Let $S=\mathbb{Z}-\{0\}$. Show the existence or nonexistence of isomorphism between $S^{-1}\prod_{1}^{\infty}\mathbb{Z}_{i}$ and $\prod_{1}^{\infty}\mathbb{Q}_{i}$ as $\mathbb{Q}$-vector spaces. (Here $\mathbb Z_i=\mathbb Z$ and $\mathbb Q_i=\mathbb Q$ for all $i\ge 1$.)

This is an example used to show that the localization does not commute with infinite products under the natural (canonical) homomorphism. I read this from a lecture note by Ravi Vakil. I think we still need to show these two are really not isomorphic as $\mathbb{Q}$-vector spaces. I tried to follow the hint to consider the element $(1,\frac{1}{2},\frac{1}{3},...,\frac{1}{n},...)$ but failed.

To user26857: I want a proof of the existence or nonexistence of isomorphism between these two vector spaces. If I get it right, your solution reduced the problem to the existence of basis of these two vector spaces. And in your sense, a basis is a subset $B$ (of the vector space $V$) such that every finite subset of $B$ is a linearly independent set and any vector in $V$ can be expressed as a finite sum of the elements in $B$. If this is what you mean, can you show the existence of such basis? Thank you!

To Ragib Zaman: Your example is a better one. Thanks!

  • $\begingroup$ Please share your thoughts so far :) $\endgroup$
    – Shaun
    Jul 8, 2014 at 13:49
  • $\begingroup$ Your edit is in fact a comment on my answer. A wrong comment! And this for two reasons: first $e_1,e_2,\dots$ is not a basis in $\mathbb Q^{\mathbb N}$ as you also observed by noticing that $(1,1/2/,\dots,1/n,\dots)$ can't be written as a finite combination of the elements $e_1,e_2,\dots$; second, if two vector spaces have bases with the same cardinality then they are isomorphic. (The more important thing is that your question is still unclear to me, although I've answered to both possible interpretations: are you asking if the two vector spaces are canonically isomorphic or not?) $\endgroup$
    – user26857
    Jul 9, 2014 at 13:32
  • $\begingroup$ Let me answer to your new edit: a basis of a vector space is exactly what you said, and this is not in "my sense", this is the definition of basis in a vector space; see en.wikipedia.org/wiki/Basis_(linear_algebra). The existence of a basis relies on the axiom of choice and you can easily find a proof in on the web; see planetmath.org/sites/default/files/texpdf/33494.pdf. $\endgroup$
    – user26857
    Jul 10, 2014 at 10:43
  • $\begingroup$ @user26857:Thanks! $\endgroup$
    – Wei Xia
    Jul 11, 2014 at 8:41
  • 1
    $\begingroup$ I think it is important to remember Ravi's universal property defining localization of the R module M at S, i.e. a map M-->S^-1M which is the initial object in the category of maps from M to S^-1R modules. Thus the point is to show that the (canonical) map from a product of R modules to the product of S^-1R modules is not isomorphic in this category, to the map from the product to the localization of the product. An isomorphism in this category is a map of S^-1R modules giving a commutative triangle, together with the given maps, hence the only possible isomorphism is the canonical one. $\endgroup$
    – roy smith
    Jul 17, 2020 at 19:39

2 Answers 2


Ravi's hint is to show that $(1 \ , \ 1/2 \ , \ 1/3 \ , \ \ldots)$ is not in the image of the natural map, so the natural map is not an isomorphism. As user26857 shows though, in this example the vector spaces are actually isomorphic, but not canonically.

It may be more satisfying to see the following example:
Take out ring to be $R=\mathbb{Z},$ $S = \mathbb{Z}\setminus \{ 0\}$ and the modules to be $M_i = \mathbb{Z}/(i) \ , \ i\geq 2.$ Then $S^{-1}M_i=0$ since for all $x\in M_i$ and $s\in S$ we have $$ \frac{x}{s} = \frac{ix}{is} = \frac{0}{is} = \frac01.$$

On the other hand, $S^{-1}\left(\prod_{i\ge2}M_i\right)$ is non-zero. To see this, note that if $\dfrac{(1,1,1,\ldots)}{1} =\dfrac01$ then there exists $s\in S$ such that $s(1,1,1,\ldots)=0.$ But this would mean that every integer $i\geq 2$ divides $s,$ and $s$ must be non-zero so this is impossible.

So we have found an example where localization does not commute with an infinite direct product through any map, not just the natural map.


They are isomorphic as $\mathbb Q$-vector spaces, but not canonically.

Let $B\subset\mathbb Q^{\mathbb N}$ a $\mathbb Q$-basis. Then $|\mathbb Q^{\mathbb N}|=|B||\mathbb Q|$. On the other side, we have $\mathbb Z^{\mathbb N}\subset S^{-1}\mathbb Z^{\mathbb N}\subset\mathbb Q^{\mathbb N}$, so $|\mathbb Z^{\mathbb N}|\le |S^{-1}\mathbb Z^{\mathbb N}|\le|\mathbb Q^{\mathbb N}|$. Since $|\mathbb Z^{\mathbb N}|=|\mathbb Q^{\mathbb N}|$ we get $|S^{-1}\mathbb Z^{\mathbb N}|=|\mathbb Q^{\mathbb N}|$. If $B'\subset S^{-1}\mathbb Z^{\mathbb N}$ is a $\mathbb Q$-basis we have $|S^{-1}\mathbb Z^{\mathbb N}|=|B'||\mathbb Q|$, and therefore $|B|=|B'|$, so $\mathbb Q^{\mathbb N}$ is isomorphic to $S^{-1}\mathbb Z^{\mathbb N}$ as $\mathbb Q$-vector spaces.

The canonical morphism sends $(k_n)_{n\ge 1}/s$ to $(k_n/s)_{n\ge 1}$. If there is such an element which is sent to $(1/n)_{n\ge 1}$, then $k_n/s=1/n$ for all $n\ge 1$, that is, $nk_n=s$ for all $n\ge 1$ (or, if you like, $n\mid s$ for all $n\ge 1$), a contradiction. Thus the canonical morphism is (injective but) not surjective.

  • $\begingroup$ I'm unfamiliar with cardinal arithmetic. Why does $|\mathbb{Q}^\mathbb{N}| = |B| |\mathbb{Q}|$? I would have thought $|\mathbb{Q}^\mathbb{N}| = |\mathbb{Q}|^{|B|}$, as in the finite case. $\endgroup$
    – Anakhand
    Jan 25 at 13:14

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