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I want to solve the following problem: show that the element $1\otimes (1,1,....)$ is not the zero element in $$\mathbb{Q}\otimes_{\mathbb{Z}} \prod^{\infty}_{n\geq 2}\mathbb{Z}/n\mathbb{Z}$$.

My approach would be to try to define a map from this tensor to a $\mathbb{Z}$-module such that the element in question is not mapped to zero. I tried to start defining a map from $\mathbb{Q}\times \prod^{\infty}_{n\geq 2}\mathbb{Z}/n\mathbb{Z}$. An idea I had was to send an element $(p/q,(x_1,x_2,...))$ to $(p/q)\sum (x_i/2^i)$. This is not well defined though. Is there a way to fix this? Or a better approach to the problem?

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  • $\begingroup$ Why exactly is it not well-defined? $\endgroup$ Dec 16, 2014 at 6:27
  • $\begingroup$ I mean, maybe it is well defined, in that case my question is why is that so. I mean, do i get the same image changing representatives for the $x_i$s? Also, it is not well behaved with the linearity I need. $\endgroup$
    – Franco
    Dec 16, 2014 at 6:47
  • $\begingroup$ well, you could try to answer those questions :-) $\endgroup$ Dec 16, 2014 at 6:50
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    $\begingroup$ If you take any element of your tensor product which is only non-zero in a finite number of the terms in the product, then you can kill it by multiplying it by a large integer from the $\mathbb Q$. Since these elements are zero in the tensor product, they are zero under any linear map, so whatever linear map you hope to define, it had better have any such element in its kernel. $\endgroup$ Dec 16, 2014 at 6:56

4 Answers 4

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If you consider any $\mathbb{Z}$-module $M$ and its torsion part $t(M)$, we have the exact sequence $$ 0\to t(M)\to M\to M/t(M)\to0 $$ that, tensored with $\mathbb{Q}$, says $$ \mathbb{Q}\otimes_{\mathbb{Z}}M\cong\mathbb{Q}\otimes_{\mathbb{Z}}M/t(M) $$ So, if $1\otimes x=0$ in $\mathbb{Q}\otimes_{\mathbb{Z}}M$ (for some $x\in M$), then also $1\otimes\pi(x)=0$ in $\mathbb{Q}\otimes_{\mathbb{Z}}M/t(M)$ (for $\pi\colon M\to M/t(M)$ the canonical map).

Now, if $N$ is torsion-free and $y\in N$, $y\ne0$, then $1\otimes y\ne0$ in $\mathbb{Q}\otimes_{\mathbb{Z}}N$, because the map $$ N\to\mathbb{Q}\otimes_{\mathbb{Z}}N,\qquad y\mapsto 1\otimes y $$ is injective.

Note that the first part uses the fact that $\mathbb{Q}$ is divisible (so tensoring with it kills the torsion part); the second part uses that $N$ is torsionfree, so flat over $\mathbb{Z}$.

In your case, the element $(1,1,\dotsc)$ has infinite order in $M=\prod_{n>0}\mathbb{Z}/n\mathbb{Z}$, so it goes to a nonzero element in the quotient $M/t(M)$.

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    $\begingroup$ If you use flatness of $ℚ$, then it already suffices to say $ℤ → \prod_{n ∈ ℕ} ℤ/nℤ$ has trivial kernel for $\bigcap_{n ∈ ℕ} nℤ = 0$, so tensoring by $ℚ$ we get an injective map $ℚ \otimes_ℤ ℤ → ℚ \otimes_ℤ \prod_{n ∈ ℕ} ℤ/nℤ$. $\endgroup$
    – k.stm
    Dec 16, 2014 at 14:36
  • $\begingroup$ @k.stm I treated a more general case. $\endgroup$
    – egreg
    Dec 16, 2014 at 15:39
  • $\begingroup$ Just mentioning it because I needed a moment to see exactly how your answer applies to the question. $\endgroup$
    – k.stm
    Dec 16, 2014 at 15:47
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    $\begingroup$ @k.stm I added a note for future readers. $\endgroup$
    – egreg
    Dec 16, 2014 at 15:53
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    $\begingroup$ @Khoata Actually it should be $N$ is torsionfree, hence flat. $\endgroup$
    – egreg
    May 26, 2021 at 8:51
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Here is an alternative approach which however only works for tensor products that can be considered as localizations: If $R$ is a commutative ring and $S\subset R$ is a multiplicative subset of $R$, then given any $R$-module $M$ the $R_S$-module $R_S\otimes_R M$ together with the map $M\to R_S\otimes_R M$, $m\mapsto 1\otimes m$, is a localization of $M$ at $S$. However, you know that the localization can also be defined as $S^{-1} M$ by taking as elements the equivalence classes of formal fractions $\frac{m}{s}$ with $s\in S$ and $m\in M$ under the relation $\frac{m_0}{s_0} = \frac{m_1}{s_1}:\Leftrightarrow t s_1 m_0 = t s_0 m_1$ for some $t\in S$, together with the map $M\to S^{-1}M$ sending $m$ to $\frac{m}{1}$. By uniqueness of localization, these two approaches are uniquely isomorphic over $R_S$ in a way compatible with the morphisms from $M$ - in particular, given $m\in M$, we have $1\otimes m=0$ in $R_S\otimes_R M$ if and only if there exists some $t\in S$ such that $tm=0$.

