mathbbZ[[x]] otimes mathbbQ to mathbbQ[[x]]$ and that it is not an isomorphism. This is the one induced by the natural inclusion $mathbbZ[[x]] to mathbbQ[[x]]$. In terms of this inclusion $mathbbZ[[x]] otimes mathbbQ$ is the subring of $mathbbQ[[x]]$ 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

$$e^x = sum_n ge 0 fracx^nn!$$

lies in $mathbbQ[[x]]$ but not in $mathbbZ[[x]] otimes mathbbQ$ 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.) $mathbbQ[[x]]$ 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 $mathbbZ[[x]] otimes mathbbQ$ 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 $S^-1(mathbb Z[[X]])notsimeq(S^-1mathbb Z)[[X]]$, where $S=mathbb Zsetminus0$.

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 $pinmathbb Z$ a (non-negative) prime number, gives rise to a maximal ideal in $S^-1(mathbb Z[[X]])$.

That the morphism induced by the inclusion $mathbbZ[[x]]tomathbbQ[[x]]$ is not an isomorphism can also be derived from group theoretical properties.

As abelian groups, $mathbbZ[[x]]$ and $mathbbQ[[x]]$ are just direct products of copies of $mathbbZ$ and $mathbbQ$. Consider the exact sequence
$$
0to mathbbZ^mathbbNto
mathbbQ^mathbbN to
(mathbbQ/mathbbZ)^mathbbN
to0
$$
Tensoring wit $mathbbQ$ is flat, we get the exact sequence
$$
mathbbZ^mathbbNotimesmathbbQto
mathbbQ^mathbbNotimesmathbbQ to
(mathbbQ/mathbbZ)^mathbbNotimesmathbbQ
to0
$$
but the group $(mathbbQ/mathbbZ)^mathbbN$ is not torsion, so
$(mathbbQ/mathbbZ)^mathbbNotimesmathbbQne0$, which means
$mathbbZ^mathbbNotimesmathbbQto mathbbQ^mathbbNotimesmathbbQ$ is not surjective.

(Note that the map $mathbbZ^mathbbNotimesmathbbQto
mathbbQ^mathbbNotimesmathbbQ$ is injective because $mathbbQ$ is flat, but it's irrelevant for the problem at hand.)

However, $mathbbZ^mathbbNotimesmathbbQ$ and $mathbbQ^mathbbNotimesmathbbQ$ are isomorphic (through a different map). Indeed they are both torsion free divisible groups (that is, $mathbbQ$-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.