Let $R=\Bbb{Z}/4\Bbb{Z}$ and let me take the ideal $(2)\subset R$. I want to show that $(2)$ is not flat in $R$.

We only had the following definition:

A module $M\subset R$ is flat if $M\otimes_R-$ is left exact.

Using this definition I want to show it.

My idea was the following. Since we know that tensoring is already right exact we only need to worry about left exactness. The problem is that I don't see what left exact sequence I need to consider such that after tensoring it isn't left exact anymore. Could someone give me a hint?

  • 2
    $\begingroup$ Isn't $(2)=(0)$? $\endgroup$
    – azif00
    Jul 2, 2022 at 21:24
  • 1
    $\begingroup$ @azif00 sorry I made a typo in the question, I edited it, so I don't think $(2)=(0)$ anymore $\endgroup$ Jul 2, 2022 at 21:29

1 Answer 1


Assuming that you intend to look at modules over the ring $R = \mathbb{Z}/4\mathbb{Z}$, consider the short exact sequence $$ 0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0, $$ where the left map is multiplication by $2$, embedding $\mathbb{Z}/2\mathbb{Z}$ into the regular module $\mathbb{Z}/4\mathbb{Z}$ as the submodule, i.e. ideal, $(2)$. Can you see how tensoring with $\mathbb{Z}/2\mathbb{Z}$ destroys injectivity?

After tensoring with $\mathbb{Z}/2\mathbb{Z}$, the map \begin{align} \mathbb{Z}/2\mathbb{Z} &\otimes \mathbb{Z}/2\mathbb{Z} &&\;\longrightarrow\; &\mathbb{Z}/2\mathbb{Z} &\otimes \mathbb{Z}/4\mathbb{Z} \\ x &\otimes y &&\;\longmapsto\; &x &\otimes 2y \end{align} sends $$ 1 \otimes 1 \;\longmapsto\; 1 \otimes 2 = 2 \otimes 1 = 0 \otimes 1 = 0 \otimes 0, $$ i.e., it's the $0$ map, which is clearly not injective.

Here it is all summarized in a commutative diagram of $\mathbb{Z}/4\mathbb{Z}$-modules, where the vertical maps are all isomorphisms and the numbers labeling the other arrows are where the class of $1$ or $1 \otimes 1$ is sent (which determines maps since they're cyclic modules): $$ \require{AMScd} \begin{CD} 0 @>>> \mathbb{Z}/2\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z} @>{1 \otimes 2}>> \mathbb{Z}/2\mathbb{Z} \otimes \mathbb{Z}/4\mathbb{Z} @>{1 \otimes 1}>> \mathbb{Z}/2\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z} @>>> 0 \\ @. @V{\sim\,}VV @V{\sim\,}VV @V{\sim\,}VV @. \\ 0 @>>> \mathbb{Z}/2\mathbb{Z} @>{\smash[t]{0}}>> \mathbb{Z}/2\mathbb{Z} @>{\smash[t]{1}}>> \mathbb{Z}/2\mathbb{Z} @>>> 0 \end{CD} $$

  • $\begingroup$ Thanks a lot for your answer. I first have a quick understanding question. Is it true that $(2)=\{\bar 0,\bar 2\}$ in $R$ since $\bar 2\cdot 2=\bar 4=\bar 0$ and $\bar 3\cdot 2=\bar 6=\bar 2$? So tensoring with $\Bbb{Z}/2\Bbb{Z}$ gives me $$0\rightarrow \Bbb{Z}/2\Bbb{Z}\otimes\Bbb{Z}/2\Bbb{Z}\rightarrow \Bbb{Z}/2\Bbb{Z}\otimes \Bbb{Z}/4\Bbb{Z}\rightarrow ...$$ Now what I don't see is what $\Bbb{Z}/2\Bbb{Z}\otimes\Bbb{Z}/2\Bbb{Z}$ gives me. Or isn't this necessary to solve the exercise? $\endgroup$ Jul 2, 2022 at 21:38
  • $\begingroup$ I would say that all the elements which maps to $0\otimes 0$ are $\Bbb{Z}/2\Bbb{Z}\otimes \bar 0$? $\endgroup$ Jul 2, 2022 at 21:42
  • $\begingroup$ Yes, but note that it's best to think of $\mathbb{Z}/4\mathbb{Z}$ as a module in the middle of the sequence. It's a tautology that a submodule of the regular module is the same set and underlying additive abelian group as an ideal of the ring (but you no longer consider multiplying elements together within the submodule). $\endgroup$ Jul 2, 2022 at 21:43
  • $\begingroup$ Sorry is your yes with respect to my claim that $(2)=\{\bar 0, \bar 2\}$ or is it with respect to my claim that I would say that the kernel of the embedding is $\Bbb{Z}/2\Bbb{Z}\otimes \bar 0$ or for both? $\endgroup$ Jul 2, 2022 at 21:45
  • 1
    $\begingroup$ I was responding to your first comment. I will edit my answer to address your further question. $\endgroup$ Jul 2, 2022 at 21:46

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