Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $R$ be a ring with unit and $I$ an ideal in $R$. I want to show that $R/I$ is need not be flat over $R$, but I do not know how to come up with a counter-example.

Any hint is appreciated.

share|improve this question
What examples of non-flat modules do you know? [Interestingly, if $I$ is finitely generated then there's really only one way for $R/I$ to be flat — $I$ has to be generated by an idempotent, and hence $R = I \times (R/I)$ as rings.] –  Dylan Moreland Aug 16 '12 at 17:09
Dear james. If you are studying commutative algebra and have not yet seen the examples given in the answers I suspect the lecture notes or the book you are following has some room left for improvement. You might want to have a look at Atiyah-Macdonald, I read most of it and I think it's excellent. –  Rudy the Reindeer Aug 16 '12 at 17:28
Alas, this is one of those questions where examples of flat $R/I$ are hard to find: if you tried just about anything at all, it would likely have been a counter-example. The question you probably should have asked is along the lines of "How do I tell if $R/I$ is flat or not?" (assuming, of course, that you really did consider various $R$ and $I$ and weren't able to tell) –  Hurkyl Aug 16 '12 at 22:20

4 Answers 4

up vote 1 down vote accepted

Consider $R = \mathbf Z$ and take the ideal $I = 2\mathbf Z$. Now take a look at the exact sequence $0\to 2\mathbf{Z}\to\mathbf{Z} \to \mathbf{Z}/2\mathbf{Z} \to 0$.

Can you see what happens when you tensor this exact sequence over $\mathbf{Z}$? You should be able to prove that the resulting sequence cannot be exact.

share|improve this answer

Analysis of the problem:
Suppose that $R/I$ is flat over $R$.
Then tensoring the short exact sequence $0\to I\to R$, by $R/I$ yields a new exact sequence $$0\to I\otimes_R R/I \to R\otimes_R R/I\quad (*)$$ Recalling the standard identification $M\otimes_R R/I\xrightarrow {\cong} M/IM:\tilde m\otimes \tilde r\mapsto \overline {rm}$ for any $R$-module $M$, we get from $(*)$ the injective map $$ 0\to I/I^2\to R/I:\tilde i \mapsto \overline {i} =\bar 0 \quad (**) $$ But the morphism $(**)$ is clearly the zero map.
It can only be injective if $I/I^2=0$ or equivalently if $I=I^2$. So we have proved $$ R/I \; \text {flat} \implies I=I^2 $$

By contraposition, if $I\neq I^2$ the $R$-module $R/I$ is guaranteed to be non-flat.
So in a non formal but very clear sense $R/I$ is practically never flat since an ideal is practically never equal to its square .
Here is a result ( a consequence of Nakayama's lemma) corroborating this informal statement :
If $I$ is finitely generated and $I=I^2$, then $I=(i)$ for some idempotent $i=i^2\in R$
If $R$ is a noetherian domain and $0\subsetneq I\subsetneq R$ an ideal, then $R/I$ is not flat.

share|improve this answer

Take the exact sequence $$ 0 \to \mathbb Z \xrightarrow{\cdot 2} \mathbb Z \to \mathbb Z / 2 \mathbb Z \to 0$$

and tensor with $\mathbb Z / 3 \mathbb Z$ to get $$ 0 \to \mathbb Z \otimes \mathbb Z / 3 \mathbb Z \xrightarrow{\cdot 2} \mathbb Z \otimes \mathbb Z / 3 \mathbb Z \to \mathbb Z / 2 \mathbb Z \otimes \mathbb Z / 3 \mathbb Z\to 0$$

which is isomorphic to $$ 0 \to \mathbb Z / 3 \mathbb Z \to \mathbb Z / 3 \mathbb Z \to 0 \to 0$$

which is no longer exact.

share|improve this answer
Could you explain where the exactness fails in the last sequence? Isn't the map $\mathbb{Z}/3\mathbb{Z}\to \mathbb{Z}/3\mathbb{Z}$ there given by $x\mapsto 2x$, which is a bijective map? –  Prism Jul 15 at 18:50

You could insist that $R$ be local, and then flatness is equivalent to projectiveness.

To thwart $R/I$ from being projective, you would just ensure that $I$ is not a summand of $R$.

So there you have it, a blueprint to find an example. Any commutative local ring with an ideal which is not a summand will work. An obvious choice would be $\mathbb{Z}/(p^2)$ for a prime $p$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.