This applies to $R := {\mathbb Z}$, $S := {\mathbb Z}\setminus\{0\}$ (so that $R_S = {\mathbb Q}$), proving that the kernel of $M\to M\otimes_{\mathbb Z}{\mathbb Q}$ is precisely the torsion subgroup of $M$. Since in your example the element $(1,1,...)\in\prod\limits_{n\geq 2}{\mathbb Z}/n{\mathbb Z}$ is not torsion, it does therefore not vanish under the localization map to ${\mathbb Q}\otimes_{\mathbb Z}\prod\limits_{n\geq 2}{\mathbb Z}/n{\mathbb Z}$.

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(This answer might be difficult.)

Let $U$ be a nonprincipal ultrafilter over the set of natural numbers which contains subsets of $\Bbb{N}$ of the form $n\Bbb{N}+1$ for each $n\in \Bbb{N}$. Consider the ultraproduct $A=\prod_{n\in \Bbb{N}} (\Bbb{Z}/n\Bbb{Z})/U$. You can check that $A$ is a $\Bbb{Z}$-module under pointwise operation. We will represent the elements of $A$ of the form $[x]_U$. More precisely, $[x]_U$ is defined as $$[x]_U=\left\{x\in\prod_{n\in \Bbb{N}} (\Bbb{Z}/n\Bbb{Z}) : \{k\in \Bbb{N}: x_k=y_k\}\in U\right\}$$ where $x=(x_1,x_2,\cdots)$ and $y=(y_1,y_2,\cdots)$

We will define a function $\phi:\Bbb{Q}\times \prod_{n\in \Bbb{N}} (\Bbb{Z}/n\Bbb{Z})\to A$, $\phi(a/b,(x_n)_{n\in\Bbb{N}})=[(ab^{-1}x_n)_{n\in \Bbb{N}}]_U$. If it is well-defined, then it gives the bilinear function and it sends $(1,(1,1,1,\cdots))$ to nonzero element of $A$. We must check that $\phi$ is well defined since $b^{-1}$ is not well-defined unless $\gcd(b,n)=1$. However for given $b$, the set of all $n$ such that $\gcd(b,n)>1$ can be ignored because $b\Bbb{N}+1\in U$ and $\gcd(bk+1,b)=1$ for every $k$ so $b^{-1}$ is well-defined for almost all $n$ (i.e. set of all $n$ with $(b,n)=1$ is an element of $U$.)

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I come a bit late but wouldn't it be easier to just say that the map :

$$ \mathbb{Q} \times \prod_{n \in \mathbb{N}} \frac{\mathbb{Z}}{n\mathbb{Z}} \longrightarrow \prod_{n \in \mathbb{N}} \mathbb{Q}$$

and say that it is a $\mathbb{Z}$-bilinear map between $\mathbb{Z}$ modules that send the wanted element to a non zero element?

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  • $\begingroup$ Sorry if I put it as an answer instead of a post, don´t know if I can change it... I think that what I proposed is slightly different from what the author wrote (I'm just multiplying the elements of the product). Plus this seems to be simpler (if correct) so you could consider it as an alternative method. $\endgroup$
    – WrabbitW
    Mar 17, 2017 at 22:37
  • $\begingroup$ Fair enough${}$. $\endgroup$
    – user228113
    Mar 18, 2017 at 10:49

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