Find an ideal $I \subset \mathbb{Z}$ such that $I\otimes \mathbb{Z}/2 \not\cong I\mathbb{Z}/2$.

We know that $\mathbb{Z}/2$ is not a flat $\mathbb{Z}$-module since the injection $\mathbb{Z} \hookrightarrow \mathbb{Q}$ will no longer be injective after the tensor as $\mathbb{Z}\otimes_{\mathbb{Z}} \mathbb{Z}/2 \cong \mathbb{Z}/2$ and $\mathbb{Q}\otimes_{\mathbb{Z}} \mathbb{Z}/2 = 0$.

We have the general statement that an $R$-module $M$ is flat if and only if for any ideal $I\subset R$, we have $I\otimes M \cong IM$.

Now what would be an ideal $I\subset \mathbb{Z}$ such that $$I \otimes_{\mathbb{Z}} \mathbb{Z}/2 \not\cong I\mathbb{Z}/2.$$

Since we know $\mathbb{Z} \otimes \mathbb{Z}/2 \cong \mathbb{Z}/2$ given by the map $a\otimes b \mapsto ab$, so we know the image of $I \otimes_{\mathbb{Z}} \mathbb{Z}/2 $ would be $ I\mathbb{Z}/2$ and this map needs to fail to be injective.

I believe $I = (2)$ will work since $(2)\mathbb{Z}/2 = 0$, and $(2)\otimes \mathbb{Z}/2 \neq 0$, since $$2\cdot 1 \otimes 1 \neq 1\otimes 2\cdot 1 = 0$$ since $1 \not\in (2)$.

I am not so sure about the last part showing $(2)\otimes \mathbb{Z}/2 \neq 0$, can we also construct a bilinear map on $(2)\times \mathbb{Z}/2$?

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    $\begingroup$ To construct the bilinear map, just map $(n,a)$ in $(2)\times Z/2$ to $(n/2)a$ in $Z/2$. $\endgroup$ – Mariano Suárez-Álvarez Jan 5 '18 at 4:57

All nonzero ideals $I$ of $Z$ are free of rank one, so $I\otimes Z/2Z$ is isomorphic to $Z/2Z$.

On the other hand, if $I=(2)$ then $IZ/2Z$ is clearly zero.


For any ring $A$, any right ideal $I$ any left $A$-module $M$, there is a canonical map $I\otimes_A M \to IM$ that assigns $a\otimes m$ to $am$. The module $M$ is flat if and only if this is an isomorphism for every ideal $I$ of $A$, in fact, it suffices it is an isomorphism for every finitely generated ideal of $A$.

Since $\mathbb Z/2$ is torsion, it is not flat, so there is at least one ideal $I$ for which the map above is not an isomorphism. In fact, since flatness fails because $\mathbb Z/2$ is torsion, it is reasonable to try with ideals that kill this module: every ideal $I$ of the form $(n)$ with $n$ divisible by $2$ works, since in this case $I \mathbb Z/2 = 0$, but, since every ideal of $\mathbb Z$ is principal, $I\otimes_\mathbb Z M \simeq M$ for any module $M$. In particular, $(2) =\text{Ann}(\mathbb Z/2)$ works.

In fact, since $\mathbb Z$ is a principal ideal domain, a $\mathbb Z$-module is flat if and only if it is torsion-free.

  • $\begingroup$ (the annihilator of a module with torsion can well be zero!) $\endgroup$ – Mariano Suárez-Álvarez Jan 5 '18 at 5:06
  • $\begingroup$ @MarianoSuárez-Álvarez I was thinking about torsion modules, heh. $\endgroup$ – Pedro Tamaroff Jan 5 '18 at 5:10
  • $\begingroup$ Consider the direct sum of all Z/n. ;-) $\endgroup$ – Mariano Suárez-Álvarez Jan 5 '18 at 5:32
  • $\begingroup$ @MarianoSuárez-Álvarez Of course you mean infinite direct sum! :D $\endgroup$ – Pedro Tamaroff Jan 5 '18 at 6:11

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