An integral domain $A$ which is also absolutely flat is a field

Question

Question:

Assume that $A$ is an integral domain such that every $A$-module is flat. Show that that $A$ is a field.

Discussion:

This seems to be very related to this question, in which it is shown that the Jacobson radical of an absolutely flat ring $A$ is $\{0\}$, using that $A$ is absolutely flat $\iff$ $A_{\mathfrak{m}}$ is a field for each maximal ideal $\mathfrak{m}\subset A$. I can't seem to draw the right conclusions specifically for the case of an integral domain. Any ideas that could help me get going on the right track would be appreciated.

Answer 1

Let $a\neq 0$ be an element of $A$ and $\cdot a: A\rightarrow A$ be the multiplication by $a$ map (i.e. $x\mapsto ax$).

Then $$0\rightarrow A\xrightarrow{\cdot a} A \rightarrow A/(a)\rightarrow 0$$ is exact, hence $$0\rightarrow A\otimes_A A/(a)\xrightarrow{\cdot a \otimes 1} A\otimes_A A/(a)\rightarrow A/(a)\otimes_A A/(a)\rightarrow 0$$

is exact (by assumption). But now $\cdot a\otimes 1$ is the $0$ map, and injectivity implies $A/(a)\cong A\otimes_A A/(a)=0$. Hence, $(a)=A$ which immediately implies $A$ is a field.

Answer 2

If $A$ is absolutely flat, then one can show that for every $x \in A$ there is some $y \in A$ such that $x = x^2 y$. This is because we have $(x)=(x^2)$ in every localization, hence globally. But if $A$ is an integral domain, this equation means for $x \neq 0$ that $1 = xy$.

Answer 3

You mentioned in the comments that you were interested in the connection with idempotents. That way is nice because it works just fine for noncommutative rings. Here is the information on that.

Once you've shown that in an "absolutely flat"(=von Neumann regular) ring every nonzero principal right ideal is generated by an idempotent, you just note "the only nonzero idempotent in a domain is $1$, so if it's also a domain, every right ideal is the whole ring. Thus $R$ is a division ring."

In Lectures on modules and rings page 126-128, you can see the two small lemmas that help you prove this equivalence (pg 128):

1. $R$ is von Neumann regular (i.e. for each $a\in R$ there exists an $x\in R$ such that $axa=a$.
2. All right $R$ modules are flat.
3. All cyclic right $R$ modules are flat.

Let's show that the first one up there is equivalent to this:

4 Every principal right ideal is generated by an idempotent.

Assuming $1$, you get that $(ax)^2=ax$, and so $aR=axaR\subseteq ax R\subseteq aR$. Thus $aR=axR$, and $ax$ is the idempotent.

Conversely, suppose $aR=eR$ for an idempotent $e$. We know that $a=er$ $as=e$ for some $r,s\in R$. The first one tells us that $ea=eer=er=a$, and then the second one combines with this to tell us that $asa=ea=a$.

Even for noncommutative rings you can see that this makes the principal right ideals idempotent. I don't know for certain that principal ideals being idempotent implies the ring is von Neumann regular, but for commutative rings it does. If $(a)=(a)^2$ for every $a\in R$, then you've got that $a^2x=a$ for some $x$, but using commutativity this is just the condition above that $axa=a$