I was doing some exercises in the book of Atiyah / MacDonald on Commutative Algebra, and I'm a little "stuck" with number 3.10 (i):

If $A$ is an absolutely flat ring and $S\subseteq A$ a multiplicatively closed subset, then the localization $S^{-1}A$ is absolutely flat.

I wanted to use a criterion shown earlier: $A$ absolutely flat $\Leftrightarrow$ every principal ideal of $A$ is idempotent.

So if $\frac{a}{s}\in S^{-1}A$, my plan was to show that there exists a $\frac{b}{t}\in S^{-1}A$ with $\frac{b}{t}\cdot(\frac{a}{s})^2=\frac{a}{s}$. From the absolute flatness of $A$, we know that there are $x,y\in A$ with $xa^2=a$ and $ys^2=s$. So choosing $b:=x$, $t:=y$ one would be done, but I don't see why $y$ should be in $S$? I guess this has to be "obvious" in some way, and I'm just being stupid!?

Then I tried to google it, but all I found was a different solution (here), and I'd like to ask one question regarding this solution, too, but I don't know if this is the appropriate place to do so? Since it's the same topic, I hope so, but if not, please correct me and I'll edit this post.

Let $P$ be an arbitrary $S^{-1}A$-module. The author says there is an obvious isomorphism $S^{-1}P\cong S^{-1}A\otimes_A P\to P$ of $S^{-1}A$-modules. The first isomorphy is clear to me (and proven in the book), but I don't get the second one. I believe the maps should look like this:

$\varphi:S^{-1}A\otimes_A P\to P, \varphi(\frac{1}{s}\otimes p)=\frac{1}{s}p$, and

$\psi:P\to S^{-1}A\otimes_A P, \psi(p)=1\otimes p$.

Then $\varphi\circ\psi=1$. As for $\psi\circ\varphi$, let $\frac{1}{s}\otimes p\in S^{-1}A\otimes_A P$, then $\psi(\varphi(\frac{1}{s}\otimes p))=\psi(\frac{1}{s}p)=1\otimes\frac{1}{s}p$, and then I don't know why I can switch over the $\frac{1}{s}$. I know that there is an $S^{-1}A$-module structure on the tensor product, but I thought it was of the form $\frac{a}{s}\cdot(\frac{1}{t}\otimes p)=\frac{a}{st}\otimes p$. It has to have something to do with the $S^{-1}A$-module structure on $P$. I guess it should be easier to show $S^{-1}M\cong M$ directly, I just had the idea to do it this way, but didn't try it out yet. I am going to do this now, but the question above still remains. It has to be a simple thing I overlook again, something I did not fully understand yet, and I hope you can help me understand it!

Thanks for your help in advance!


2 Answers 2


Hint: for any $S^{-1} A$-module $M$, $x,y \in M$ and $s \in S$, $sx = y \iff x = \frac{1}{s} y$.

  • $\begingroup$ So since $s(1\otimes\frac{1}{s}p)=1\otimes p$, $1\otimes\frac{1}{s}p=\frac{1}{s}(1\otimes p)=\frac{1}{s}\otimes p$, correct? Thank you very much for your help! How can I mark this as "answered" by the way? $\endgroup$ Feb 6, 2012 at 13:14
  • $\begingroup$ @Rand al'Thor: yes, you got it. (And it seems that you found the check mark as well...) $\endgroup$ Feb 6, 2012 at 13:17

I am too late with the answer, but I am doing the same exercise now.

Is it possible to use the following argument, since for any $a\in A$ we may find $x\in A$ such that $a = a^2x$ we may try the following trick, take any $a/s\in S^{-1}A$

$$ \frac{xs}{1}\left(\frac{a}{s}\right)^2 = \frac{a^2sx}{s^2}=\frac{a^2x}{s}=\frac{a}{s}, $$ where of course $xa^2=a$, note that $xs/1$ is a correct element of $S^{-1}A$. It seems that this argument shows that every principal ideal of $S^{-1}A$ is idempotent, doesn't it?


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