# If $A$ is absolutely flat, so is $S^{-1}A$ (A&M, Problem 3.10)

If $$A$$ is absolutely flat,$$S$$ is a multiplicatively closed subset of $$A$$, then $$S^{-1}A$$ is also absolutely flat. (Atiyah & MacDonald, Problem 3.10)

I'm trying to prove this problem using a previous result proved in the same book, claiming that absolute flatness is equivalent to the fact that every principal ideal is also idempotent.

My idea is that since every element of the principal ideal $$(\frac{a}{s})=(a) \subset S^{-1}A$$ is equal to $$a \cdot \frac{b}{t}$$, where $$\frac{b}{t} \in S^{-1}A$$, we may show that this ideal is equal to the extended ideal of $$(a)\subset A$$, while the latter one is idempotent, hence $$\exists x_i, y_i \in (a)$$, such that $$\sum_{i\in I} x_i y_i=a$$(by definition of idempotent ideals), and then we can prove the same result for $$(a) \subset S^{-1}A$$ using the previous identity, hence $$S^{-1}A$$ is absolutely flat.

In some solutions, various authors have given different solutions using modules and exact sequences, but none has used the fact that principal ideals should be idempotent under this condition, which makes me feel concerned about my solution. So could anyone tell me if my solution is correct? Thanks.

Is this criterion for absolute flatness not given in your book?

A commutative ring $$A$$ is absolutely flat (a.k.a. a von Neumann regular) iff for every $$a\in A$$, there exists a $$b\in A$$ such that $$aba=a$$.

Given that, the following solution seems the most straightforward to me.

So let $$\frac as\in S^{-1}A$$. Since $$A$$ is absolutely flat, there is $$b$$ such that $$aba=a$$.

Then $$\frac as \frac{bs}{1}\frac as=\frac as$$.

Hence $$S^{-1}A$$ is absolutely flat as well.

I think you can also prove it along the lines you were using using the fact that $$I^{ce}=I$$ for every ideal $$I\lhd S^{-1}A$$ and that extending ideals commutes with products. In fact, being absolutely flat is equivalent to having all ideals idempotent.

Using these facts, one could begin with $$I\lhd S^{-1}A$$ and then compute that $$I=I^{ce}=(I^cI^c)^e = I^{ce}I^{ce}=I^2$$.

• I think this criterion is not mentioned in A&M, but I can see that it can be deduced from the equivalent statement that every principal ideal is also idempotent. This criterion is very useful, thanks!
– Ubik
Commented Aug 9, 2023 at 1:01
• @RHspqr I added a second thing more along the lines you were wanting to follow. Let me know if I misremembered how extension and contraction behave. I get these things mixed up sometimes. Commented Aug 9, 2023 at 3:44