TITLE: Nilradical of absolutely flat ring
QUESTION [7 upvotes]: Suppose $A$ is an absolutely flat ring (i.e. every $A$-module is flat). Is it true that nilradical of $A$ is trivial, i.e. $\mathfrak{N}(A)=\{0\}$? 
I believe the answer is yes. Here is my attempted proof:

We use the following characterization of absolutely flat rings (Chapter
  2, Exercise 27 in Atiyah & Macdonald). $A$ is absolutely flat ring $\Leftrightarrow$
  $(x)=(x^2)$ for each $x\in A$. Now, suppose $x^{n}=0$ for
  $n\in\mathbb{N}$, where $n$ is minimal. If $x\neq 0$, we necessarily
  have $n\ge 2$. Since $(x)=(x^2)$, we have $x=ax^2$ for some $a\in A$. Thus,
  $x^{n-1}=ax^{n}=0$, contradicting minimality of $n$. Hence, $x=0$ is
  the only nilpotent element of $A$.

Is my proof correct?
I have done Chapter 2, Exercise 28 in Atiyah & Macdonald, which lists important properties of absolutely flat rings, but this simple property was not mentioned there.

REPLY [4 votes]: Here is another bit of information that you can get out of absolutely flat rings that is very useful. 


If $A$ is absolutely flat then for any multiplicative subset $S \subseteq A$, we have $S^{-1}A$ being absolutely flat.


Thus in particular for any maximal ideal $\mathfrak{m} \subseteq A$, we have $A_{\mathfrak{m}}$ being absolutely flat. However a local ring that is absolutely flat is a field and so this means that $\operatorname{Jac}(A)_{\mathfrak{m}} \subseteq A_{\mathfrak{m}}$ is zero for all maximal ideals $\mathfrak{m}$. The Jacobson radical is an ideal in the localization because $\operatorname{Jac}(A) \subseteq \mathfrak{m}$ for any maximal ideal of $A$. Since being zero is a local property this means that $\operatorname{Jac}(A) = 0$. The same proof mutadis mutants also works to show that the nilradical of an absolutely flat ring is zero.


Exercise: Proof the result mentioned above. Hint: You will not believe it, but for any two $S^{-1}A$ - modules $M$ and $N$, the modules $M \otimes_A N $ and $M \otimes_{S^{-1}A} N$ are isomorphic as $S^{-1}A$ - modules!! Now why is $M \otimes_A N$ a $S^{-1}A$ - module?
Proof the a local ring that is absolutely flat is a field: Let $A$ be a local ring with maximal ideal $\mathfrak{m}$. Choose any $x \in \mathfrak{m}$. Then we have $\mathfrak{m}(x) = (x)$. Indeed, $\mathfrak{m}(x) \subseteq (x)$ and the other inclusion follows from $(x) = (x^2)$. By the Nakayama Lemma it follows $x = 0$ and since this was any element of $\mathfrak{m}$, we have $\mathfrak{m} = 0$. In other words, $A$ is a field.